Assignment 7

library(tidyverse)
library(openintro)
library(ISLR)
library(ISLR2)
library(tree)
library(rpart)
library(caret)
library(randomForest)
library(BART)

Exercise 3

Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of \(\hat{p}_{m1}\). The x-axis should display \(\hat{p}_{m1}\), ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.

Hint: In a setting with two classes, \(\hat{p}_{m1}\) = 1 − \(\hat{p}_{m2}\). You could make this plot by hand, but it will be much easier to make in R.

p <- seq(0, 1, 0.001)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, 
        cbind(gini.index, class.error, cross.entropy), 
        ylab = "Value", 
        col = c("blue", "gray", "red"), 
        type = "l", 
        lwd = 1)

legend("topright", 
       legend = c("Gini Index", "Classification Error", "Cross Entropy"), 
       col = c("blue", "gray", "red"), 
       lty = 1, 
       lwd = 1, 
       cex = 0.7)

Exercise 8

In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.

\((a)\) Split the data set into a training set and a test set.

set.seed(123)
train <- sample(1:nrow(Carseats), nrow(Carseats)/2)
strain <- Carseats[train, ]
stest <- Carseats[-train, ]

\((b)\) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

set.seed(123)
tree.seats <- tree(Sales ~ ., data = strain)
summary(tree.seats)

## 
## Regression tree:
## tree(formula = Sales ~ ., data = strain)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Income"      "Age"         "Population" 
## [6] "Education"   "CompPrice"   "Advertising"
## Number of terminal nodes:  18 
## Residual mean deviance:  2.132 = 388.1 / 182 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -4.08000 -0.92870  0.06244  0.00000  0.87020  3.71700

plot(tree.seats)
text(tree.seats, pretty = 0, cex = 0.6)

treeseat.pred <- predict(tree.seats, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)

## [1] 4.395357

The test MSE is 4.395357.

\((c)\) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?

set.seed(123)
cv.seats <- cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b")

prune.car <- prune.tree(tree.seats, best = cv.seats$size[which.min(cv.seats$dev)])
plot(prune.car)
text(prune.car,pretty=0)

treeseat.pred <- predict(prune.car, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)

## [1] 4.658628

Pruning the tree to 14 increased the test MSE from 4.395357 to 4.658628.

\((d)\) Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.

set.seed(123)
bag.seats <- randomForest(Sales~., data = strain, mtry = 10, ntree = 500, 
                          importance = TRUE)
bagseat.pred <- predict(bag.seats, newdata = stest)
mean((bagseat.pred - stest$Sales)^2)

## [1] 2.76144

randomForest::importance(bag.seats)

##                %IncMSE IncNodePurity
## CompPrice   20.3414969    158.911610
## Income       6.6237140     90.369331
## Advertising  5.7777253     72.793558
## Population  -2.2001506     55.786278
## Price       44.3578602    380.255094
## ShelveLoc   48.3345635    387.886972
## Age         18.6296851    187.107660
## Education    2.6619834     55.987493
## Urban        0.9276070      8.152320
## US           0.4202302      5.900097

The test MSE is 2.76144. The most important variables are ShelveLoc, Price, CompPrice, and Age.

\((e)\) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of \(m\), the number of variables considered at each split, on the error rate obtained.

mtry_vals <- c(3, 5, 7, 9)

for (m in mtry_vals) {
  set.seed(123)
  rf.seats <- randomForest(Sales ~ ., data = strain, mtry = m, ntree = 500, importance = TRUE)
  
  pred <- predict(rf.seats, newdata = stest)
  mse <- mean((pred - stest$Sales)^2)
  
  cat("mtry =", m, "- Test MSE =", round(mse, 3), "\n")
  cat("Variable Importance (based on %IncMSE):\n")
  print(randomForest::importance(rf.seats)[, "%IncMSE"])
  cat("------------------------------------------------------------\n")
}

## mtry = 3 - Test MSE = 3.533 
## Variable Importance (based on %IncMSE):
##   CompPrice      Income Advertising  Population       Price   ShelveLoc 
## 11.95200717  5.20336607  7.04990655  2.81702037 31.00971810 31.20191070 
##         Age   Education       Urban          US 
## 16.38791734  0.70334051  0.04221302  1.67637748 
## ------------------------------------------------------------
## mtry = 5 - Test MSE = 3.017 
## Variable Importance (based on %IncMSE):
##   CompPrice      Income Advertising  Population       Price   ShelveLoc 
##  15.6874802   6.1782093   7.4822778   0.1519680  35.6880806  42.1336259 
##         Age   Education       Urban          US 
##  18.3059283   1.8670124  -0.1467837   0.3849865 
## ------------------------------------------------------------
## mtry = 7 - Test MSE = 2.832 
## Variable Importance (based on %IncMSE):
##   CompPrice      Income Advertising  Population       Price   ShelveLoc 
##  18.1621394   6.5536181   5.5340650  -0.1318886  40.2052050  44.2028844 
##         Age   Education       Urban          US 
##  19.7420288   2.4063248   1.2543161   2.1656941 
## ------------------------------------------------------------
## mtry = 9 - Test MSE = 2.746 
## Variable Importance (based on %IncMSE):
##   CompPrice      Income Advertising  Population       Price   ShelveLoc 
## 21.05555659  6.65634770  6.89407868 -0.07229015 44.96551093 47.51448378 
##         Age   Education       Urban          US 
## 20.84240270  2.80047063 -2.15163575  1.90015802 
## ------------------------------------------------------------

The test MSE decreases as \(m\) increases. This suggests that increasing \(m\) reduces the bias of the model, leading to better predictions on the test set. ShelveLoc, Price, CompPrice, and Age continue to have the most variable importance.

\((f)\) Now analyze the data using BART, and report your results.

set.seed(123)

x_train <- strain[, setdiff(names(strain), "Sales")]
y_train <- strain$Sales
x_test  <- stest[, setdiff(names(stest), "Sales")]
y_test  <- stest$Sales

bart_model <- gbart(x.train = x_train, y.train = y_train, x.test = x_test)

## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 200
## y1,yn: 3.230000, 3.070000
## x1,x[n*p]: 104.000000, 1.000000
## xp1,xp[np*p]: 138.000000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 63 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.260569,3,0.191523,7.43
## *****sigma: 0.991574
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****printevery: 100
## 
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 2s
## trcnt,tecnt: 1000,1000

pred_bart <- bart_model$yhat.test.mean
mse_bart <- mean((y_test - pred_bart)^2)
mse_bart

## [1] 1.622453

BART produces a test MSE of 1.622453.

Exercise 9

This problem involves the OJ data set which is part of the ISLR2 package.

\((a)\) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

set.seed(1)
train2 <- sample(1:nrow(OJ), 800)
OJtrain <- OJ[train2, ]
OJtest <- OJ[-train2, ]

\((b)\) Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?

set.seed(1)
tree.OJ <- tree(Purchase ~ ., data = OJtrain)
summary(tree.OJ)

## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "SpecialCH"     "ListPriceDiff"
## [5] "PctDiscMM"    
## Number of terminal nodes:  9 
## Residual mean deviance:  0.7432 = 587.8 / 791 
## Misclassification error rate: 0.1588 = 127 / 800

5 variables were used to construct the tree (LoyalCH, PriceDiff, SpecialCH, ListPriceDiff and PctDiscMM). The training error rate is 0.1588. There are 9 terminal nodes on the tree.

\((c)\) Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed.

tree.OJ

## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 800 1073.00 CH ( 0.60625 0.39375 )  
##    2) LoyalCH < 0.5036 365  441.60 MM ( 0.29315 0.70685 )  
##      4) LoyalCH < 0.280875 177  140.50 MM ( 0.13559 0.86441 )  
##        8) LoyalCH < 0.0356415 59   10.14 MM ( 0.01695 0.98305 ) *
##        9) LoyalCH > 0.0356415 118  116.40 MM ( 0.19492 0.80508 ) *
##      5) LoyalCH > 0.280875 188  258.00 MM ( 0.44149 0.55851 )  
##       10) PriceDiff < 0.05 79   84.79 MM ( 0.22785 0.77215 )  
##         20) SpecialCH < 0.5 64   51.98 MM ( 0.14062 0.85938 ) *
##         21) SpecialCH > 0.5 15   20.19 CH ( 0.60000 0.40000 ) *
##       11) PriceDiff > 0.05 109  147.00 CH ( 0.59633 0.40367 ) *
##    3) LoyalCH > 0.5036 435  337.90 CH ( 0.86897 0.13103 )  
##      6) LoyalCH < 0.764572 174  201.00 CH ( 0.73563 0.26437 )  
##       12) ListPriceDiff < 0.235 72   99.81 MM ( 0.50000 0.50000 )  
##         24) PctDiscMM < 0.196196 55   73.14 CH ( 0.61818 0.38182 ) *
##         25) PctDiscMM > 0.196196 17   12.32 MM ( 0.11765 0.88235 ) *
##       13) ListPriceDiff > 0.235 102   65.43 CH ( 0.90196 0.09804 ) *
##      7) LoyalCH > 0.764572 261   91.20 CH ( 0.95785 0.04215 ) *

Node 9 (LoyalCH) has 118 observations. It also shows that it has a value of LoyalCH < 0.0356415. Over 80% of the observations in this node take the value of MM and just under 20% of the observations take the value of CH.

\((d)\) Create a plot of the tree, and interpret the results.

plot(tree.OJ)
text(tree.OJ, pretty = 0)

LoyalCH, SpecialCH, PriceDiff, PctDiscMM, and ListPriceDiff are the most important variables.

\((e)\) Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate?

set.seed(1)
treeOJ.pred <- predict(tree.OJ, newdata = OJtest, type = "class")
confusionMatrix(OJtest$Purchase, treeOJ.pred)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  38  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.7333          
##     P-Value [Acc > NIR] : 0.0001259       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.8081          
##             Specificity : 0.8889          
##          Pos Pred Value : 0.9524          
##          Neg Pred Value : 0.6275          
##              Prevalence : 0.7333          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.8485          
##                                           
##        'Positive' Class : CH              
## 

1 - confusionMatrix(OJtest$Purchase, treeOJ.pred)$overall['Accuracy']

##  Accuracy 
## 0.1703704

0.1703704 is the test error rate.

\((f)\) Apply the cv.tree() function to the training set in order to determine the optimal tree size.

set.seed(1)
OJcv <- cv.tree(tree.OJ, FUN = prune.misclass)
OJcv

## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 145 145 146 146 167 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"

\((g)\) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

plot(OJcv$size, OJcv$dev, type = "b", xlab = "Tree Size", ylab = "cv classification error rate")

\((h)\) Which tree size corresponds to the lowest cross-validated classification error rate?

OJcv$size[which.min(OJcv$dev)]

## [1] 9

The tree size of 9 corresponds to the lowest cross-validated classification error rate. As this is the same as the number of terminal nodes from the original tree, the size suggested by cross-validation is not considered.

\((i)\) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.

set.seed(1)
prune.OJ <- prune.tree(tree.OJ,best=5)

\((j)\) Compare the training error rates between the pruned and unpruned trees. Which is higher?

summary(tree.OJ)

## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "SpecialCH"     "ListPriceDiff"
## [5] "PctDiscMM"    
## Number of terminal nodes:  9 
## Residual mean deviance:  0.7432 = 587.8 / 791 
## Misclassification error rate: 0.1588 = 127 / 800

summary(prune.OJ)

## 
## Classification tree:
## snip.tree(tree = tree.OJ, nodes = c(4L, 12L, 5L))
## Variables actually used in tree construction:
## [1] "LoyalCH"       "ListPriceDiff"
## Number of terminal nodes:  5 
## Residual mean deviance:  0.8239 = 655 / 795 
## Misclassification error rate: 0.205 = 164 / 800

The pruned tree has a higher training error rate (0.205) than the unpruned tree (0.1588).

\((k)\) Compare the test error rates between the pruned and unpruned trees. Which is higher?

set.seed(1)
treeOJ.pred <- predict(tree.OJ, newdata = OJtest, type = "class")
confusionMatrix(OJtest$Purchase, treeOJ.pred)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  38  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.7333          
##     P-Value [Acc > NIR] : 0.0001259       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.8081          
##             Specificity : 0.8889          
##          Pos Pred Value : 0.9524          
##          Neg Pred Value : 0.6275          
##              Prevalence : 0.7333          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.8485          
##                                           
##        'Positive' Class : CH              
## 

test_error_OJ <- 1 - confusionMatrix(OJtest$Purchase, treeOJ.pred)$overall['Accuracy']
cat("Test error rate for unpruned tree:", test_error_OJ, "\n")

## Test error rate for unpruned tree: 0.1703704

pruneOJ.pred <- predict(prune.OJ, newdata = OJtest, type = "class")
confusionMatrix(OJtest$Purchase, pruneOJ.pred)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 136  32
##         MM  21  81
##                                           
##                Accuracy : 0.8037          
##                  95% CI : (0.7512, 0.8494)
##     No Information Rate : 0.5815          
##     P-Value [Acc > NIR] : 7.709e-15       
##                                           
##                   Kappa : 0.5911          
##                                           
##  Mcnemar's Test P-Value : 0.1696          
##                                           
##             Sensitivity : 0.8662          
##             Specificity : 0.7168          
##          Pos Pred Value : 0.8095          
##          Neg Pred Value : 0.7941          
##              Prevalence : 0.5815          
##          Detection Rate : 0.5037          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.7915          
##                                           
##        'Positive' Class : CH              
## 

test_error_prune <- 1 - confusionMatrix(OJtest$Purchase, pruneOJ.pred)$overall['Accuracy']
cat("Test error rate for pruned tree:", test_error_prune, "\n")

## Test error rate for pruned tree: 0.1962963

The unpruned tree has a lower test error rate (0.1703704) than the pruned tree (0.1962963).