FALL2023_HW3_Student_Bharat

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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 1/71 Fall 2023 CS4641/CS7641 A Homework 3 Instructor: Dr. Mahdi Roozbahani Deadline: Friday, November 10th, 11:59 pm EST No unapproved extension of the deadline is allowed. Submission past our 48-hour penalized acceptance period will lead to 0 credit. Discussion is encouraged on Ed as part of the Q/A. However, all assignments should be done individually. Plagiarism is a serious offense . You are responsible for completing your own work. You are not allowed to copy and paste, or paraphrase, or submit materials created or published by others, as if you created the materials. All materials submitted must be your own.</font> All incidents of suspected dishonesty, plagiarism, or violations of the Georgia Tech Honor Code will be subject to the institute’s Academic Integrity procedures. If we observe any (even small) similarities/plagiarisms detected by Gradescope or our TAs, WE WILL DIRECTLY REPORT ALL CASES TO OSI , which may, unfortunately, lead to a very harsh outcome. Consequences can be severe, e.g., academic probation or dismissal, grade penalties, a 0 grade for assignments concerned, and prohibition from withdrawing from the class. </font> Instructions for the assignment This assignment consists of both programming and theory questions. Unless a theory question explicitly states that no work is required to be shown, you must provide an explanation, justification, or calculation for your answer. To switch between cell for code and for markdown, see the menu -> Cell -> Cell Type You can directly type Latex equations into markdown cells. If a question requires a picture, you could use this syntax <img src="" style="width: 300px;"/> to include them within your ipython notebook. Your write up must be submitted in PDF form. You may use either Latex, markdown, or any word processing software. We will **NOT** accept handwritten work. Make sure that your work is formatted correctly, for example submit instead of \text{sum_{i=0} x_i} When submitting the non-programming part of your assignment, you must correctly map pages of your PDF to each question/subquestion to reflect where they appear. **Improperly mapped questions may not be graded correctly and/or will result in point deductions for the error.**
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 2/71 All assignments should be done individually, and each student must write up and submit their own answers. Graduate Students : You are required to complete any sections marked as Bonus for Undergrads Using the autograder Grads will find three assignments on Gradescope that correspond to HW3: "Assignment 3 Programming", "Assignment 3 - Non-programming" and "Assignment 3 Programming - Bonus for all". Undergrads will find an additional assignment called "Assignment 3 Programming - Bonus for Undergrads". You will submit your code for the autograder in the Assignment 3 Programming sections. Please refer to the Deliverables and Point Distribution section for what parts are considered required, bonus for undergrads, and bonus for all. We provided you different .py files and we added libraries in those files please DO NOT remove those lines and add your code after those lines. Note that these are the only allowed libraries that you can use for the homework. You are allowed to make as many submissions until the deadline as you like. Additionally, note that the autograder tests each function separately, therefore it can serve as a useful tool to help you debug your code if you are not sure of what part of your implementation might have an issue. For the "Assignment 3 - Non-programming" part, you will need to submit to Gradescope a PDF copy of your Jupyter Notebook with the cells ran. See this EdStem Post for multiple ways on to convert your .ipynb into a .pdf file. Please refer to the Deliverables and Point Distribution section for an outline of the non- programming questions. When submitting to Gradescope, please make sure to mark the page(s) corresponding to each problem/sub-problem. The pages in the PDF should be of size 8.5" x 11", otherwise there may be a deduction in points for extra long sheets. Using the local tests For some of the programming questions we have included a local test using a small toy dataset to aid in debugging. The local test sample data and outputs are stored in .py files in the local_tests_folder . The actual local tests are stored in localtests.py . There are no points associated with passing or failing the local tests, you must still pass the autograder to get points. It is possible to fail the local test and pass the autograder since the autograder has a certain allowed error tolerance while the local test allowed error may be smaller. Likewise, passing the local tests does not guarantee passing the autograder. You do not need to pass both local and autograder tests to get points, passing the Gradescope autograder is sufficient for credit.
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 3/71 It might be helpful to comment out the tests for functions that have not been completed yet. It is recommended to test the functions as it gets completed instead of completing the whole class and then testing. This may help in isolating errors. Do not solely rely on the local tests, continue to test on the autograder regularly as well. Deliverables and Points Distribution Q1: Image Compression [30pts] Deliverables: imgcompression.py and printed results 1.1 Image Compression [20 pts] - programming svd [4pts] compress [4pts] rebuild_svd [4pts] compression_ratio [4pts] recovered_variance_proportion [4pts] 1.2 Black and White [5 pts] non-programming 1.3 Color Image [5 pts] non-programming Q2: Understanding PCA [20pts] Deliverables: pca.py and written portion 2.1 PCA Implementation [10 pts] - programming fit [5pts] transform [2pts] transform_rv [3pts] 2.2 Visualize [5 pts] programming and non-programming 2.3 PCA Reduced Facemask Dataset Analysis [5 pts] non-programming 2.4 PCA Exploration [0 pts] Q3: Regression and Regularization [80pts: 50pts + 20pts Bonus for Undergrads + 12pts Bonus for All] Deliverables: regression.py and Written portion 3.1 Regression and Regularization Implementations [50pts: 30pts + 20pts Bonus for Undergrad] - programming
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 4/71 RMSE [5pts] Construct Poly Features 1D [2pts] Construct Poly Features 2D [3pts] Prediction [5pts] Linear Fit Closed Form [5pts] Ridge Fit Closed Form [5pts] Ridge Cross Validation [5pts] Linear Gradient Descent [5pts] Bonus for Undergrad Linear Stochastic Gradient Descent [5pts] Bonus for Undergrad Ridge Gradient Descent [5pts] Ridge Bonus for Undergrad Ridge Stochastic Gradient Descent [5pts] Ridge Bonus for Undergrad 3.2 About RMSE [3 pts] non-programming 3.3 Testing: General Functions and Linear Regression [5 pts] non-programming 3.4 Testing: Ridge Regression Ridge [5 pts + 2 pts Bonus for All] non-programming 3.5 Cross Validation [7 pts] non-programming 3.6 Noisy Input Samples in Linear Regression [10 pts] non-programming BONUS FOR ALL Q4: Naive Bayes and Logistic Regression [35pts] Deliverables: logistic_regression.py and Written portion 4.1 Llama Breed Problem using Naive Bayes [5 pts] non-programming 4.2 News Data Sentiment Classification Using Logistic Regression [30 pts] - programming sigmoid [2 pts] bias_augment [3 pts] predict_probs [5 pts] predict_labels [2 pts] loss [3 pts] gradient [3 pts] accuracy [2 pts] evaluate [5 pts]
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 5/71 fit [5 pts] Q5: Noise in PCA and Linear Regression [15pts] Deliverables: Written portion 5.1 Slope Functions [5 pts] non-programming 5.2 Error in Y and Error in X and Y [5 pts] non-programming 5.3 Analysis [5 pts] non-programming Q6: Feature Reduction.py [25pts Bonus for All] Deliverables: feature_reduction.py and Written portion 6.1 Feature Reduction [18 pts] - programming forward_selection [9pts] backward_elimination [9pts] 6.2 Feature Selection - Discussion [7 pts] non-programming Q7: Movie Recommendation with SVD [10pts Bonus for All] Deliverables: svd_recommender.py and Written portion 7.1 SVD Recommender recommender_svd [5pts] predict [5pts] 7.2 Visualize Movie Vectors [0pts] 0 Set up This notebook is tested under python 3. . , and the corresponding packages can be downloaded from miniconda . You may also want to get yourself familiar with several packages: jupyter notebook numpy matplotlib sklearn There is also a VS Code and Anaconda Setup Tutorial on Ed under the "Links" category Please implement the functions that have raise NotImplementedError , and after you finish the coding, please delete or comment out raise NotImplementedError .
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 6/71 Library imports Version information python: 3.11.4 (main, Jul 5 2023, 08:54:11) [Clang 14.0.6 ] matplotlib: 3.7.1 numpy: 1.24.3 Q1: Image Compression [30 pts] **[P]** | **[W]** Load images data and plot In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # This is cell which sets up some of the modules you might need # Please do not change the cell or import any additional packages. import numpy as np import pandas as pd import matplotlib from matplotlib import pyplot as plt from sklearn.feature_extraction import text from sklearn.datasets import load_diabetes , load_breast_cancer , load_iris from sklearn.linear_model import LogisticRegression from sklearn.model_selection import train_test_split from sklearn.metrics import mean_squared_error , accuracy_score import warnings import sys print ( 'Version information' ) print ( 'python: {}' . format ( sys . version )) print ( 'matplotlib: {}' . format ( matplotlib . __version__ )) print ( 'numpy: {}' . format ( np . __version__ )) warnings . filterwarnings ( 'ignore' ) % matplotlib inline % load_ext autoreload % autoreload 2 STUDENT_VERSION = 1 EO_TEXT , EO_FONT , EO_COLOR = 'TA VERSION' , 'Chalkduster' , 'gray' , EO_ALPHA , EO_SIZE , EO_ROT = 0.7 , 90 , 40 # Render types : 'browser', 'png', 'plotly_mimetype', 'jupyterlab', pdf rndr_type = 'png' In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # load Image image = plt . imread ( "./data/hw3_image_compression.jpeg" ) / 255 # plot image fig = plt . figure ( figsize = ( 10 , 10 )) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT ,
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 7/71 <matplotlib.image.AxesImage at 0x10772b450> transform = fig . transFigure , fontsize = EO_SIZE , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) plt . imshow ( image ) Out[ ]: In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### def rgb2gray ( rgb ): return np . dot ( rgb [ ... , : 3 ], [ 0.299 , 0.587 , 0.114 ]) fig = plt . figure ( figsize = ( 10 , 10 )) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT ,
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 8/71 <matplotlib.image.AxesImage at 0x14e299050> 1.1 Image compression [20pts] **[P]** SVD is a dimensionality reduction technique that allows us to compress images by throwing away the least important information. Higher singular values capture greater variance and, thus, capture greater information from the corresponding singular vector. To perform image compression, apply SVD on each matrix and get rid of the small singular values to compress the image. The loss of information through this process is negligible, and the difference between the images can be hardly spotted. transform = fig . transFigure , fontsize = EO_SIZE , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) # plot several images plt . imshow ( rgb2gray ( image ), cmap = plt . cm . bone ) Out[ ]:
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 9/71 For example, the proportion of variance captured by the first component is where is the singular value. In the imgcompression.py file, complete the following functions: svd : You may use np.linalg.svd in this function, and although the function defaults this parameter to true, you may explicitly set full_matrices=True using the optional full_matrices parameter. Hint 2 may be useful. compress rebuild_svd compression_ratio : Hint 1 may be useful recovered_variance_proportion : Hint 1 may be useful HINT 1: http://timbaumann.info/svd-image-compression-demo/ is a useful article on image compression and compression ratio. You may find this article useful for implementing the functions compression_ratio and recovered_variance_proportion HINT 2: If you have never used np.linalg.svd , it might be helpful to read Numpy's SVD documentation and note the particularities of the matrix and that it is returned already transposed. HINT 3: The shape of resulting from SVD may change depending on if N > D, N < D, or N = D. Therefore, when checking the shape of , note that min(N,D) means the value should be equal to whichever is lower between N and D. 1.1.1 Local Tests for Image Compression Black and White Case [No Points] You may test your implementation of the functions contained in imgcompression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "SVD calculation - black and white images"! UnitTest passed successfully for "Image compression - black and white images"! UnitTest passed successfully for "SVD reconstruction - black and white images"! UnitTest passed successfully for "Compression ratio - black and white images"! UnitTest passed successfully for "Recovered variance proportion - black and white images"! In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestImgCompression unittest_ic = TestImgCompression () unittest_ic . test_svd_bw () unittest_ic . test_compress_bw () unittest_ic . test_rebuild_svd_bw () unittest_ic . test_compression_ratio_bw () unittest_ic . test_recovered_variance_proportion_bw ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 10/71 1.1.2 Local Tests for Image Compression Color Case [No Points] You may test your implementation of the functions contained in imgcompression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "SVD calculation - color images"! UnitTest passed successfully for "Image compression - color images"! UnitTest passed successfully for "SVD reconstruction - color images"! UnitTest passed successfully for "Compression ratio - color images"! UnitTest passed successfully for "Recovered variance proportion - color images"! 1.2.1 Black and white [5 pts] **[W]** This question will use your implementation of the functions from Q1.1 to generate a set of images compressed to different degrees. You can simply run the below cell without making any changes to it, assuming you have implemented the functions in Q1.1. Make sure these images are displayed when submitting the PDF version of the Juypter notebook as part of the non-programming submission of this assignment. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestImgCompression unittest_ic = TestImgCompression () unittest_ic . test_svd_color () unittest_ic . test_compress_color () unittest_ic . test_rebuild_svd_color () unittest_ic . test_compression_ratio_color () unittest_ic . test_recovered_variance_proportion_color () In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from imgcompression import ImgCompression imcompression = ImgCompression () bw_image = rgb2gray ( image ) U , S , V = imcompression . svd ( bw_image ) component_num = [ 1 , 2 , 5 , 10 , 20 , 40 , 80 , 160 , 256 ] fig = plt . figure ( figsize = ( 18 , 18 )) # plot several images i = 0 for k in component_num : U_compressed , S_compressed , V_compressed = imcompression . compress ( U , S , V , k ) img_rebuild = imcompression . rebuild_svd ( U_compressed , S_compressed , V_compresse c = np . around ( imcompression . compression_ratio ( bw_image , k ), 4 ) r = np . around ( imcompression . recovered_variance_proportion ( S , k ), 3 ) ax = fig . add_subplot ( 3 , 3 , i + 1 , xticks = [], yticks = []) ax . imshow ( img_rebuild , cmap = plt . cm . bone ) ax . set_title ( f"{ k } Components" ) if not STUDENT_VERSION : ax . text (
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 11/71 1.2.2 Black and White Compression Savings [No Points] This question will use your implementation of the functions from Q1.1 to compare the number of bytes required to represent the SVD decomposition for the original image to the compressed image using different degrees of compression. You can simply run the below cell without making any changes to it, assuming you have implemented the functions in Q1.1. 0.5 , 0.5 , EO_TEXT , transform = ax . transAxes , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) ax . set_xlabel ( f"Compression: { c },\nRecovered Variance: { r }" ) i = i + 1
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 12/71 Running this cell is primarily for your own understanding of the compression process. 1 components: Original Image: 10.986 MB -> Compressed Image: 18.758 KB, Savings: 1 0.968 MB, Compression Ratio 599.8:1 2 components: Original Image: 10.986 MB -> Compressed Image: 37.516 KB, Savings: 1 0.95 MB, Compression Ratio 299.9:1 5 components: Original Image: 10.986 MB -> Compressed Image: 93.789 KB, Savings: 1 0.895 MB, Compression Ratio 120.0:1 10 components: Original Image: 10.986 MB -> Compressed Image: 187.578 KB, Savings: 10.803 MB, Compression Ratio 60.0:1 20 components: Original Image: 10.986 MB -> Compressed Image: 375.156 KB, Savings: 10.62 MB, Compression Ratio 30.0:1 40 components: Original Image: 10.986 MB -> Compressed Image: 750.312 KB, Savings: 10.254 MB, Compression Ratio 15.0:1 80 components: Original Image: 10.986 MB -> Compressed Image: 1.465 MB, Savings: 9.521 MB, Compression Ratio 7.5:1 160 components: Original Image: 10.986 MB -> Compressed Image: 2.931 MB, Savings: 8.055 MB, Compression Ratio 3.7:1 256 components: Original Image: 10.986 MB -> Compressed Image: 4.689 MB, Savings: 6.297 MB, Compression Ratio 2.3:1 1.3.1 Color image [5 pts] **[W]** This section will use your implementation of the functions from Q1.1 to generate a set of images compressed to different degrees. You can simply run the below cell without making any changes to it, assuming you have implemented the functions in Q1.1. Make sure these images are displayed when submitting the PDF version of the Juypter notebook as part of the non-programming submission of this assignment. NOTE: You might get warning "Clipping input data to the valid range for imshow with RGB data ([0..1] for floats or [0..255] for integers)." This warning is acceptable since some of the pixels may go above 1.0 while rebuilding. You should see similar images to original even with such clipping. HINT 1: Make sure your implementation of recovered_variance_proportion returns an array of 3 floats for a color image. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from imgcompression import ImgCompression imcompression = ImgCompression () bw_image = rgb2gray ( image ) U , S , V = imcompression . svd ( bw_image ) component_num = [ 1 , 2 , 5 , 10 , 20 , 40 , 80 , 160 , 256 ] # Compare memory savings for BW image for k in component_num : og_bytes , comp_bytes , savings = imcompression . memory_savings ( bw_image , U , S , V , comp_ratio = og_bytes / comp_bytes og_bytes = imcompression . nbytes_to_string ( og_bytes ) comp_bytes = imcompression . nbytes_to_string ( comp_bytes ) savings = imcompression . nbytes_to_string ( savings ) print ( f"{ k } components: Original Image: { og_bytes } -> Compressed Image: { comp_byt )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 13/71 HINT 2: Try performing SVD on the individual color channels and then stack the individual channel , , matrices. HINT 3: You may need separate implementations for a color or grayscale image in the same function. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from imgcompression import ImgCompression imcompression = ImgCompression () image_rolled = np . moveaxis ( image , - 1 , 0 ) U , S , V = imcompression . svd ( image_rolled ) component_num = [ 1 , 2 , 5 , 10 , 20 , 40 , 80 , 160 , 256 ] fig = plt . figure ( figsize = ( 18 , 18 )) # plot several images i = 0 for k in component_num : U_compressed , S_compressed , V_compressed = imcompression . compress ( U , S , V , k ) img_rebuild = np . clip ( imcompression . rebuild_svd ( U_compressed , S_compressed , V_compressed ), 0 , 1 ) img_rebuild = np . moveaxis ( img_rebuild , 0 , - 1 ) c = np . around ( imcompression . compression_ratio ( image_rolled , k ), 4 ) r = np . around ( imcompression . recovered_variance_proportion ( S , k ), 3 ) ax = fig . add_subplot ( 3 , 3 , i + 1 , xticks = [], yticks = []) ax . imshow ( img_rebuild ) ax . set_title ( f"{ k } Components" ) if not STUDENT_VERSION : ax . text ( 0.5 , 0.5 , EO_TEXT , transform = ax . transAxes , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) ax . set_xlabel ( f"Compression: { np . around ( c , 4 ) },\nRecovered Variance: R: { r [ 0 ] } G: { r [ 1 ] } ) i = i + 1
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 14/71 1.3.2 Color Compression Savings [No Points] This question will use your implementation of the functions from Q1.1 to compare the number of bytes required to represent the SVD decomposition for the original image to the compressed image using different degrees of compression. You can simply run the below cell without making any changes to it, assuming you have implemented the functions in Q1.1. Running this cell is primarily for your own understanding of the compression process. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from imgcompression import ImgCompression imcompression = ImgCompression () U , S , V = imcompression . svd ( image_rolled ) component_num = [ 1 , 2 , 5 , 10 , 20 , 40 , 80 , 160 , 256 ] # Compare the memory savings of the color image i = 0
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 15/71 1 components: Original Image: 32.959 MB -> Compressed Image: 56.273 KB, Savings: 3 2.904 MB, Compression Ratio 599.8:1 2 components: Original Image: 32.959 MB -> Compressed Image: 112.547 KB, Savings: 32.849 MB, Compression Ratio 299.9:1 5 components: Original Image: 32.959 MB -> Compressed Image: 281.367 KB, Savings: 32.684 MB, Compression Ratio 120.0:1 10 components: Original Image: 32.959 MB -> Compressed Image: 562.734 KB, Savings: 32.409 MB, Compression Ratio 60.0:1 20 components: Original Image: 32.959 MB -> Compressed Image: 1.099 MB, Savings: 3 1.86 MB, Compression Ratio 30.0:1 40 components: Original Image: 32.959 MB -> Compressed Image: 2.198 MB, Savings: 3 0.761 MB, Compression Ratio 15.0:1 80 components: Original Image: 32.959 MB -> Compressed Image: 4.396 MB, Savings: 2 8.563 MB, Compression Ratio 7.5:1 160 components: Original Image: 32.959 MB -> Compressed Image: 8.793 MB, Savings: 24.166 MB, Compression Ratio 3.7:1 256 components: Original Image: 32.959 MB -> Compressed Image: 14.068 MB, Savings: 18.891 MB, Compression Ratio 2.3:1 Q2: Understanding PCA [20 pts] **[P]** | **[W]** Principal Component Analysis (PCA) is another dimensionality reduction technique that reduces dimensions or features while still preserving the maximum (or close-to) amount of information. This is useful when analyzing large datasets that contain a high number of dimensions or features that may be correlated. PCA aims to eliminate features that are highly correlated and only retain the important/uncorrelated ones that can describe most or all the variance in the data. This enables better interpretability and visualization of the multi- dimensional data. In this problem, we will investigate how PCA can be used to improve features for regression and classification tasks and how the data itself affects the behavior of PCA. Here, we will employ Singular Value Decomposition (SVD) for PCA. In PCA, we first center the data by subtracting the mean of each feature. SVD is well suited for this task since each singular value tells us the amount of variance captured in each component for a given matrix (e.g. image). Hence, we can use SVD to extract data only in directions with high variances using either a threshold of the amount of variance or the number of bases/components. Here, we will reduce the data to a set number of components. Recall from class that in PCA, we project the original matrix into new components, each one corresponding to an eigenvector of the covariance matrix . We know that SVD decomposes X into three matrices U, S, and V^T. We can find the SVD decomposition of X^T*X using the decomposition for X as follows: for k in component_num : og_bytes , comp_bytes , savings = imcompression . memory_savings ( image_rolled , U , S , V , k ) comp_ratio = og_bytes / comp_bytes og_bytes = imcompression . nbytes_to_string ( og_bytes ) comp_bytes = imcompression . nbytes_to_string ( comp_bytes ) savings = imcompression . nbytes_to_string ( savings ) print ( f"{ k } components: Original Image: { og_bytes } -> Compressed Image: { comp_byt )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 16/71 This means two important things for us: The matrix , often referred to as the right singular vectors of , is equivalent to the eigenvectors of . is equivalent to the eigenvalues of . So the first -principal components are obtained by projecting by the first vectors from . Similarly, gives a measure of the variance retained. 2.1 Implementation [10 pts] **[P]** Implement PCA. In the pca.py file, complete the following functions: fit : You may use np.linalg.svd . Set full_matrices=False . Hint 1 may be useful. transform transform_rv : You may find np.cumsum helpful for this function. Assume a dataset is composed of N datapoints, each of which has D features with D < N. The dimension of our data would be D. However, it is possible that many of these dimensions contain redundant information. Each feature explains part of the variance in our dataset, and some features may explain more variance than others. HINT 1: Make sure you remember to first center your data by subtracting the mean of each feature. 2.1.1 Local Tests for PCA [No Points] You may test your implementation of the functions contained in pca.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "PCA fit"! UnitTest passed successfully for "PCA transform"! UnitTest passed successfully for "PCA transform with recovered variance"! 2.2 Visualize [5 pts] **[W]** PCA is used to transform multivariate data tables into smaller sets so as to observe the hidden trends and variations in the data. It can also be used as a feature extractor for images. Here you will visualize two datasets using PCA, first is the iris dataset and then a dataset of masked and unmasked images. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestPCA unittest_pca = TestPCA () unittest_pca . test_pca () unittest_pca . test_transform () unittest_pca . test_transform_rv ()
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 17/71 In the pca.py , complete the following function: visualize : Use your implementation of PCA and reduce the datasets such that they contain only two features. Using Plotly's Express make a 2d and 3d scatterplot of the data points using these features. Make sure to differentiate the data points according to their true labels using color. We recommend converting the data to a pandas dataframe before plotting. The datasets have already been loaded for you in the subsequent cells. NOTE: Here, we won't be testing for accuracy. Even with correct implementations of PCA, the accuracy can differ from the TA solution. That is fine as long as the visualizations come out similar. Iris Dataset In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Use PCA for visualization of iris dataset from pca import PCA iris_data = load_iris ( return_X_y = True ) X = iris_data [ 0 ] y = iris_data [ 1 ] fig_title = "Iris Dataset with Dimensionality Reduction" PCA () . visualize ( X , y , fig_title )
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 18/71 0.5 1 Iris Dataset with Dimensionality Reduction (2D)
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 19/71 2.3 PCA Reduced Facemask Dataset Analysis [5 pts] **[W]** Facemask Dataset The masked and unmasked dataset is made up of grayscale images of human faces facing forward. Half of these images are faces that are completely unmasked, and the remaining images show half of the face covered with an artificially generated face mask. The images have already been preprocessed, they are also reduced to a small size of 64x64 pixels and then reshaped into a feature vector of 4096 pixels. Below is a sample of some of the images in the dataset. Iris Dataset with Dimensionality Reduction (3D) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### X = np . load ( "./data/smallflat_64.npy" ) y = np . load ( "./data/masked_labels.npy" ) . astype ( "int" ) i = 0 fig = plt . figure ( figsize = ( 18 , 18 )) for idx in [ 0 , 1 , 2 , 150 , 151 , 152 ]: ax = fig . add_subplot ( 6 , 6 , i + 1 , xticks = [], yticks = []) image = ( np . rot90 ( X [ idx ] . reshape ( 64 , 64 ), k = 1 ) if idx % 2 == 1 and idx < 150 else X [ idx ] . reshape ( 64 , 64 ) )
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 20/71 m_status = "Unmasked" if idx < 150 else "Masked" ax . imshow ( image , cmap = "gray" ) ax . set_title ( f"{ m_status } Image at i = { idx }" ) i += 1 In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Use PCA for visualization of masked and unmasked images X = np . load ( "./data/smallflat_64.npy" ) y = np . load ( "./data/masked_labels.npy" ) fig_title = "Facemask Dataset Visualization with Dimensionality Reduction" PCA () . visualize ( X , y , fig_title ) print ( "*In this plot, the 0 points are unmasked images and the 1 points are masked im ) 4 6 8 Facemask Dataset Visualization with Dimensionality Reduct
11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 21/71 *In this plot, the 0 points are unmasked images and the 1 points are masked image s. What do you think of this 2 dimensional plot, knowing that the original dataset was originally a set of flattened image vectors that had 4096 pixels/features?. 1. Look at the 2-dimensional plot above. If the facemask dataset that has been reduced to 2 features was fed into a classifier, do you think the classifier would produce high accuracy or low accuracy in comparison to the original dataset which had 4096 pixels/features? Why? You can refer to the 2D visualization made above (One or two sentences will suffice for this question) (3 pts) Answer - The 2D plot shows some distinct clustering, which suggests that the two features derived from dimensionality reduction may capture significant aspects for classification. However, because dimensionality reduction from 4096 to 2 features loses a considerable amount of detail, a classifier using only these 2 features will have lower accuracy compared to one using all original features. 1. Assuming an equal rate of accuracy, what do you think is the main advantage in feeding a classifier a dataset with 2 features vs a dataset with 4096 features? (One sentence will suffice for this question.) (2 pts) Answer - The main advantage of feeding a classifier a dataset with 2 features versus 4096 features is the significant reduction in computational complexity and resource Facemask Dataset Visualization with Dimensionality Reduct
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 22/71 usage, leading to faster training and prediction times. 2.4 PCA Exploration [No Points] Note The accuracy can differ from the TA solution and this section is not graded. Emotion Dataset [No Points] Now you will use PCA on an actual real-world dataset. We will use your implementation of PCA function to reduce the dataset with 99% retained variance and use it to obtain the reduced features. On the reduced dataset, we will use logistic and linear regression to compare results between PCA and non-PCA datasets. Run the following cells to see how PCA works on regression and classification tasks. The first dataset we will use is an emotion dataset made up of grayscale images of human faces faces that are visibly happy and visibly sad. Note how Accuracy increases after reducing the number of features used. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### X = np . load ( "./data/emotion_features.npy" ) y = np . load ( "./data/emotion_labels.npy" ) . astype ( "int" ) i = 0 fig = plt . figure ( figsize = ( 18 , 18 )) for idx in [ 0 , 1 , 2 , 150 , 151 , 152 ]: ax = fig . add_subplot ( 6 , 6 , i + 1 , xticks = [], yticks = []) image = ( np . rot90 ( X [ idx ] . reshape ( 64 , 64 ), k = 1 ) if idx % 2 == 1 and idx < 150 else X [ idx ] . reshape ( 64 , 64 ) ) m_status = "Unmasked" if idx < 150 else "Masked" ax . imshow ( image , cmap = "gray" ) m_status = "Sad" if idx < 150 else "Happy" ax . set_title ( f"{ m_status } Image at i = { idx }" ) i += 1 In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### X = np . load ( "./data/emotion_features.npy" ) y = np . load ( "./data/emotion_labels.npy" ) . astype ( "int" ) print ( "Not Graded - Data shape before PCA " , X . shape ) pca = PCA () pca . fit ( X ) X_pca = pca . transform_rv ( X , retained_variance = 0.99 )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 23/71 Not Graded - Data shape before PCA (600, 4096) Not Graded - Data shape with PCA (600, 150) Not Graded - Accuracy before PCA: 0.95000 Not Graded - Accuracy after PCA: 0.95556 Now we will explore sklearn's Diabetes dataset using PCA dimensionality reduction and regression. Notice the RMSE score reduction after we apply PCA. print ( "Not Graded - Data shape with PCA " , X_pca . shape ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Train, test splits X_train , X_test , y_train , y_test = train_test_split ( X , y , test_size = 0.3 , stratify = y , random_state = 42 ) # Use logistic regression to predict classes for test set clf = LogisticRegression () clf . fit ( X_train , y_train ) preds = clf . predict_proba ( X_test ) print ( "Not Graded - Accuracy before PCA: {:.5f}" . format ( accuracy_score ( y_test , preds . argmax ( axis = 1 )) ) ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Train, test splits X_train , X_test , y_train , y_test = train_test_split ( X_pca , y , test_size = 0.3 , stratify = y , random_state = 42 ) # Use logistic regression to predict classes for test set clf = LogisticRegression () clf . fit ( X_train , y_train ) preds = clf . predict_proba ( X_test ) print ( "Not Graded - Accuracy after PCA: {:.5f}" . format ( accuracy_score ( y_test , preds . argmax ( axis = 1 )) ) ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from sklearn.linear_model import RidgeCV Ridge def apply_regression ( X_train , y_train , X_test ): ridge ridge = RidgeCV Ridge ( alphas = [ 1e-3 , 1e-2 , 1e-1 , 1 ]) clf = ridge ridge . fit ( X_train , y_train ) y_pred = ridge ridge . predict ( X_test ) return y_pred In [ ]: ############################### ### DO NOT CHANGE THIS CELL ###
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 24/71 (442, 10) (442,) Not Graded - data shape with PCA (442, 7) Not Graded - RMSE score using Ridge Regression before PCA: 53.101 Ridge Not Graded - RMSE score using Ridge Regression after PCA: 53.024 Ridge Q3 Polynomial regression and regularization [80pts: 50pts + 20pts Bonus for Undergrads + 10pts Bonus for All] **[P]** | **[W]** ############################### # load the dataset diabetes = load_diabetes () X = diabetes . data y = diabetes . target print ( X . shape , y . shape ) pca = PCA () pca . fit ( X ) X_pca = pca . transform_rv ( X , retained_variance = 0.9 ) print ( "Not Graded - data shape with PCA " , X_pca . shape ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Train, test splits X_train , X_test , y_train , y_test = train_test_split ( X , y , test_size = 0.3 , random_state = 42 ) # Ridge regression without PCA Ridge y_pred = apply_regression ( X_train , y_train , X_test ) # calculate RMSE rmse_score = np . sqrt ( mean_squared_error ( y_pred , y_test )) print ( "Not Graded - RMSE score using Ridge Regression before PCA: Ridge {:.5}" . format ( rmse_score ) ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Ridge regression with PCA Ridge X_train , X_test , y_train , y_test = train_test_split ( X_pca , y , test_size = 0.3 , random_state = 42 ) # use Ridge Regression for getting predicted labels Ridge y_pred = apply_regression ( X_train , y_train , X_test ) # calculate RMSE rmse_score = np . sqrt ( mean_squared_error ( y_pred , y_test )) print ( "Not Graded - RMSE score using Ridge Regression after PCA: Ridge {:.5}" . format ( rmse_s )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 25/71 3.1 Regression and regularization implementations [50pts: 30 pts + 20 pts bonus for CS 4641] **[P]** We have three methods to fit linear and ridge regression models: 1) closed form solution; 2) ridge gradient descent (GD); 3) stochastic gradient descent (SGD). Some of the functions are bonus, see the below function list on what is required to be implemented for graduate and undergraduate students. We use the term weight in the following code. Weights and parameters ( ) have the same meaning here. We used parameters ( ) in the lecture slides. In the regression.py file, complete the Regression class by implementing the listed functions below. We have provided the Loss function, , associated with the GD and SGD function for Linear and Ridge Regression for deriving the gradient update. Ridge rmse construct_polynomial_feats predict linear_fit_closed : You should use np.linalg.pinv in this function linear_fit_GD (bonus for undergrad, required for grad ): linear_fit_SGD (bonus for undergrad, required for grad ): ridge_fit_closed ridge : You should adjust your I matrix to handle the bias term differently than the rest of the terms ridge_fit_GD ridge (bonus for undergrad, required for grad ): ridge_fit_SGD ridge (bonus for undergrad, required for grad ): ridge_cross_validation ridge : Use ridge_fit_closed ridge for this function IMPORTANT NOTE: Use your RMSE function to calculate actual loss when coding GD and SGD, but use the loss listed above to derive the gradient update. In ridge_fit_GD ridge and ridge_fit_SGD ridge , you should avoid applying regularization to the bias term in the gradient update. The points for each function is in the Deliverables and Points Distribution section. 3.1.1 Local Tests for Helper Regression Functions [No Points]
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 26/71 You may test your implementation of the functions contained in regression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "RMSE"! UnitTest passed successfully for "Polynomial feature construction"! UnitTest passed successfully for "Linear regression prediction"! 3.1.2 Local Tests for Linear Regression Functions [No Points] You may test your implementation of the functions contained in regression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "Closed form linear regression"! 3.1.3 Local Tests for Ridge Regression Functions [No Points] Ridge You may test your implementation of the functions contained in regression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "Closed form ridge regression"! ridge UnitTest passed successfully for "Ridge regression cross validation"! Ridge 3.1.4 Local Tests for Gradient Descent and SGD (Bonus for Undergrad Tests) [No Points] In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestRegression unittest_reg = TestRegression () unittest_reg . test_rmse () unittest_reg . test_construct_polynomial_feats () unittest_reg . test_predict () In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestRegression unittest_reg = TestRegression () unittest_reg . test_linear_fit_closed () In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestRegression unittest_reg = TestRegression () unittest_reg . test_ridge_fit_closed ridge () unittest_reg . test_ridge_cross_validation ridge ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 27/71 You may test your implementation of the functions contained in regression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. UnitTest passed successfully for "Gradient descent linear regression"! UnitTest passed successfully for "Stochastic gradient descent linear regression"! UnitTest passed successfully for "Gradient descent ridge regression"! ridge 3.2 About RMSE [3 pts] **[W]** What is a good RMSE value? If we normalize our labels such that the true labels and the model outputs can only be between 0 and 1, what does it mean when the RMSE = 1? Please provide an example with your explanation. Answer - In general, a "good" RMSE value is one that is low relative to the range of the dependent variable in your dataset. It's ideally closer to 0, which would indicate that the predicted values are very close to the actual values. When both the true labels and the model outputs are normalized to be between 0 and 1, an RMSE value of 1 indicates the worst possible prediction accuracy. This means that on average, the predicted values are at the maximum possible error distance from the actual values. For example, if the true value of an observation is 0 (the minimum in the normalized scale), and the model predicts 1 (the maximum in the normalized scale), the error for this prediction is 1 - 0 = 1. If every prediction made by the model is off by the maximum amount (1 in this case), then the RMSE would be 1. This scenario would occur if the model always predicts the exact opposite of the true value, which indicates that the model has essentially learned nothing and is performing as poorly as possible. 3.3 Testing: General Functions and Linear Regression [5 pts] ** [W]** In this section. we will test the performance of the linear regression. As long as your test RMSE score is close to the TA's answer (TA's answer ), you can get full points. Let's first construct a dataset for polynomial regression. In this case, we construct the polynomial features up to degree 5. Each data sample consists of two features . We compute the polynomial features of both and in order to yield In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestRegression unittest_reg = TestRegression () unittest_reg . test_linear_fit_GD () unittest_reg . test_linear_fit_SGD () unittest_reg . test_ridge_fit_GD ridge () #unittest_reg.test_ridge_fit_SGD() ridge
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 28/71 the vectors and . We train our model with the cartesian product of these polynomial features. The cartesian product generates a new feature vector consisting of all polynomial combinations of the features with degree less than or equal to the specified degree. For example, if degree = 2, we will have the polynomial features and for the datapoint . The cartesian product of these two vectors will be . We do not generate and since their degree is greater than 2 (specified degree). x_all: 700 (rows/samples) 2 (columns/features) y_all: 700 (rows/samples) 1 (columns/features) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from regression import Regression from plotter import Plotter In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Generate a sample regression dataset with polynomial features # using the student's regression implementation. POLY_DEGREE = 8 reg = Regression () plotter = Plotter ( regularization = reg , poly_degree = POLY_DEGREE , student_version = STUDENT_VERSION , eo_params = ( EO_TEXT , EO_FONT , EO_COLOR , EO_ALPHA , EO_SIZE , EO_ROT ), ) x_all , y_all , p , x_all_feat , x_cart_flat = plotter . create_data () In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Visualize simulated regression data plotter . plot_all_data ( x_all , y_all , p )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 29/71 In the figure above, the red curve is the true fuction we want to learn, while the blue dots are the noisy data points. The data points are generated by , where are i.i.d. generated noise. Now let's split the data into two parts, the training set and testing set. The yellow dots are for training, while the black dots are for testing. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### xtrain , ytrain , xtest , ytest , train_indices , test_indices = plotter . split_data ( x_all , y_all ) plotter . plot_split_data ( xtrain , xtest , ytrain , ytest )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 30/71 Now let us train our model using the training set and see how our model performs on the testing set. Observe the red line, which is our model's learned function. Linear (closed) RMSE: 1.0072 In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for both Grad and Undergrad weight = reg . linear_fit_closed ( x_all_feat [ train_indices ], y_all [ train_indices ]) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all [ test_indices ]) y_pred = reg . predict ( x_all_feat , weight ) print ( "Linear (closed) RMSE: %.4f" % test_rmse ) plotter . plot_linear_closed ( xtrain , xtest , ytrain , ytest , x_all , y_pred )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 31/71 HINT: If your RMSE is off, make sure to follow the instruction given for linear_fit_closed in the list of functions to implement above. Now let's use our linear gradient descent function with the same setup. Observe that the trendline is now less optimal, and our RMSE increased. Do not be alarmed. Linear (GD) RMSE: 5.8381 In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for Grad Only # This cell may take more than 1 minute weight , _ = reg . linear_fit_GD ( x_all_feat [ train_indices ], y_all [ train_indices ], epochs = 50000 , learning_rate = 1e ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all [ test_indices ]) print ( "Linear (GD) RMSE: %.4f" % test_rmse ) y_pred = reg . predict ( x_all_feat , weight ) y_pred = np . reshape ( y_pred , ( y_pred . size ,)) plotter . plot_linear_gd ( xtrain , xtest , ytrain , ytest , x_all , y_pred )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 32/71 We must tune our epochs and learning_rate. As we tune these parameters our trendline will approach the trendline generated by the linear closed form solution. Observe how we slowly tune (increase) the epochs and learning_rate below to create a better model. Note that the closed form solution will always give the most optimal/overfit results. We cannot outperform the closed form solution with GD. We can only approach closed forms level of optimality/overfitness. We leave the reasoning behind this as an exercise to the reader. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for Grad Only # This cell may take more than 1 minute learning_rates = [ 1e-8 , 1e-6 , 1e-4 ] weights = np . zeros (( 3 , POLY_DEGREE ** 2 + 2 )) for ii in range ( len ( learning_rates )): weights [ ii , :] = reg . linear_fit_GD ( x_all_feat [ train_indices ], y_all [ train_indices ], epochs = 50000 , learning_rate = learning_rates [ ii ], )[ 0 ] . ravel () y_test_pred = reg . predict ( x_all_feat [ test_indices ], weights [ ii , :] . reshape (( POLY_DEGREE ** 2 + 2 , 1 ))
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 33/71 Linear (GD) RMSE: 5.8381 (learning_rate=1e-08) Linear (GD) RMSE: 2.7259 (learning_rate=1e-06) Linear (GD) RMSE: 1.2373 (learning_rate=0.0001) And what if we just use the first 10 data points to train? Linear Closed 10 Samples ) test_rmse = reg . rmse ( y_test_pred , y_all [ test_indices ]) print ( "Linear (GD) RMSE: %.4f (learning_rate=%s)" % ( test_rmse , learning_rates [ plotter . plot_linear_gd_tuninglr ( xtrain , xtest , ytrain , ytest , x_all , x_all_feat , learning_rates , weights ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### rng = np . random . RandomState ( seed = 3 ) y_all_noisy = np . dot ( x_cart_flat , np . zeros (( POLY_DEGREE ** 2 + 2 , 1 ))) + rng . randn ( x_all_feat . shape [ 0 ], 1 ) sub_train = train_indices [ 10 : 20 ] In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for both Grad and Undergrad
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 34/71 Linear (closed) 10 Samples RMSE: 2207393.1407 Did you see a worse performance? Let's take a closer look at what we have learned. 3.4 Testing: Testing ridge regression [5 pts] ridge **[W]** 3.4.1 [3pts] **[W]** Now let's try ridge regression. Like before, undergraduate students need to implement the ridge closed form, and graduate students need to implement all three methods. We will call the prediction function from linear regression part. As long as your test RMSE score is close to the TA's answer (TA's answer ), you can get full points. Again, let's see what we have learned. You only need to run the cell corresponding to your specific implementation. weight = reg . linear_fit_closed ( x_all_feat [ sub_train ], y_all_noisy [ sub_train ]) y_pred = reg . predict ( x_all_feat , weight ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all_noisy [ test_indices ]) print ( "Linear (closed) 10 Samples RMSE: %.4f" % test_rmse ) plotter . plot_linear_closed_10samples ( x_all , y_all_noisy , sub_train , y_pred ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ###############################
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 35/71 Ridge Regression (closed) RMSE: 1.8283 Ridge HINT: Make sure to follow the instruction given for ridge_fit_closed ridge in the list of functions to implement above. # Required for both Grad and Undergrad weight = reg . ridge_fit_closed ridge ( x_all_feat [ sub_train ], y_all_noisy [ sub_train ], c_lambda = 10 ) y_pred = reg . predict ( x_all_feat , weight ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all_noisy [ test_indices ]) print ( "Ridge Regression (closed) RMSE: Ridge %.4f" % test_rmse ) plotter . plot_ridge_closed_10samples ridge ( x_all , y_all_noisy , sub_train , y_pred ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for Grad Only weight , _ = reg . ridge_fit_GD ridge ( x_all_feat [ sub_train ], y_all_noisy [ sub_train ], c_lambda = 20 , learning_rate = 1e-5 ) y_pred = reg . predict ( x_all_feat , weight ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all_noisy [ test_indices ]) print ( "Ridge Regression (GD) RMSE: Ridge %.4f" % test_rmse ) plotter . plot_ridge_gd_10samples ridge ( x_all , y_all_noisy , sub_train , y_pred )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 36/71 Ridge Regression (GD) RMSE: 1.0416 Ridge Ridge Regression (SGD) RMSE: 1.0433 Ridge In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Required for Grad Only weight , _ = reg . ridge_fit_SGD ridge ( x_all_feat [ sub_train ], y_all_noisy [ sub_train ], c_lambda = 20 , learning_rate = 1e-5 ) y_pred = reg . predict ( x_all_feat , weight ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all_noisy [ test_indices ]) print ( "Ridge Regression (SGD) RMSE: Ridge %.4f" % test_rmse ) plotter . plot_ridge_sgd_10samples ridge ( x_all , y_all , sub_train , y_pred )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 37/71 Linear vs. Ridge Regression Ridge Regression technique comparison Analyze the difference in performance between the linear and ridge regression methods ridge given the output RMSE from the testing on 10 samples and their corresponding approximation plots. 3.4.2 Why does ridge regression achieve a lower RMSE than linear regression on 10 ridge sample points? [1pts] **[W]** 3.4.3 Describe and contrast two scenarios (real life applications): One where linear is more suitable than ridge, and one in which ridge is better choice than linear. Explain why. [1 pts] ridge ridge **[W]** 3.4.4 [Bonus pts] What is the impact of having some highly correlated features on the data set in terms of linear algebra? Mathematically explain (include expressions) how ridge has an advantage on this in comparison to linear regression. Include the idea of ridge numerical stability. [2pts Bonus For All] **[W]** 3.4.2. Answer - Ridge regression achieves a lower RMSE than linear regression on a small Ridge dataset of 10 sample points because it includes a regularization term that prevents overfitting. This regularization makes the model more robust to noise and fluctuations in the data, leading to better generalization and lower error on unseen data. 3.4.3. Answer - Linear Regression Suitable Scenario : Linear regression is well-suited for predicting a company's sales based on straightforward, linearly related variables such as advertising
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 38/71 spend and market size. In this scenario, the relationship is direct and clear, with minimal risk of overfitting due to the simplicity of the factors involved. Ridge Regression Suitable Scenario Ridge : Ridge regression is more appropriate for financial risk Ridge modeling where many correlated variables (like various market indicators and economic factors) might influence the outcome. Here, the regularization in ridge regression helps ridge manage the multicollinearity and complexity of the data, providing more reliable predictions. 3.4.4. Answer Linear Regression In linear regression, the coefficient calculation uses the formula: where: : Feature matrix. : Target vector. : Coefficients. Issues arise when features in are highly correlated, leading to becoming nearly singular (determinant close to zero). This causes numerical instability in computing and results in unreliable coefficient estimates. Ridge Regression Ridge Ridge regression modifies this formula to: Ridge where: : Regularization term ( is the regularization parameter, is the identity matrix). This addition prevents from being singular by ensuring that remains invertible. It increases numerical stability and reduces the impact of multicollinearity. So in summary, Ridge regression's regularization term Ridge enhances numerical stability in the presence of correlated features, making the model estimation more robust compared to standard linear regression. 3.5 Cross validation [7 pts] **[W]** Let's use Cross Validation to search for the best value for c_lambda in ridge regression. ridge Imagine we have a dataset of 10 points [1,2,3,4,5,6,7,8,9,10] and we want to do 5-fold cross validation. The first iteration we would train with [3,4,5,6,7,8,9,10] and test (validate) with [1,2]
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 39/71 The second iteration we would train with [1,2,5,6,7,8,9,10] and test (validate) with [3,4] The third iteration we would train with [1,2,3,4,7,8,9,10] and test (validate) with [5,6] The fourth iteration we would train with [1,2,3,4,5,6,9,10] and test (validate) with [7,8] The fifth iteration we would train with [1,2,3,4,5,6,7,8] and test (validate) with [9,10] We provided a list of possible values for , and you will use them in cross validation. For cross validation, use 10-fold method and only use it for your training data (you already have the train_indices to get training data). For the training data, split them in 10 folds which means that use 10 percent of training data for test and 90 percent for training. For each , you will have calculated 10 RMSE values. Compute the mean of the 10 RMSE values. Then pick the with the lowest mean RMSE. HINTS: np.concatenate is your friend Make sure to follow the instruction given for ridge_fit_closed ridge in the list of functions to implement above. To use the 10-fold method, that would include looping over all the data 10 times, where we split a different 10% of the data at every iteration. So the first iteration extracts the first 10% to testing and the remaining 90% for training.The second iteration splits the second 10% of data for testing and the (different) remaining 90% for testing. If we have the array of elements 1 - 10, the second iteration would extract the number "2" because that's in the second 10% of the array. The hyperparameter_search function will handle averaging the errors, so don't average the errors in ridge_cross_validation ridge . We've done this so you can see your error across every fold when using the gradescope tests. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### lambda_list = [ 0.0001 , 0.001 , 0.1 , 1 , 5 , 10 , 50 , 100 , 1000 , 10000 ] kfold = 10 best_lambda , best_error , error_list = reg . hyperparameter_search ( x_all_feat [ train_indices ], y_all [ train_indices ], lambda_list , kfold ) for lm , err in zip ( lambda_list , error_list ): print ( "Lambda: %.4f" % lm , "RMSE: %.6f" % err ) print ( "Best Lambda: %.4f" % best_lambda ) weight = reg . ridge_fit_closed ridge ( x_all_feat [ train_indices ], y_all_noisy [ train_indices ], c_lambda = best_lambda ) y_test_pred = reg . predict ( x_all_feat [ test_indices ], weight ) test_rmse = reg . rmse ( y_test_pred , y_all_noisy [ test_indices ]) print ( "Best Test RMSE: %.4f" % test_rmse )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 40/71 Lambda: 0.0001 RMSE: 0.957365 Lambda: 0.0010 RMSE: 0.955850 Lambda: 0.1000 RMSE: 0.953591 Lambda: 1.0000 RMSE: 0.951137 Lambda: 5.0000 RMSE: 0.949576 Lambda: 10.0000 RMSE: 0.949279 Lambda: 50.0000 RMSE: 0.949647 Lambda: 100.0000 RMSE: 0.951808 Lambda: 1000.0000 RMSE: 1.162351 Lambda: 10000.0000 RMSE: 3.065832 Best Lambda: 10.0000 Best Test RMSE: 1.0463 3.6 Noisy Input Samples in Linear Regression [10 pts Bonus for All] **[W]** Consider a linear model of the form: where and weights . Given the the D- dimension input sample set with corresponding target value , the sum-of-squares error function is: Now, suppose that Gaussian noise is added independently to each of the input sample to generate a new sample set . Here, (an entry of ) has zero mean and variance . For each sample , let , where and is independent across both and indices. 1. (3pts) Show that 2. (7pts) Assume the sum-of-squares error function of the noise sample set is . Prove the expectation of is equivalent to the sum-of-squares error for noise-free input samples with the addition of a weight-decay regularization term (e.g. norm) , in which the bias parameter is omitted from the regularizer. In other words, show that N.B. You should be incorporating your solution from the first part of this problem into the given sum of squares equation for the second part. Write your responses below using LaTeX in Markdown. HINT: During the class, we have discussed how to solve for the weight for ridge regression, ridge the function looks like this:
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 41/71 where the first term is the sum-of-squares error and the second term is the regularization term. N is the number of samples. In this question, we use another form of the ridge regression, which is: ridge For the Gaussian noise , we have Assume the noise are independent to each other, we have 1. Answer After adding Guassian noise into input sample , we have 1. Answer: According to the previous question, we have For simplicity, put , then we can get Q4: Naive Bayes and Logistic Regression [35pts] ** [P]** | **[W]** In Bayesian classification, we're interested in finding the probability of a label given some observed feature vector , which we can write as . Bayes's theorem tells us how to express this in terms of quantities we can compute more directly: The main assumption in Naive Bayes is that, given the label, the observed features are conditionally independent i.e.
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 42/71 Therefore, we can rewrite Bayes rule as Training Naive Bayes One way to train a Naive Bayes classifier is done using frequentist approach to calculate probability, which is simply going over the training data and calculating the frequency of different observations in the training set given different labels. For example, Testing Naive Bayes During the testing phase, we try to estimate the probability of a label given an observed feature vector. We combine the probabilities computed from training data to estimate the probability of a given label. For example, if we are trying to decide between two labels and , then we compute the ratio of the posterior probabilities for each label: All we need now is to compute and for each label by plugging in the numbers we got during training. The label with the higher posterior probabilities is the one that is selected. 4.1 Llama Breed Problem using Naive Bayes [5pts] [W] Above are images of two different breeds of llamas – the Suri and the Wooly. The difference between these two breeds is subtle, as these two breeds are often mixed up. However the Suri Llama is vastly more valuable than the Wooly llama. You devise a way to determine with some confidence, which is which – without the need for expensive genetic testing. You look at four key features of the llama: {curly hair, over 14 inch tail, over 400 pounds, extremely shy}. You only have 7 randomly chosen llamas to work with, and their breed as the ground truth. You record the data as vectors with the entry 1 if true and 0 if false. For example a llama with vector {1,1,0,1} would have curly hair, a tail over 14 inches, be less than 400 pounds, and be extremely shy. The Suri Llamas yield the following data: {1, 0, 1, 0}, {0, 1, 0, 1}, {1, 1, 1, 1}, {0, 0, 0, 1}
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 43/71 The Wooly Llamas yield the following data: {0, 0, 1, 0}, {1, 1, 0, 0}, {1, 0, 1, 1}. Now is the time to test your method! You see a new llama you are interested in that has curly hair, does have a tail over 14 inches, is more than 400 pounds, and is not shy. Using Naive Bayes, is this a Suri or a Wooly Llama? NOTE: We expect students to show their work (Naive Bayes calculations) and not just the final answer. Answer From the dataset, we can calculate the probabilities of Suri Llamas and Wooly Llamas: Also, we can calculate the conditional probabilities of four features. For Suri Llamas: Therefore, the probability that the test llama is a Suri Llama can be calculated as follow: For Wooly Llamas: Therefore, the probability that the test llama is a Wooly Llama can be calculated as follow:
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 44/71 Since , this new llama is classified as a Wooly Llama. 4.2 News Data Sentiment Classification via Logistic Regression [30pts] **[P]** This dataset contains the sentiments for financial news headlines from the perspective of a retail investor. The sentiment of news has 3 classes, negative, positive and neutral. In this problem, we only use the negative (class label = 0) and positive (class label = 1) classes for binary logistic regression. For data preprocessing, we remove the duplicate headlines and remove the neutral class to get 1967 unique news headlines. Then we randomly split the 1967 headlines into training set and evaluation set with 8:2 ratio. We use the training set to fit a binary logistic regression model. The code which is provided loads the documents, preprocess the data, builds a “bag of words” representation of each document. Your task is to complete the missing portions of the code in logisticRegression.py to determine whether a news headline is negative or positive. In logistic_regression.py file, complete the following functions: sigmoid : transform to probability of being positive using sigmoid function, which is . bias_augment : augment with 1's to account for bias term in predict_probs : predicts the probability of positive label predict_labels : predicts labels loss : calculates binary cross-entropy loss gradient : calculate the gradient of the loss function with respect to the parameters . accuracy : calculate the accuracy of predictions evaluate : gives loss and accuracy for a given set of points fit : fit the logistic regression model on the training data. Logistic Regression Overview: 1. In logistic regression, we model the conditional probability using parameters , which includes a bias term b. where is the sigmoid function as follows: 1. The conditional probabilities of the positive class and the negative class of the sample attributes are combined into one equation as follows:
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 45/71 1. Assuming that the samples are independent of each other, the likelihood of the entire dataset is the product of the probabilities of all samples. We use maximum likehood estimation to estimate the model parameters . The negative log likelihood (scaled by the dataset size ) is given by: where: number of training samples bag of words features of the i-th training sample label of the i-th training sample Note that this will be our model's loss function 1. Then calculate the gradient and use gradient descent to optimize the loss function: where is the learning rate and the gradient is given by: 4.2.1 Local Tests for Logistic Regression [No Points] You may test your implementation of the functions contained in logistic_regression.py in the cell below. Feel free to comment out tests for functions that have not been completed yet. See Using the Local Tests for more details. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestLogisticRegression unittest_lr = TestLogisticRegression () unittest_lr . test_sigmoid () unittest_lr . test_bias_augment () unittest_lr . test_loss () unittest_lr . test_predict_probs () unittest_lr . test_predict_labels () unittest_lr . test_loss () unittest_lr . test_accuracy () unittest_lr . test_evaluate () unittest_lr . test_fit ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 46/71 UnitTest passed successfully for "Logistic Regression sigmoid"! UnitTest passed successfully for "Logistic Regression bias_augment"! UnitTest passed successfully for "Logistic Regression loss"! UnitTest passed successfully for "Logistic Regression predict_probs"! UnitTest passed successfully for "Logistic Regression predict_labels"! UnitTest passed successfully for "Logistic Regression loss"! UnitTest passed successfully for "Logistic Regression accuracy"! UnitTest passed successfully for "Logistic Regression evaluate"! Epoch 0: train loss: 0.675 train acc: 0.7 val loss: 0.675 val acc: 0.7 UnitTest passed successfully for "Logistic Regression fit"! 4.2.2 Logistic Regression Model Training [No Points] Fit the model to the training data Try different learning rates lr and number of epochs to achieve >80% test accuracy. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from logistic_regression import LogisticRegression as LogReg In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### news_data = pd . read_csv ( "./data/news-data.csv" , encoding = "cp437" , header = None ) class_to_label_mappings = { "negative" : 0 , "positive" : 1 } label_to_class_mappings = { 0 : "negative" , 1 : "positive" } news_data . columns = [ "Sentiment" , "News" ] news_data . drop_duplicates ( inplace = True ) news_data = news_data [ news_data . Sentiment != "neutral" ] news_data [ "Sentiment" ] = news_data [ "Sentiment" ] . map ( class_to_label_mappings ) vectorizer = text . CountVectorizer ( stop_words = "english" ) X = news_data [ "News" ] . values y = news_data [ "Sentiment" ] . values . reshape ( - 1 , 1 ) RANDOM_SEED = 5 BOW = vectorizer . fit_transform ( X ) . toarray () indices = np . arange ( len ( news_data )) X_train , X_test , y_train , y_test , indices_train , indices_test = train_test_split ( BOW , y , indices , test_size = 0.2 , random_state = RANDOM_SEED ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### model = LogReg () lr = 0.05 epochs = 10000 theta = model . fit ( X_train , y_train , X_test , y_test , lr , epochs )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 47/71 Epoch 0: train loss: 0.69 train acc: 0.7 val loss: 0.691 val acc: 0.665 Epoch 1000: train loss: 0.436 train acc: 0.794 val loss: 0.532 val acc: 0.701 Epoch 2000: train loss: 0.364 train acc: 0.846 val loss: 0.484 val acc: 0.746 Epoch 3000: train loss: 0.318 train acc: 0.873 val loss: 0.456 val acc: 0.761 Epoch 4000: train loss: 0.286 train acc: 0.896 val loss: 0.438 val acc: 0.772 Epoch 5000: train loss: 0.262 train acc: 0.914 val loss: 0.425 val acc: 0.782 Epoch 6000: train loss: 0.242 train acc: 0.926 val loss: 0.416 val acc: 0.789 Epoch 7000: train loss: 0.226 train acc: 0.933 val loss: 0.409 val acc: 0.797 Epoch 8000: train loss: 0.212 train acc: 0.943 val loss: 0.404 val acc: 0.802 Epoch 9000: train loss: 0.2 train acc: 0.95 val loss: 0.4 val acc: 0.799 4.2.3 Logistic Regression Model Evaluation [No Points] Evaluate the model on the test dataset Test Dataset Accuracy: 0.807 Plotting the loss function on the training data and the test data for every 100th epoch In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### test_loss , test_acc = model . evaluate ( X_test , y_test , theta ) print ( f"Test Dataset Accuracy: { round ( test_acc , 3 ) }" ) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### model . plot_loss ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 48/71 Plotting the accuracy function on the training data and the test data for each epoch In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### model . plot_accuracy () In [ ]: np . reshape ( X_test [ 0 ], ( 1 , X_test . shape [ 1 ])) . shape
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 49/71 (1, 5286) Check out sample evaluations from the test set. Out[ ]: In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### num_samples = 10 for i in range ( 10 ): rand_index = np . random . randint ( 0 , len ( X_test )) x_test = np . reshape ( X_test [ rand_index ], ( 1 , X_test . shape [ 1 ])) prob = model . predict_probs ( model . bias_augment ( x_test ), theta ) pred = model . predict_labels ( prob ) print ( f"Input News: { X [ indices_test [ rand_index ]] }\n" ) print ( f"Predicted Sentiment: { label_to_class_mappings [ pred [ 0 ][ 0 ]] }" ) print ( f"Actual Sentiment: { label_to_class_mappings [ y_test [ rand_index ][ 0 ]] }\n" )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 50/71 Input News: Operating profit for the 12-month period decreased from EUR5 .4 m whil e net sales increased from EUR62 .0 m , as compared to the financial year 2004 . Predicted Sentiment: negative Actual Sentiment: negative Input News: It is a disappointment to see the plan folded . Predicted Sentiment: positive Actual Sentiment: negative Input News: Finnish Bore that is owned by the Rettig family has grown recently thr ough the acquisition of smaller shipping companies . Predicted Sentiment: positive Actual Sentiment: positive Input News: Meanwhile , Nokia said that it will be able to deliver a complete rang e of services from deployment operations to consulting and integration to managed services as a result of the buyout . Predicted Sentiment: positive Actual Sentiment: positive Input News: Mika Stahlberg , VP F-Secure Labs , said , `` We are excited and proud that F-Secure has been recognized by AV-Comparatives as the Product of the Year . Predicted Sentiment: positive Actual Sentiment: positive Input News: Besides we have increased the share of meat in various sausages and ar e offering a number of new tastes in the grill products and shish kebabs segment , '' Paavel said . Predicted Sentiment: positive Actual Sentiment: positive Input News: Consumption is forecast to grow by about 2 % . Predicted Sentiment: positive Actual Sentiment: positive Input News: In September alone , the market declined by 10.2 percent year-on-year to 19.28 million liters . Predicted Sentiment: positive Actual Sentiment: negative Input News: However , the suspect stole his burgundy Nissan Altima . Predicted Sentiment: positive Actual Sentiment: negative Input News: ADP News - May 29 , 2009 - Bank of America BofA downgraded today its r atings on Swedish-Finnish paper maker Stora Enso Oyj HEL : STERV and on Finnish se ctor player UPM-Kymmene Oyj HEL : UPM1V to `` underperf Predicted Sentiment: negative Actual Sentiment: negative Q5: Noise in PCA and Linear Regression [15pts] ** [W]**
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 51/71 Both PCA and least squares regression can be viewed as algorithms for inferring (linear) relationships among data variables. In this part of the assignment, you will develop some intuition for the differences between these two approaches and develop an understanding of the settings that are better suited to using PCA or better suited to using the least squares fit. The high level bit is that PCA is useful when there is a set of latent (hidden/underlying) variables, and all the coordinates of your data are linear combinations (plus noise) of those variables. The least squares fit is useful when you have direct access to the independent variables, so any noisy coordinates are linear combinations (plus noise) of known variables. 5.1 Slope Functions [5 pts] **[W]** In the following cell , complete the following: 1. pca_slope : For this function, assume that is the first feature and is the second feature for the data. Write a function, that takes in the first feature vector and the second feature vector . Stack these two feature vectors into a single N x 2 matrix and use this to determine the first principal component vector of this dataset. Be careful of how you are stacking the two vectors. You can check the output by printing it which should help you debug. Finally, return the slope of this first component. You should use the PCA implementation from Q2. 2. lr_slope : Write a function that takes and and returns the slope of the least squares fit. You should use the Linear Regression implementation from Q3 but do not use any kind of regularization. Think about how weight could relate to slope. In later subparts, we consider the case where our data consists of noisy measurements of and . For each part, we will evaluate the quality of the relationship recovered by PCA, and that recovered by standard least squares regression. As a reminder, least squares regression minimizes the squared error of the dependent variable from its prediction. Namely, given pairs, least squares returns the line that minimizes . In [ ]: import numpy as np from pca import PCA from regression import Regression def pca_slope ( X , y ): """ Calculates the slope of the first principal component given by PCA Args: x: N x 1 array of feature x y: N x 1 array of feature y Return: slope: (float) scalar slope of the first principal component """ pca = PCA ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 52/71 We will consider a simple example with two variables, and , where the true relationship between the variables is . Our goal is to recover this relationship—namely, recover the coefficient “3”. We set and . Make sure both functions return 3. data = np . concatenate (( X , y ), axis = 1 ) pca . fit ( data ) pc_1 = pca . get_V () slope = pc_1 [ 0 , 1 ] / pc_1 [ 0 , 0 ] return slope def lr_slope ( X , y ): """ Calculates the slope of the best fit returned by linear_fit_closed() For this function don't use any regularization Args: X: N x 1 array corresponding to a dataset y: N x 1 array of labels y Return: slope: (float) slope of the best fit """ reg = Regression () weight = reg . linear_fit_closed ( X , y ) return weight [ 0 , 0 ] In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### x = np . arange ( 0 , 1.02 , 0.02 ) . reshape ( - 1 , 1 ) y = 3 * np . arange ( 0 , 1.02 , 0.02 ) . reshape ( - 1 , 1 ) print ( "Slope of first principal component" , pca_slope ( x , y )) print ( "Slope of best linear fit" , lr_slope ( x , y )) fig = plt . figure () plt . scatter ( x , y ) plt . xlabel ( "x" ) plt . ylabel ( "y" ) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT * 0.8 , ) plt . show ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 53/71 Slope of first principal component 2.9999999999999987 Slope of best linear fit 3.0 5.2 Analysis Setup [5 pts] **[W]** Error in y In this subpart, we consider the setting where our data consists of the actual values of , and noisy estimates of . Run the following cell to see how the data looks when there is error in . In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### base = np . arange ( 0.001 , 1.001 , 0.001 ) . reshape ( - 1 , 1 ) c = 0.5 X = base y = 3 * base + np . random . normal ( loc = [ 0 ], scale = c , size = base . shape ) fig = plt . figure () plt . scatter ( X , y ) plt . xlabel ( "x" ) plt . ylabel ( "y" ) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA , fontname = EO_FONT ,
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 54/71 In following cell , you will implement the addNoise function: 1. Create a vector where . 2. For a given noise level , set , and . You can use the np.random.normal function, where scale is equal to noise level, to add noise to your points. 3. Notice the parameter x*noise in the addNoise function. When this parameter is set to , you will have to add noise to . For a given noise level c, let , and 4. Return the pca_slope and lr_slope values of this and dataset you have created where has noise ( or depending on the problem). Hint 1: Refer to the above example on how to add noise to or Hint 2: Be careful not to add double noise to or ha = "center" , va = "center" , rotation = EO_ROT , ) plt . show () In [ ]: def addNoise ( c , x_noise = False , seed = 1 ): """ Creates a dataset with noise and calculates the slope of the dataset using the pca_slope and lr_slope functions implemented in this class.
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 55/71 A scatter plot with on the horizontal axis and the output of pca_slope and lr_slope on the vertical axis has already been implemented for you. A sample has been taken for each in . The output of pca_slope is plotted as a red dot, and the output of lr_slope as a blue dot. This has been repeated 30 times, you can see that we end up with a plot of 1260 dots, in 21 columns of 60, half red and half blue. Note that the plot you get might not look exactly like the TA version and that is fine because you might have randomized the noise slightly differently than how we did it. NOTE : Here, x_noise = False since we only want Y to have noise. Args: c: (float) scalar, a given noise level to be used on Y and/or X x_noise: (Boolean) When set to False, X should not have noise added When set to True, X should have noise. Note that the noise added to X should be different from the noise added to Y. You should NOT use the same noise you add to Y here. seed: (int) Random seed Return: pca_slope_value: (float) slope value of dataset created using pca_slope lr_slope_value: (float) slope value of dataset created using lr_slope """ np . random . seed ( seed ) #### DO NOT CHANGE THIS #### ############# START YOUR CODE BELOW ############# X = np . arange ( 0.001 , 1 + 0.001 , 0.001 ) Y_cap = 4 * X + np . random . normal ( loc = 0.0 , scale = c , size = ( X . shape )) if x_noise : X = X + np . random . normal ( loc = 0.0 , scale = c , size = ( X . shape )) X = X . reshape (( - 1 , 1 )) Y_cap = Y_cap . reshape (( - 1 , 1 )) #print(X.shape) #print(Y_cap.shape) pca_slope_value = pca_slope ( X , Y_cap ) lr_slope_value = lr_slope ( X , Y_cap ) ############# END YOUR CODE ABOVE ############# return pca_slope_value , lr_slope_value In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### pca_slope_values = [] linreg_slope_values = [] c_values = [] s_idx = 0 for i in range ( 30 ): for c in np . arange ( 0 , 1.05 , 0.05 ): # Calculate pca_slope_value (psv) and lr_slope_value (lsv) psv , lsv = addNoise ( c , seed = s_idx ) # Append pca and lr slope values to list for plot function pca_slope_values . append ( psv ) linreg_slope_values . append ( lsv ) # Append c value to list for plot function
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 56/71 Error in and We will now examine the case where our data consists of noisy estimates of both and . Run the following cell to see how the data looks when there is error in both. c_values . append ( c ) # Increment random seed index s_idx += 1 fig = plt . figure () plt . scatter ( c_values , pca_slope_values , c = "r" ) plt . scatter ( c_values , linreg_slope_values , c = "b" ) plt . xlabel ( "c" ) plt . ylabel ( "slope" ) if not STUDENT_VERSION : fig . text ( 0.6 , 0.4 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA * 0.5 , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) plt . show ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 57/71 In the below cell, we graph the predicted PCA and LR slopes on the vertical axis against the value of c on the horizontal axis. Note that the graph you get might not look exactly like the TA version and that is fine because you might have randomized the noise slightly differently than how we did it. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### base = np . arange ( 0.001 , 1 , 0.001 ) . reshape ( - 1 , 1 ) c = 0.5 X = base + np . random . normal ( loc = [ 0 ], scale = c , size = base . shape ) y = 3 * base + np . random . normal ( loc = [ 0 ], scale = c , size = base . shape ) fig = plt . figure () plt . scatter ( X , y ) plt . xlabel ( "x" ) plt . ylabel ( "y" ) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA * 0.8 , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) plt . show ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 58/71 In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### pca_slope_values = [] linreg_slope_values = [] c_values = [] s_idx = 0 for i in range ( 30 ): for c in np . arange ( 0 , 1.05 , 0.05 ): # Calculate pca_slope_value (psv) and lr_slope_value (lsv), notice x_noise psv , lsv = addNoise ( c , x_noise = True , seed = s_idx ) # Append pca and lr slope values to list for plot function pca_slope_values . append ( psv ) linreg_slope_values . append ( lsv ) # Append c value to list for plot function c_values . append ( c ) # Increment random seed index s_idx += 1 fig = plt . figure () plt . scatter ( c_values , pca_slope_values , c = "r" ) plt . scatter ( c_values , linreg_slope_values , c = "b" ) plt . xlabel ( "c" ) plt . ylabel ( "slope" ) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA * 0.5 , fontname = EO_FONT , ha = "center" , va = "center" , rotation = EO_ROT , ) plt . show ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 59/71 5.3. Analysis [5 pts] **[W]** Based on your observations from previous subsections answer the following questions about the two cases (error in and error in both and ) in 2-3 lines. NOTE: 1. You don't need to provide a mathematical proof for this question. 2. Understanding how PCA and Linear Regression work should help you decipher which case was better for which algorithm. Base your answer on this understanding of how either algorithms works. QUESTIONS: 1. Based on the obtained plots, how can you determine which technique (PCA/Linear regression) is performing better in comparison? (1 Pt) 2. In case-1 where there is error in which technique gave better performance and why do you think it performed better in this case? (2 Pts) 3. In case-2 where there is error in both and which technique gave better performance and why do you think it performed better in this case? (2 Pts) Answer 1. The closer the value of the calculated slope to actual slope ("4" here) the better the algorithm is performing.In the above 2 cases, Linear Regression performs better when we have ground truth X data and noisy Y data, and PCA performs better when X and Y are both noisy.
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 60/71 2. In case-1 where there is error in , Linear Regression performs better than PCA because because it is specifically designed to predict the dependent variable (in this case, ) from the independent variable (here, ), and is robust to noise in the dependent variable. PCA, on the other hand, may be misled by the variance introduced by the noise in , as it treats all variables equally and aims to capture the direction of maximum variance in the data, regardless of whether this variance reflects the underlying relationship between the variables. 3. In case-2 where there is error in both and , PCA performs better than Linear Regression, because the noise affects both dimensions in a similar way, allowing PCA to identify the principal component that captures the core variance in the original and data. In contrast, linear regression, which relies on the assumption of a less noisy independent variable, struggled under these conditions. Q6 Feature Reduction Implementation [25pts Bonus for All] **[P]** | **[W]** 6.1 Implementation [18 Points] **[P]** Feature selection is an integral aspect of machine learning. It is the process of selecting a subset of relevant features that are to be used as the input for the machine learning task. Feature selection may lead to simpler models for easier interpretation, shorter training times, avoidance of the curse of dimensionality, and better generalization by reducing overfitting. In the feature_reduction.py file, complete the following functions: forward_selection backward_elimination These functions should each output a list of features. Forward Selection: In forward selection, we start with a null model, start fitting the model with one individual feature at a time, and select the feature with the minimum p-value. We continue to do this until we have a set of features where one feature's p-value is less than the confidence level. Steps to implement it: 1. Choose a significance level (given to you). 2. Fit all possible simple regression models by considering one feature at a time. 3. Select the feature with the lowest p-value. 4. Fit all possible models with one extra feature added to the previously selected feature(s). 5. Select the feature with the minimum p-value again. if p_value < significance, go to Step 4. Otherwise, terminate. Backward Elimination:
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 61/71 In backward elimination, we start with a full model, and then remove the insignificant feature with the highest p-value (that is greater than the significance level). We continue to do this until we have a final set of significant features. Steps to implement it: 1. Choose a significance level (given to you). 2. Fit a full model including all the features. 3. Select the feature with the highest p-value. If (p-value > significance level), go to Step 4, otherwise terminate. 4. Remove the feature under consideration. 5. Fit a model without this feature. Repeat entire process from Step 3 onwards. HINT 1: The p-value is known as the observed significance value for a null hypothesis. In our case, the p-value of a feature is associated with the hypothesis . If , then this feature contributes no predictive power to our model and should be dropped. We reject the null hypothesis if the p-value is smaller than our significance level. More briefly, a p- value is a measure of how much the given feature significantly represents an observed change. A lower p-value represents higher significance. Some more information about p- values can be found here: https://towardsdatascience.com/what-is-a-p-value-b9e6c207247f HINT 2: For this function, you will have to install statsmodels if not installed already. To do this, run pip install statsmodels in command line/terminal. In the case that you are using an Anaconda environment, run conda install -c conda-forge statsmodels in the command line/terminal. For more information about installation, refer to https://www.statsmodels.org/stable/install.html . The statsmodels library is a Python module that provides classes and functions for the estimation of many different statistical models, as well as for conducting statistical tests, and statistical data exploration. You will have to use this library to choose a regression model to fit your data against. Some more information about this module can be found here: https://www.statsmodels.org/stable/index.html HINT 3: For step 2 in each of the forward and backward selection functions, you can use the sm.OLS function as your regression model. Also, do not forget to add a bias to your regression model. A function that may help you is the sm.add_constants function. TIP 4: You should be able to implement these function using only the libraries provided in the cell below. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from feature_reduction import FeatureReduction In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### bc_dataset = load_breast_cancer () bc = pd . DataFrame ( bc_dataset . data , columns = bc_dataset . feature_names ) bc [ "Diagnosis" ] = bc_dataset . target # print(bc) X = bc . drop ( "Diagnosis" , axis = 1 )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 62/71 Features selected by forward selection: ['worst concave points', 'worst radius', 'worst texture', 'worst area', 'smoothness error', 'worst symmetry', 'compactness error', 'radius error', 'worst fractal dimension', 'mean compactness', 'mean conca ve points', 'worst concavity', 'concavity error', 'area error'] Features selected by backward elimination: ['mean radius', 'mean compactness', 'me an concave points', 'radius error', 'smoothness error', 'concavity error', 'concav e points error', 'worst radius', 'worst texture', 'worst area', 'worst concavity', 'worst symmetry', 'worst fractal dimension'] 6.2 Feature Selection - Discussion [7pts] **[W]** Question 6.2.1: We have seen two regression methods namely Lasso and Ridge regression earlier in this Ridge assignment. Another extremely important and common use-case of these methods is to perform feature selection. Considering there are no restrictions set on the dataset, according to you, which of these two methods is more appropriate for feature selection generally (choose one method)? Why? (3 pts) Answer - For feature selection purposes, Lasso regression is typically the preferred method. Its key advantage lies in its capability to reduce the coefficients of less influential features to zero, thereby excluding them from the equation. This characteristic is especially valuable for pinpointing key features in a large dataset. Ridge regression, on the other hand, usually only Ridge decreases the coefficients close to zero but doesn't set any of them exactly to zero, making it less suitable for direct feature reduction. Question 6.2.2: We have seen that we use different subsets of features to get different regression models. These models depend on the relevant features that we have selected. Using forward selection, what fraction of the total possible models can we explore? Assume that the total number of features that we have at our disposal is . Remember that in stepwise feature selection (like forward selection and backward elimination), we always include an intercept in our model, so you only need to consider the features. (4 pts) Answer In forward selection with features, the fraction of total possible models explored is determined by the number of models considered in forward selection divided by the total possible models. The total number of models with features is . Now if we consider the number of models we take into account at each step, we would have one less model to choose from in every subsequent step: a. At first step, we will have all N features, hence N models to evaluate and pick one. y = bc [ "Diagnosis" ] featureselection = FeatureReduction () # Run the functions to make sure two lists are generated, one for each method print ( "Features selected by forward selection:" , FeatureReduction . forward_selection ( X ) print ( "Features selected by backward elimination:" , FeatureReduction . backward_elimination ( X , y ), )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 63/71 b. At second step, now we have N-1 features, hence N-1 models now c. This process goes until Nth step, where we consider just 1 model that's left Therefore, total number of models explored Therefore, fraction of total models we explore would be: fraction = Q7: Netflix Movie Recommendation Problem Solved using SVD [10pts Bonus for All] **[P]** Let us try to tackle the famous problem of movie recommendation using just our SVD functions that we have implemented. We are given a table of reviews that 600+ users have provided for close to 10,000 different movies. Our challenge is to predict how much a user would rate a movie that they have not seen (or rated) yet. Once we have these ratings, we would then be able to predict which movies to recommend to that user. Understanding How SVD Helps in Movie Recommendation We are given a dataset of user-movie ratings ( ) that looks like the following: Ratings in the matrix range from 1-5. In addition, the matrix contains nan wherever there is no rating provided by the user for the corresponding movie. One simple way to utilize this matrix to predict movie ratings for a given user-movie pair would be to fill in each row / column with the average rating for that row / column. For example: For each movie, if any rating is missing, we could just fill in the average value of all available ratings and expect this to be around the actual / expected rating. While this may sound like a good approximation, it turns out that by just using SVD we can improve the accuracy of the predicted rating. How does SVD fit into this picture? Recall how we previously used SVD to compress images by throwing out less important information. We could apply the same idea to our above matrix ( ) to generate another matrix ( ) which will provide the same information, i.e ratings for any user-movie pairs but by combining only the most important features.
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 64/71 Let's look at this with an example: Assume that decomposition of matrix looks like: We can re-write this decomposition as follows: If we were to take only the top K singular values from this matrix, we could again write this as: Thus we have now effectively separated our ratings matrix into two matrices given by: and There are many ways to visualize the importance of and matrices but with respect to our context of movie ratings, we can visualize these matrices as follows: We can imagine each row of to be holding some information how much each user likes a particular feature (feature1, feature2, feature 3...feature ). On the contrary, we can imagine each column of to be holding some information about how much each movie relates to the given features (feature 1, feature 2, feature 3 ... feature ). Lets denote the row of by and the column of by . Then the dot-product: can provide us with information on how much a user i likes movie j . What have we achieved by doing this? Starting with a matrix containing very few ratings, we have been able to summarize the sparse matrix of ratings into matrices and which each contain feature vectors about the Users and the Movies. Since these feature vectors are summarized from only the most important K features (by our SVD), we can predict any User-Movie rating that is closer to the
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 65/71 actual value than just taking any average rating of a row / column (recall our brute force solution discussed above). Now this method in practice is still not close to the state-of-the-art but for a naive and simple method we have used, we can still build some powerful visualizations as we will see in part 3. We have divided the task into 3 parts: 1. Implement recommender_svd to return matrices and 2. Implement predict to predict top 3 movies a given user would watch 3. (Ungraded) Feel free to run the final cell labeled to see some visualizations of the feature vectors you have generated Hint: Movie IDs are IDs assigned to the movies in the dataset and can be greater than the number of movies. This is why we have given movies_index and users_index as well that map between the movie IDs and the indices in the ratings matrix. Please make sure to use this as well. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from svd_recommender import SVDRecommender from regression import Regression In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### recommender = SVDRecommender () recommender . load_movie_data () regression = Regression () # Read the data into the respective train and test dataframes train , test = recommender . load_ratings_datasets () print ( "---------------------------------------------" ) print ( "Train Dataset Stats:" ) print ( "Shape of train dataset: {}" . format ( train . shape )) print ( "Number of unique users (train): {}" . format ( train [ "userId" ] . unique () . shape [ 0 ] print ( "Number of unique users (train): {}" . format ( train [ "movieId" ] . unique () . shape [ 0 print ( "Sample of Train Dataset:" ) print ( "------------------------------------------" ) print ( train . head ()) print ( "------------------------------------------" ) print ( "Test Dataset Stats:" ) print ( "Shape of test dataset: {}" . format ( test . shape )) print ( "Number of unique users (test): {}" . format ( test [ "userId" ] . unique () . shape [ 0 ])) print ( "Number of unique users (test): {}" . format ( test [ "movieId" ] . unique () . shape [ 0 ]) print ( "Sample of Test Dataset:" ) print ( "------------------------------------------" ) print ( test . head ()) print ( "------------------------------------------" ) # We will first convert our dataframe into a matrix of Ratings: R # R[i][j] will indicate rating for movie:(j) provided by user:(i) # users_index, movies_index will store the mapping between array indices and actual R , users_index , movies_index = recommender . create_ratings_matrix ( train )
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 66/71 --------------------------------------------- Train Dataset Stats: Shape of train dataset: (88940, 4) Number of unique users (train): 671 Number of unique users (train): 8370 Sample of Train Dataset: ------------------------------------------ userId movieId rating timestamp 0 1 2294 2.0 1260759108 1 1 2455 2.5 1260759113 2 1 3671 3.0 1260759117 3 1 1339 3.5 1260759125 4 1 1343 2.0 1260759131 ------------------------------------------ Test Dataset Stats: Shape of test dataset: (10393, 4) Number of unique users (test): 671 Number of unique users (test): 4368 Sample of Test Dataset: ------------------------------------------ userId movieId rating timestamp 0 1 2968 1.0 1260759200 1 1 1405 1.0 1260759203 2 1 1172 4.0 1260759205 3 2 52 3.0 835356031 4 2 314 4.0 835356044 ------------------------------------------ Shape of Ratings Matrix (R): (671, 8370) 7.1.1 Implement the recommender_svd method to use SVD for Recommendation [5pts] **[P]** In svd_recommender.py file, complete the following function: recommender_svd : Use the above equations to output and . You can utilize the svd and compress methods from imgcompression.py to retrieve your initial , and matrices. Then, calculate and based on the decomposition example above. Local Test for recommender_svd Function [No Points] You may test your implementation of the function in the cell below. See Using the Local Tests for more details. print ( "Shape of Ratings Matrix (R): {}" . format ( R . shape )) # Replacing `nan` with average rating given for the movie by all users # Additionally, zero-centering the array to perform SVD mask = np . isnan ( R ) masked_array = np . ma . masked_array ( R , mask ) r_means = np . array ( np . mean ( masked_array , axis = 0 )) R_filled = masked_array . filled ( r_means ) R_filled = R_filled - r_means In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### from utilities.localtests import TestSVDRecommender
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 67/71 UnitTest passed successfully for "recommender_svd() function"! RMSE for k = 2 --> 1.0223035413708281 RMSE for k = 3 --> 1.022526649417955 RMSE for k = 8 --> 1.0182709203352787 RMSE for k = 15 --> 1.017307118738714 RMSE for k = 18 --> 1.0166562048687973 RMSE for k = 25 --> 1.0182856984912254 RMSE for k = 30 --> 1.0186282488126601 Plot the Test Error over the different values of k unittest_svd_rec = TestSVDRecommender () unittest_svd_rec . test_recommender_svd () In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### # Implement the method `recommender_svd` and run it for the following values of fea no_of_features = [ 2 , 3 , 8 , 15 , 18 , 25 , 30 ] test_errors = [] for k in no_of_features : U_k , V_k = recommender . recommender_svd ( R_filled , k ) pred = [] # to store the predicted ratings for _ , row in test . iterrows (): user = row [ "userId" ] movie = row [ "movieId" ] u_index = users_index [ user ] # If we have a prediction for this movie, use that if movie in movies_index : m_index = movies_index [ movie ] pred_rating = np . dot ( U_k [ u_index , :], V_k [:, m_index ]) + r_means [ m_inde # Else, use an average of the users ratings else : pred_rating = np . mean ( np . dot ( U_k [ u_index ], V_k )) + r_means [ m_index ] pred . append ( pred_rating ) test_error = regression . rmse ( test [ "rating" ], pred ) test_errors . append ( test_error ) print ( "RMSE for k = {} --> {}" . format ( k , test_error )) In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### fig = plt . figure () plt . plot ( no_of_features , test_errors , "bo" ) plt . plot ( no_of_features , test_errors ) plt . xlabel ( "Value for k" ) plt . ylabel ( "RMSE on Test Dataset" ) plt . title ( "SVD Recommendation Test Error with Different k values" ) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA * 0.5 , fontname = EO_FONT , ha = "center" , va = "center" ,
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 68/71 7.1.2 Implement the predict method to find which movie a user is interested in watching next [5pts] ** [P]** Our goal here is to predict movies that a user would be interested in watching next. Since our dataset contains a large list of movies and our model is very naive, filtering among this huge set for top 3 movies can produce results that we may not correlate immediately. Therefore, we'll restrict this prediction to only movies among a subset as given by movies_pool. Let us consider a user (ID: 660) who has already watched and rated well (>3) on the following movies: Iron Man (2008) Thor: The Dark World (2013) Avengers, The (2012) The following cell tries to predict which among the movies given by the list below, the user would be most interested in watching next: movies_pool : Ant-Man (2015) rotation = EO_ROT , ) plt . show ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 69/71 Iron Man 2 (2010) Avengers: Age of Ultron (2015) Thor (2011) Captain America: The First Avenger (2011) Man of Steel (2013) Star Wars: Episode IV - A New Hope (1977) Ladybird Ladybird (1994) Man of the House (1995) Jungle Book, The (1994) In svd_recommender.py file, complete the following function: predict : Predict the next 3 movies that the user would be most interested in watching among the ones above. HINT: You can use the method get_movie_id_by_name to convert movie names into movie IDs and vice-versa. NOTE: The user may have already watched and rated some of the movies in movies_pool . Remember to filter these out before returning the output. The original Ratings Matrix, might come in handy here along with np.isnan Local Test for predict Functions [No Points] You may test your implementation of the function in the cell below. See Using the Local Tests for more details. Top 3 Movies the User would want to watch: Captain America: The First Avenger (2011) Ant-Man (2015) Avengers: Age of Ultron (2015) -------------------------------------------------------------- UnitTest passed successfully for "predict() function"! 7.2 Visualize Movie Vectors [No Points] Our model is still a very naive model, but it can still be used for some powerful analysis such as clustering similar movies together based on user's ratings. We have said that our matrix that we have generated above contains information about movies. That is, each column in contains (feature 1, feature 2, .... feature ) for each movie. We can also say this in other terms that gives us a feature vector (of length k) for each movie that we can visualize in a -dimensional space. For example, using this feature vector, we can find out which movies are similar or vary. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### unittest_svd_rec . test_predict ()
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 70/71 While we would love to visualize a -dimensional space, the constraints of our 2D screen wouldn't really allow us to do so. Instead let us set and try to plot the feature vectors for just a couple of these movies. As a fun activity run the following cell to visualize how our model separates the two sets of movies given below. NOTE: There are 2 possible visualizations. Your plot could be the one that's given on the expected PDF or the one where the y-coordinates are inverted. In [ ]: ############################### ### DO NOT CHANGE THIS CELL ### ############################### marvel_movies = [ "Thor: The Dark World (2013)" , "Avengers: Age of Ultron (2015)" , "Ant-Man (2015)" , "Iron Man 2 (2010)" , "Avengers, The (2012)" , "Thor (2011)" , "Captain America: The First Avenger (2011)" , ] marvel_labels = [ "Blue" ] * len ( marvel_movies ) star_wars_movies = [ "Star Wars: Episode IV - A New Hope (1977)" , "Star Wars: Episode V - The Empire Strikes Back (1980)" , "Star Wars: Episode VI - Return of the Jedi (1983)" , "Star Wars: Episode I - The Phantom Menace (1999)" , "Star Wars: Episode II - Attack of the Clones (2002)" , "Star Wars: Episode III - Revenge of the Sith (2005)" , ] star_wars_labels = [ "Green" ] * len ( star_wars_movies ) movie_titles = star_wars_movies + marvel_movies genre_labels = star_wars_labels + marvel_labels movie_indices = [ movies_index [ recommender . get_movie_id_by_name ( str ( x ))] for x in movie_titles ] _ , V_k = recommender . recommender_svd ( R_filled , k = 2 ) x , y = V_k [ 0 , movie_indices ], V_k [ 1 , movie_indices ] fig = plt . figure () plt . scatter ( x , y , c = genre_labels ) for i , movie_name in enumerate ( movie_titles ): plt . annotate ( movie_name , ( x [ i ], y [ i ])) if not STUDENT_VERSION : fig . text ( 0.5 , 0.5 , EO_TEXT , transform = fig . transFigure , fontsize = EO_SIZE / 2 , color = EO_COLOR , alpha = EO_ALPHA * 0.5 , fontname = EO_FONT , ha = "center" , va = "center" ,
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11/10/23, 11:49 PM FALL2023_HW3_Student_Bharat file:///C:/Users/Arpit/Downloads/FALL2023_HW3_Student_Bharat.html 71/71 rotation = EO_ROT , )
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