tutorial_regression2

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Tutorial 9: Regression Continued Lecture and Tutorial Learning Goals: By the end of the week, you will be able to: Recognize situations where a simple regression analysis would be appropriate for making predictions. Explain the -nearest neighbour ( -nn) regression algorithm and describe how it differs from k-nn classification. Interpret the output of a -nn regression. In a dataset with two variables, perform -nearest neighbour regression in R using tidymodels to predict the values for a test dataset. Using R, execute cross-validation in R to choose the number of neighbours. Using R, evaluate -nn regression prediction accuracy using a test data set and an appropriate metric ( e.g. , root means square prediction error). In a dataset with > 2 variables, perform -nn regression in R using tidymodels to predict the values for a test dataset. In the context of -nn regression, compare and contrast goodness of fit and prediction properties (namely RMSE vs RMSPE). Describe advantages and disadvantages of the -nearest neighbour regression approach. Perform ordinary least squares regression in R using tidymodels to predict the values for a test dataset. Compare and contrast predictions obtained from -nearest neighbour regression to those obtained using simple ordinary least squares regression from the same dataset. This tutorial covers parts of the Regression II chapter of the online textbook. You should read this chapter before attempting the worksheet. ### Run this cell before continuing. library ( tidyverse ) library ( repr ) library ( tidymodels ) library ( GGally ) library ( ISLR ) options ( repr.matrix.max.rows = 6 ) source ( "tests.R" ) source ( "cleanup.R" ) Predicting credit card balance In [ ]:

Source: https://media.giphy.com/media/LCdPNT81vlv3y/giphy-downsized- large.gif Here in this worksheet we will work with a simulated data set that contains information that we can use to create a model to predict customer credit card balance. A bank might use such information to predict which customers might be the most profitable to lend to (customers who carry a balance, but do not default, for example). Specifically, we wish to build a model to predict credit card balance ( Balance column) based on income ( Income column) and credit rating ( Rating column). We access this data set by reading it from an R data package that we loaded at the beginning of the worksheet, ISLR . Loading that package gives access to a variety of data sets, including the Credit data set that we will be working with. We will rename this data set credit_original to avoid confusion later in the worksheet. credit_original <- Credit credit_original Question 1.1 {points: 1} Select only the columns of data we are interested in using for our prediction (both the predictors and the response variable) and use the as_tibble function to convert it to a tibble (it is currently a base R data frame). Name the modified data frame credit (using a lowercase c). Note: We could alternatively just leave these variables in and use our recipe formula below to specify our predictors and response. But for this worksheet, let's select the relevant columns first. ### BEGIN SOLUTION credit <- credit_original |> select ( Balance , Income , Rating ) |> as_tibble () In [ ]: In [ ]:

### END SOLUTION credit test_1.1 () Question 1.2 {points: 1} Before we perform exploratory data analysis, we should create our training and testing data sets. First, split the credit data set. Use 60% of the data and set the variables we want to predict as the strata argument. Assign your answer to an object called credit_split . Assign your training data set to an object called credit_training and your testing data set to an object called credit_testing . set.seed ( 2000 ) ### BEGIN SOLUTION credit_split <- initial_split ( credit , prop = 0.6 , strata = Balance ) credit_training <- training ( credit_split ) credit_testing <- testing ( credit_split ) ### END SOLUTION test_1.2 () Question 1.3 {points: 1} Using only the observations in the training data set, use the ggpairs library create a pairplot (also called "scatter plot matrix") of all the columns we are interested in including in our model. Since we have not covered how to create these in the textbook, we have provided you with most of the code below and you just need to provide suitable options for the size of the plot. The pairplot contains a scatter plot of each pair of columns that you are plotting in the lower left corner, the diagonal contains smoothed histograms of each individual column, and the upper right corner contains the correlation coefficient (a quantitative measure of the relation between two variables) Name the plot object credit_pairplot . # options(...) # credit_pairplot <- credit_training |> # ggpairs( # lower = list(continuous = wrap('points', alpha = 0.4)), # diag = list(continuous = "barDiag") # ) + # theme(text = element_text(size = 20)) ### BEGIN SOLUTION options ( repr.plot.height = 8 , repr.plot.width = 9 ) credit_pairplot <- credit_training |> ggpairs ( mapping = aes ( alpha = 0.4 )) + In [ ]: In [ ]: In [ ]: In [ ]:

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theme ( text = element_text ( size = 20 )) ### END SOLUTION credit_pairplot test_1.3 () Question 1.4 Multiple Choice: {points: 1} Looking at the ggpairs plot above, which of the following statements is incorrect ? A. There is a strong positive relationship between the response variable ( Balance ) and the Rating predictor B. There is a strong positive relationship between the two predictors ( Income and Rating ) C. There is a strong positive relationship between the response variable ( Balance ) and the Income predictor D. None of the above statements are incorrect Assign your answer to an object called answer1.4 . Make sure your answer is an uppercase letter and is surrounded by quotation marks (e.g. "F" ). ### BEGIN SOLUTION answer1.4 <- "C" ### END SOLUTION answer1.4 test_1.4 () Question 1.5 {points: 1} Now that we have our training data, we will fit a linear regression model. Create and assign your linear regression model specification to an object called lm_spec . Create a recipe for the model. Assign your answer to an object called credit_recipe . set.seed ( 2020 ) #DO NOT REMOVE ### BEGIN SOLUTION lm_spec <- linear_reg () |> set_engine ( "lm" ) |> set_mode ( "regression" ) credit_recipe <- recipe ( Balance ~ . , data = credit_training ) ### END SOLUTION In [ ]: In [ ]: In [ ]: In [ ]:

print ( lm_spec ) print ( credit_recipe ) test_1.5 () Question 1.6 {points: 1} Now that we have our model specification and recipe, let's put them together in a workflow, and fit our simple linear regression model. Assign the fit to an object called credit_fit . set.seed ( 2020 ) # DO NOT REMOVE ### BEGIN SOLUTION credit_fit <- workflow () |> add_recipe ( credit_recipe ) |> add_model ( lm_spec ) |> fit ( data = credit_training ) ### END SOLUTION credit_fit test_1.6 () Question 1.7 Multiple Choice: {points: 1} Looking at the slopes/coefficients above from each of the predictors, which of the following mathematical equations is correct for your prediction model? A. B. C. D. Assign your answer to an object called answer1.7 . Make sure your answer is an uppercase letter and is surrounded by quotation marks (e.g. "F" ). ### BEGIN SOLUTION answer1.7 <- "A" ### END SOLUTION answer1.7 test_1.7 () In [ ]: In [ ]: In [ ]: In [ ]: In [ ]:

Question 1.8 {points: 1} Calculate the to assess goodness of fit on credit_fit (remember this is how well it predicts on the training data used to fit the model). Return a single numerical value named lm_rmse . set.seed ( 2020 ) # DO NOT REMOVE #... <- credit_fit |> # predict(...) |> # bind_cols(...) |> # ...(truth = ..., estimate = ...) |> # filter(.metric == ...) |> # select(...) |> # pull() ### BEGIN SOLUTION lm_rmse <- credit_fit |> predict ( credit_training ) |> bind_cols ( credit_training ) |> metrics ( truth = Balance , estimate = .pred ) |> filter ( .metric == 'rmse' ) |> select ( .estimate ) |> pull () ### END SOLUTION lm_rmse test_1.8 () Question 1.9 {points: 1} Calculate using the test data. Return a single numerical value named lm_rmspe . set.seed ( 2020 ) # DO NOT REMOVE ### BEGIN SOLUTION lm_rmspe <- credit_fit |> predict ( credit_testing ) |> bind_cols ( credit_testing ) |> metrics ( truth = Balance , estimate = .pred ) |> filter ( .metric == 'rmse' ) |> select ( .estimate ) |> pull () ### END SOLUTION lm_rmspe test_1.9 () Question 1.9.1 {points: 3} In [ ]: In [ ]: In [ ]: In [ ]:

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Redo this analysis using -nn regression instead of linear regression. Use set.seed(2000) at the beginning of this code cell to make it reproducible. Use the same predictors and train - test data splits as you used for linear regression, and use 5-fold cross validation to choose from the range 1-10 . Remember to scale and shift your predictors on your training data, and to apply that same standardization to your test data! Assign a single numeric value for for your k-nn model as your answer, and name it knn_rmspe . set.seed ( 2000 ) # DO NOT REMOVE ### BEGIN SOLUTION # NOTE TO TAs: the recipe step uses randomness, so if the student # does recipe AFTER vfoldcv, they'll get a different answer (184.26) than # if they do recipe BEFORE vfoldcv (179.88). They should get # **full marks in either case** credit_knn_recipe <- recipe ( Balance ~ . , data = credit_training ) |> step_center ( all_predictors ()) |> step_scale ( all_predictors ()) credit_knn_spec <- nearest_neighbor ( weight_func = "rectangular" , neighbors = tun set_engine ( "kknn" ) |> set_mode ( "regression" ) credit_vfold <- vfold_cv ( credit_training , v = 5 , strata = Balance ) credit_knn_workflow <- workflow () |> add_recipe ( credit_knn_recipe ) |> add_model ( credit_knn_spec ) gridvals <- tibble ( neighbors = seq ( 1 , 10 )) credit_knn_results <- credit_knn_workflow |> tune_grid ( resamples = credit_vfold , grid = gridvals ) |> collect_metrics () #select the value of k resulting in best RMSE kmin <- credit_knn_results |> filter ( .metric == 'rmse' ) |> filter ( mean == min ( mean )) |> pull ( neighbors ) #retrain the model using that final k, predict on held-out data credit_spec <- nearest_neighbor ( weight_func = "rectangular" , neighbors = kmin ) | set_engine ( "kknn" ) |> set_mode ( "regression" ) credit_fit <- workflow () |> add_recipe ( credit_knn_recipe ) |> add_model ( credit_spec ) |> fit ( data = credit_training ) knn_rmspe <- credit_fit |> predict ( credit_testing ) |> bind_cols ( credit_testing ) |> metrics ( truth = Balance , estimate = .pred ) |> In [ ]:

filter ( .metric == 'rmse' ) |> pull ( .estimate ) ### END SOLUTION knn_rmspe Question 1.9.2 {points: 3} Discuss which model, linear regression versus -nn regression, gives better predictions and why you think that might be happening. BEGIN SOLUTION Linear regression is giving better predictions as measured by . The for linear regression is ~155 and the for k-nn regression is ~175. This is likely happening because of one or two of the following reasons: Even with the best 𝑘 we can pick, -nn regression could have slightly overfit the training data which would explain that it doesn't generalize as well to data that wasn't used to train it. There is a fairly linear relationship between most/all of the predictors and the target/outcome variable, so linear regression is an appropriate model and fits well. END SOLUTION 2. Ames Housing Prices Source: https://media.giphy.com/media/xUPGGuzpmG3jfeYWIg/giphy.gif If we take a look at the Business Insider report What do millenials want in a home? , we can see that millenials like newer houses that have their own defined

spaces. Today we are going to be looking at housing data to understand how the sale price of a house is determined. Finding highly detailed housing data with the final sale prices is very hard, however researchers from Truman State Univeristy have studied and made available a dataset containing multiple variables for the city of Ames, Iowa. The data set describes the sale of individual residential property in Ames, Iowa from 2006 to 2010. You can read more about the data set here . Today we will be looking at 5 different variables to predict the sale price of a house. These variables are: Lot Area: lot_area Year Built: year_built Basement Square Footage: bsmt_sf First Floor Square Footage: first_sf Second Floor Square Footage: second_sf First, load the data with the script given below. # run this cell ames_data <- read_csv ( 'data/ames.csv' , col_types = cols ()) |> select ( lot_area = Lot.Area , year_built = Year.Built , bsmt_sf = Total.Bsmt.SF , first_sf = `X1st.Flr.SF` , second_sf = `X2nd.Flr.SF` , sale_price = SalePrice ) |> filter ( ! is.na ( bsmt_sf )) ames_data Question 2.1 {points: 3} Split the data into a train dataset and a test dataset, based on a 70%-30% train- test split. Use set.seed(2019) . Remember that we want to predict the sale_price based on all of the other variables. Assign the objects to ames_split , ames_training , and ames_testing , respectively. Use 2019 as your seed for the split. set.seed ( 2019 ) # DO NOT CHANGE! ### BEGIN SOLUTION ames_split <- initial_split ( ames_data , prop = 0.7 , strata = sale_price ) ames_training <- training ( ames_split ) ames_testing <- testing ( ames_split ) ### END SOLUTION # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create objects named ames_split, ames_training and ames_testi In [ ]: In [ ]: In [ ]:

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expect_true ( exists ( "ames_split" )) expect_true ( exists ( "ames_training" )) expect_true ( exists ( "ames_testing" )) }) ### BEGIN HIDDEN TESTS test_that ( 'ames_split should be a rsplit object.' , { expect_true ( 'rsplit' %in% class ( ames_split )) }) test_that ( 'ames_training is not a tibble.' , { expect_true ( 'tbl' %in% class ( ames_training )) }) test_that ( 'ames_training does not contain the correct number of rows and/or colu expect_equal ( dim ( ames_training ), c ( 2048 , 6 )) expect_equal ( digest ( int_round ( sum ( ames_training $ lot_area ), 2 )), '0f473284653 expect_equal ( digest ( int_round ( sum ( ames_training $ first_sf ), 2 )), '46b1007aee0 }) test_that ( 'ames_testing is not a tibble.' , { expect_true ( 'tbl' %in% class ( ames_testing )) }) test_that ( 'ames_testing does not contain the correct number of rows and/or colum expect_equal ( dim ( ames_testing ), c ( 881 , 6 )) expect_equal ( digest ( int_round ( sum ( ames_testing $ lot_area ), 2 )), 'ef74702fa3ef expect_equal ( digest ( int_round ( sum ( ames_testing $ first_sf ), 2 )), '2b626f5c6c11 }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.2 {points: 3} Let's start by exploring the training data. Use the ggpairs() function from the GGally package to explore the relationships between the different variables. Assign your plot object to a variable named answer2.2 . set.seed ( 2020 ) # DO NOT REMOVE ### BEGIN SOLUTION options ( repr.plot.height = 10 , repr.plot.width = 20 ) answer2.2 <- ames_training |> ggpairs ( mapping = aes ( alpha = 0.05 )) + theme ( text = element_text ( size = 17 )) ### END SOLUTION answer2.2 # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create a plot named answer2.2' , { expect_true ( exists ( "answer2.2" )) }) ### BEGIN HIDDEN TESTS test_that ( 'answer2.2 should be using data from ames_training' , { expect_equal ( int_round ( nrow ( answer2.2 $ data ), 0 ), 2048 ) expect_equal ( int_round ( ncol ( answer2.2 $ data ), 0 ), 6 ) }) test_that ( 'answer2.2 should be a pairwise plot matrix.' , { In [ ]: In [ ]:

expect_true ( 'ggmatrix' %in% c ( class ( answer2.2 ))) }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.3 Multiple Choice: {points: 1} Now that we have seen all the relationships between the variables, which of the following variables would not be a strong predictor for sale_price ? A. bsmt_sf B. year_built C. first_sf D. lot_area E. second_sf F. It isn't clear from these plots Assign your answer to an object called answer2.3 . Make sure your answer is an uppercase letter and is surrounded by quotation marks (e.g. "F" ). ### BEGIN SOLUTION answer2.3 <- "F" ### END SOLUTION answer2.3 # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object called answer2.3' , { expect_true ( exists ( 'answer2.3' )) }) ### BEGIN HIDDEN TESTS test_that ( 'Solution is incorrect' , { expect_equal ( digest ( answer2.3 ), 'f76b651ab8fcb8d470f79550bf2af53a' ) }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.4 - Linear Regression {points: 3} Fit a linear regression model using tidymodels with ames_training using all the variables in the data set. create a model specification called lm_spec create a recipe called ames_recipe create a workflow with your model spec and recipe, and then create the model fit and name it ames_fit In [ ]: In [ ]:

set.seed ( 2020 ) # DO NOT REMOVE ### BEGIN SOLUTION lm_spec <- linear_reg () |> set_engine ( "lm" ) |> set_mode ( "regression" ) ames_recipe <- recipe ( sale_price ~ . , data = ames_training ) ames_fit <- workflow () |> add_recipe ( ames_recipe ) |> add_model ( lm_spec ) |> fit ( data = ames_training ) ### END SOLUTION ames_fit # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object named lm_spec' , { expect_true ( exists ( "lm_spec" )) }) test_that ( 'Did not create an object named ames_recipe' , { expect_true ( exists ( "ames_recipe" )) }) test_that ( 'Did not create an object named ames_fit' , { expect_true ( exists ( "ames_fit" )) }) ### BEGIN HIDDEN TESTS test_that ( 'lm_spec is not a linear regression model' , { expect_true ( 'linear_reg' %in% class ( lm_spec )) }) test_that ( 'lm_spec does not contain the correct specifications' , { expect_equal ( digest ( as.character ( lm_spec $ mode )), 'b8bdd7015e0d1c6037512fd139 expect_equal ( digest ( as.character ( lm_spec $ engine )), '0995419f6f003f701c545d05 }) test_that ( 'ames_recipe is not a recipe' , { expect_true ( 'recipe' %in% class ( ames_recipe )) }) test_that ( 'ames_recipe does not contain the correct variables' , { expect_equal ( digest ( int_round ( sum ( ames_recipe $ template $ lot_area ), 2 )), '0f47 expect_equal ( digest ( int_round ( sum ( ames_recipe $ template $ first_sf ), 2 )), '46b1 }) test_that ( 'ames_fit is not a workflow' , { expect_true ( 'workflow' %in% class ( ames_fit )) }) test_that ( 'ames_fit does not contain the correct data' , { expect_equal ( digest ( int_round ( sum ( ames_fit $ pre $ actions $ recipe $ recipe $ templat expect_equal ( digest ( int_round ( sum ( ames_fit $ pre $ actions $ recipe $ recipe $ templat }) test_that ( 'ames_fit coefficients are incorrect' , { expect_equal ( digest ( int_round ( sum ( ames_fit $ fit $ fit $ fit $ coefficients ), 2 )), ' }) print ( "Success!" ) ### END HIDDEN TESTS In [ ]: In [ ]:

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Question 2.5 True or False: {points: 1} Aside from the intercept, all the variables have a positive relationship with the sale_price . This can be interpreted as the value of the variables decrease, the prices of the houses increase. Assign your answer to an object called answer2.5 . Make sure your answer is in lowercase letters and is surrounded by quotation marks (e.g. "true" or "false" ). # run this cell ames_fit $ fit $ fit $ fit $ coefficients ### BEGIN SOLUTION answer2.5 <- "false" ### END SOLUTION answer2.5 # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object named answer2.5' , { expect_true ( exists ( "answer2.5" )) }) ### BEGIN HIDDEN TESTS test_that ( 'Solution is incorrect' , { expect_equal ( digest ( answer2.5 ), 'd2a90307aac5ae8d0ef58e2fe730d38b' ) }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.6 {points: 3} Looking at the coefficients and intercept produced from the cell block above, write down the equation for the linear model. Make sure to use correct math typesetting syntax (surround your answer with dollar signs, e.g. $0.5 * a$ ) BEGIN SOLUTION As long as the coefficients are right and the variable names are right, don't be too picky about math syntax. END SOLUTION In [ ]: In [ ]: In [ ]:

Question 2.7 Multiple Choice: {points: 1} Why can we not easily visualize the model above as a line or a plane in a single plot? A. This is not true, we can actually easily visualize the model B. The intercept is much larger (6 digits) than the coefficients (single/double digits) C. There are more than 2 predictors D. None of the above Assign your answer to an object called answer2.7 . Make sure your answer is an uppercase letter and is surrounded by quotation marks (e.g. "F" ). ### BEGIN SOLUTION answer2.7 <- "C" ### END SOLUTION answer2.7 # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object named answer2.7' , { expect_true ( exists ( "answer2.7" )) }) ### BEGIN HIDDEN TESTS test_that ( 'Solution is incorrect' , { expect_equal ( digest ( answer2.7 ), '475bf9280aab63a82af60791302736f6' ) }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.8 {points: 3} We need to evaluate how well our model is doing. For this question, calculate the (a single numerical value) of the linear regression model using the test data set and assign it to an object named ames_rmspe . set.seed ( 2020 ) # DO NOT REMOVE ### BEGIN SOLUTION ames_rmspe <- ames_fit |> predict ( ames_testing ) |> bind_cols ( ames_testing ) |> metrics ( truth = sale_price , estimate = .pred ) |> filter ( .metric == 'rmse' ) |> select ( .estimate ) |> pull () In [ ]: In [ ]: In [ ]:

### END SOLUTION ames_rmspe # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object named ames_rmspe' , { expect_true ( exists ( "ames_rmspe" )) }) ### BEGIN HIDDEN TESTS test_that ( 'ames_rmspe is incorrect' , { expect_equal ( digest ( int_round ( ames_rmspe , 2 )), '449c6dc6cc4df30b73051b58cabe }) print ( "Success!" ) ### END HIDDEN TESTS Question 2.9 Multiple Choice: {points: 1} Which of the following statements is incorrect ? A. is a measure of goodness of fit B. measures how well the model predicts on data it was trained with C. measures how well the model predicts on data it was not trained with D. measures how well the model predicts on data it was trained with Assign your answer to an object called answer2.9 . Make sure your answer is an uppercase letter and is surrounded by quotation marks (e.g. "F" ). ### BEGIN SOLUTION answer2.9 <- "D" ### END SOLUTION answer2.9 # We check that you've created objects with the right names below # But all other tests were intentionally hidden so that you can practice decidin # when you have the correct answer. test_that ( 'Did not create an object named answer2.9' , { expect_true ( exists ( "answer2.9" )) }) ### BEGIN HIDDEN TESTS test_that ( 'Solution is incorrect' , { expect_equal ( digest ( answer2.9 ), 'c1f86f7430df7ddb256980ea6a3b57a4' ) }) print ( "Success!" ) ### END HIDDEN TESTS source ( "cleanup.R" ) In [ ]: In [ ]: In [ ]: In [ ]:

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