Homework2Final

Homework 2 2023-09-06 Question 3.1 Using the same data set (credit_card_data.txt or credit_card_data-headers.txt) as in Question 2.2, use the ksvm or kknn function to find a good classifier: (a) using cross-validation (do this for the k-nearest-neighbors model;SVM is optional);and (b) splitting the data into training, validation, and test data sets(pick either KNN or SVM;the other is optional). Question 3.1.a For part a, I used cross validation to identify the optimal k value to be 12 with an associated accuracy of 0.8517. I essentially built a function that ran a value of k through the cv.kknn function to produce a vector with decimal outputs showing the confidence of that value to the actual results from the original dataset. Once I had that vector, I used if if-else statement to round values equal to or above 0.5 to be 1 and values below 0.5 to be 0. Then, I checked the values against the results from the dataset to get the number of matches. I divided by the number of rows, to get the optimal probability. Final Answer: k=12 / Accuracy: 0.8517 rm ( list= ls ()) #importing data & loading kknn function data1 <- read.table ( "credit_card_data.txt" , stringsAsFactors = FALSE , header = FALSE ) library (kknn) # Question 3.1.a #Creating a function to test different values of K using KNN cross validation #and find optimal k value kOptimal <- function (kTest){ set.seed ( 1 ) testArray = c () model1 <- cv.kknn (V11 ~ ., data1, kcv= 10 , k= kTest, scale= TRUE ) results1 <- model1[[ 1 ]][, 2 ] #creating variable to pull fitted data from function #using for loop to make each value >=.5 to a 1, and <.5 to a zero. This is #essentially rounding the data to best fit predicted value and later match #to the V11 or results column in the data set for (i in 1 : nrow (data1)) { if (results1[i] >= . 5 ){ i = 1 1

} else { i = 0 } testArray = c (testArray,i) #storing outputs into a new vector } #printing the sum of the number of times the model outputs match those of the #data set. Dividing this by number of rows in data set to get percentage of #correct matches kOptVal <- sum (testArray == data1[, 11 ]) / nrow (data1) return (kOptVal) } #creating for loop to test k values from 2 to 20 in function built above kOptTestValStorage <- c () for (kTest in 2 : 20 ) { save1 <- kOptimal (kTest) kOptTestValStorage <- c (kOptTestValStorage,save1) } #identifying the best k to be 12, the output is 0.8517 Question 3.1.b In this problem, I split the data using a 60/20/20 Training/Validation/Testing breakdown, with points sampled randomly without replacement, until each quota was met. The for loop you will see tests values of k from 2 to 20 using a kknn function with train and validation data. The outputs returned 116 (out of 131 validation values) as the highest value, corresponding to k = 2,3, and 4. I chose 3. Using this value for k, I ran the kknn function again with the optimal k value to get an accuracy of ~89%. I then completed the problem by running the kknn function using the train and test data sets, and got an accuracy of ~90%. Final answer: k=3, Accuracy=~90% set.seed ( 2 ) #Using 60/20/20 Train/Validate/Test Split, sampling randomly w/o replacement train1 <- data1[ sample ( nrow (data1), 392 ),] validate1 <- data1[ sample ( nrow (data1), 131 ),] test1 <- data1[ sample ( nrow (data1), 131 ),] #testing optimal value of k nearest neighbors, using for loop to test values of #k from 2 to 20. The outputs are rounded to the nearest 0 or 1 value then #compared to the values in the validate set response column. The highest value #was 116 which corresponded to k=2,3,4. I chose 3. kTest2 = 2 kTest2Storage <- c () for (kTest2 in 2 : 20 ) { kknn1 <- kknn (V11 ~ ., train1, validate1, k= kTest2, distance = 2 , kernel = "optimal" , scale = TRUE ) kknn1Eval <- round ( fitted (kknn1)) == validate1 $ V11 kknn1EvalVal <- sum (kknn1Eval) kTest2Storage <- c (kTest2Storage,kknn1EvalVal) 2

rm ( list= ls ()) data2 <- read.csv ( "iris.data" , header= FALSE ) table (data2 $ V5) # previewing the amount of each type of flower ## ## Iris-setosa Iris-versicolor Iris-virginica ## 50 50 50 library (ggplot2) #previewing the data comparing V1 to V2 and V3 to V4 ggplot (data2, aes (V1,V2, color= V5)) + geom_point () 2.0 2.5 3.0 3.5 4.0 4.5 5 6 7 8 V1 V2 V5 Iris-setosa Iris-versicolor Iris-virginica ggplot (data2, aes (V3,V4, color= V5)) + geom_point () 4

0.0 0.5 1.0 1.5 2.0 2.5 2 4 6 V3 V4 V5 Iris-setosa Iris-versicolor Iris-virginica set.seed ( 3 ) #Using a for loop to test for the optimal number of cluster centers starting #from 1 to 10. The main goal of this is to build an elbow diagram, with the #number of centers as the x axis and WSS value as the Y axis cent <- 1 centarray <- c () centVal <- c () for (cent in 1 : 10 ) { clusterTest <- kmeans (data2[, 1 : 4 ], centers= cent, nstart= 5 ) wssVal <- clusterTest $ tot.withinss centVal <- c (centVal,cent) centarray <- c (centarray,wssVal ) } plot (centVal,centarray) #optimal k is 3 5