Assignment 7
Alexandra Holen
2026-07-17
Chapter 8
Question 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 pm1. The xaxis should display pm1, 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, pm1 = 1- pm2. You could make this plot by hand, but it will be much easier to make in R.
p=seq(0,1,0.0001)
#Gini
G=2*p*(1-p)
#Classification Error
E=1-pmax(p,1-p)
#Entropy
D=-(p*log(p) + (1-p)*log(1-p))
plot(p,D, col="red",ylab="")
lines(p,E,col='green')
lines(p,G,col='blue')
legend(0.3,0.15,c("Entropy", "Misclassification","Gini"),lty=c(1,1,1),lwd=c(2.5,2.5,2.5),col=c('red','green','blue'))
Question 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.
library(ISLR2)
library(tree)## Warning: package 'tree' was built under R version 4.6.1library(randomForest)## Warning: package 'randomForest' was built under R version 4.6.1## randomForest 4.7-1.2## Type rfNews() to see new features/changes/bug fixes.library(BART)## Warning: package 'BART' was built under R version 4.6.1## Loading required package: nlme## Loading required package: survival- Split the data set into a training set and a test set.
train_idx <- sample(1:nrow(Carseats), nrow(Carseats) * 0.7)
carseats.train <- Carseats[train_idx, ]
carseats.test <- Carseats[-train_idx, ]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.carseats <- tree(Sales ~ ., data = carseats.train)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "CompPrice" "Advertising"
## [6] "Income"
## Number of terminal nodes: 19
## Residual mean deviance: 2.284 = 596.1 / 261
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.86100 -1.01200 -0.09753 0.00000 1.09000 3.88200plot(tree.carseats)
text(tree.carseats, pretty = 0, cex = 0.8)
pred.tree <- predict(tree.carseats, newdata = carseats.test)
test_mse_tree <- mean((pred.tree - carseats.test$Sales)^2)
cat("Test MSE for the Single Regression Tree:", test_mse_tree, "\n")## Test MSE for the Single Regression Tree: 4.813943- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b",
xlab = "Tree Size", ylab = "Deviance (RSS)")
best_size <- cv.carseats$size[which.min(cv.carseats$dev)]
cat("Optimal tree size selected by CV:", best_size, "\n")## Optimal tree size selected by CV: 19prune.carseats <- prune.tree(tree.carseats, best = best_size)
pred.prune <- predict(prune.carseats, newdata = carseats.test)
test_mse_prune <- mean((pred.prune - carseats.test$Sales)^2)
cat("Test MSE for the Pruned Tree:", test_mse_prune, "\n")## Test MSE for the Pruned Tree: 4.813943- 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.carseats <- randomForest(Sales ~ ., data = carseats.train, mtry = 10, importance = TRUE)
pred.bag <- predict(bag.carseats, newdata = carseats.test)
test_mse_bag <- mean((pred.bag - carseats.test$Sales)^2)
cat("Test MSE for Bagging:", test_mse_bag, "\n")## Test MSE for Bagging: 2.528781importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 24.603481 189.274612
## Income 9.652677 129.305699
## Advertising 19.423534 138.692538
## Population -1.000631 70.267856
## Price 62.020526 585.983120
## ShelveLoc 67.855830 645.735751
## Age 25.090758 222.756237
## Education 1.215032 48.740091
## Urban 2.268797 11.507730
## US 3.650981 5.785804varImpPlot(bag.carseats)
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(123)
rf.carseats <- randomForest(Sales ~ ., data = carseats.train, mtry = 3, importance = TRUE)
pred.rf <- predict(rf.carseats, newdata = carseats.test)
test_mse_rf <- mean((pred.rf - carseats.test$Sales)^2)
cat("Test MSE for Random Forest:", test_mse_rf, "\n")## Test MSE for Random Forest: 3.026197importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 15.229528 195.25268
## Income 5.590697 182.38525
## Advertising 11.877026 170.57448
## Population -1.457289 120.91234
## Price 38.471823 460.11516
## ShelveLoc 43.246974 481.41708
## Age 14.338502 250.10551
## Education 2.255961 81.71434
## Urban 1.160462 18.48680
## US 2.884426 21.35574Bagging uses m = p of 10 variables, giving an MSE of 2.000448. Reducing m to 3 de-correlates the trees, but in this specific instance, it actually increased the test error rate to 2.6392. Typically, lowering m reduces variance at the expense of bias, but if there are dominant predictors like Price and ShelveLoc, forcing splits on weaker variables can sometimes degrade regression performance.
- Now analyze the data using BART, and report your results.
x.train <- model.matrix(Sales ~ ., data = carseats.train)[, -1]
x.test <- model.matrix(Sales ~ ., data = carseats.test)[, -1]
y.train <- carseats.train$Sales
set.seed(123)
bart.carseats <- gbart(x.train, y.train, x.test = x.test)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 280, 11, 120
## y1,yn: -1.757071, 1.372929
## x1,x[n*p]: 141.000000, 0.000000
## xp1,xp[np*p]: 117.000000, 0.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 67 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.287616,3,0.193657,7.48707
## *****sigma: 0.997084
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,11,0
## *****printevery: 100
##
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 4s
## trcnt,tecnt: 1000,1000pred.bart <- bart.carseats$yhat.test.mean
test_mse_bart <- mean((pred.bart - carseats.test$Sales)^2)
cat("Test MSE for BART:", test_mse_bart, "\n")## Test MSE for BART: 1.391552Question 9
This problem involves the OJ data set which is part of the ISLR2 package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train_idx <- sample(1:nrow(OJ), 800)
oj.train <- OJ[train_idx, ]
oj.test <- OJ[-train_idx, ]- 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?
tree.oj <- tree(Purchase ~ ., data = oj.train)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## 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- 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 ) *- Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)
- 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?
pred.oj <- predict(tree.oj, newdata = oj.test, type = "class")
conf_matrix <- table(pred.oj, oj.test$Purchase)
print(conf_matrix)##
## pred.oj CH MM
## CH 160 38
## MM 8 64test_error_rate <- mean(pred.oj != oj.test$Purchase)
cat("Unpruned Test Error Rate:", test_error_rate, "\n")## Unpruned Test Error Rate: 0.1703704Apply the cv.tree() function to the training set in order to determine the optimal tree size.
Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
cv.oj <- cv.tree(tree.oj, FUN = prune.misclass)
plot(cv.oj$size, cv.oj$dev, type = "b",
xlab = "Tree Size", ylab = "Cross-Validated Misclassification Classification Error Count")
- Which tree size corresponds to the lowest cross-validated classification error rate?
best_oj_size <- cv.oj$size[which.min(cv.oj$dev)]
cat("The lowest cross-validated classification error corresponds to a size of:", min(best_oj_size), "\n")## The lowest cross-validated classification error corresponds to a size of: 7- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.misclass(tree.oj, best = 7)
plot(prune.oj)
text(prune.oj, pretty = 0)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
cat("Unpruned Summary:\n")## Unpruned Summary:summary(tree.oj)$misclass## [1] 127 800cat("Pruned Summary:\n")## Pruned Summary:summary(prune.oj)$misclass## [1] 130 800The pruned tree’s training misclassification count 130/800 is higher than the unpruned tree’s 127/800. This makes complete sense because pruning reduces the model’s flexibility on the training data.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.pruned.oj <- predict(prune.oj, newdata = oj.test, type = "class")
pruned_test_error <- mean(pred.pruned.oj != oj.test$Purchase)
cat("Unpruned Test Error Rate:", test_error_rate, "\n")## Unpruned Test Error Rate: 0.1703704cat("Pruned Test Error Rate:", pruned_test_error, "\n")## Pruned Test Error Rate: 0.162963The unpruned tree’s test error rate 0.1704 is higher than the pruned tree’s 0.1630. This shows that pruning successfully reduced overfitting, improving generalization performance on unseen data