Assignment 7
Valerie Ceciliano
2026-07-27
Chapter 08 (page 361): 3, 8, 9
Q3
Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that dis- plays each of these quantities as a function ofˆ pm1. The x-axis should displayˆ 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,ˆ 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')
p <- seq(0, 1, length.out = 100)
Entropy <- -p * log2(p) - (1 - p) * log2(1 - p) # Example calculation
G <- 2 * p * (1 - p)
plot(p, Entropy, type = 'l', col = 'red', xlab = "p", ylab = "Criteria", ylim = c(0, 1))
lines(p, G, col = 'blue')
legend(0.3,0.15,c("Entropy", "Missclassification","Gini"),lty=c(1,1,1),lwd=c(2.5,2.5,2.5),col=c('red','green','blue'))
Q8
In the lab, a classification tree was applied to the Carseats data set af- ter 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.
- Split the data set into a training set and a test set.
library(ISLR2)
set.seed(1)
train_idx <- sample(nrow(Carseats), nrow(Carseats) / 2)
train_set <- Carseats[train_idx, ]
test_set <- Carseats[-train_idx, ]- Fit a regression tree to the training set. Plot the tree, and inter- pret the results. What test MSE do you obtain?
library(tree)
tree.fit <- tree(Sales ~ ., data = train_set)
plot(tree.fit)
text(tree.fit, pretty = 0)
tree.pred <- predict(tree.fit, test_set)
tree.mse <- mean((test_set$Sales - tree.pred)^2)
cat("Regression Tree Test MSE:", tree.mse, "\n")## Regression Tree Test MSE: 4.922039- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(tree.fit)
plot(cv.carseats$size, cv.carseats$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
best_size <- cv.carseats$size[which.min(cv.carseats$dev)]
cat("Optimal Tree Size by CV:", best_size, "\n")## Optimal Tree Size by CV: 18prune.carseats <- prune.tree(tree.fit, best = best_size)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
prune.pred <- predict(prune.carseats, test_set)
prune.mse <- mean((test_set$Sales - prune.pred)^2)
cat("Pruned Tree Test MSE:", prune.mse, "\n")## Pruned Tree Test MSE: 4.922039Answer: Cross-validation selects a tree size of 18. Since the optimal size is the full tree, pruning does not alter the tree and the test MSE remains 4.92.
- Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to de- termine which variables are most important.
library(randomForest)## randomForest 4.7-1.2## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
bag.fit <- randomForest(Sales ~ ., data = train_set, mtry = 10, importance = TRUE)
bag.pred <- predict(bag.fit, test_set)
bag.mse <- mean((test_set$Sales - bag.pred)^2)
cat("Bagging Test MSE:", bag.mse, "\n")## Bagging Test MSE: 2.605253importance(bag.fit)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863Answer: Test MSE with bagging is 2.61. Price and ShelveLoc are the two most important variables.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which vari- ables are # most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(1)
# Fit Random Forest (mtry = p/3 = 10/3 approx 3)
rf.fit <- randomForest(Sales ~ ., data = train_set, mtry = 3, importance = TRUE)
rf.pred <- predict(rf.fit, test_set)
rf.mse <- mean((test_set$Sales - rf.pred)^2)
cat("Random Forest Test MSE:", rf.mse, "\n")## Random Forest Test MSE: 2.960559importance(rf.fit)## %IncMSE IncNodePurity
## CompPrice 14.8840765 158.82956
## Income 4.3293950 125.64850
## Advertising 8.2215192 107.51700
## Population -0.9488134 97.06024
## Price 34.9793386 385.93142
## ShelveLoc 34.9248499 298.54210
## Age 14.3055912 178.42061
## Education 1.3117842 70.49202
## Urban -1.2680807 17.39986
## US 6.1139696 33.98963- Now analyze the data using BART, and report your results.
library(gbm)## Loaded gbm 2.3.1## This version of gbm is no longer under development. Consider transitioning to gbm3, https://github.com/gbm-developers/gbm3set.seed(1)
boost.fit <- gbm(Sales ~ ., data = train_set, distribution = "gaussian", n.trees = 1000, shrinkage = 0.01)
boost.pred <- predict(boost.fit, test_set, n.trees = 1000)
boost.mse <- mean((test_set$Sales - boost.pred)^2)
cat("Boosting Test MSE:", boost.mse, "\n")## Boosting Test MSE: 2.033312Answer: Test MSE with Boosting is 2.03.
Q9
- This problem involves the OJ data set which is part of the ISLR2 package.
- Create a training set containing a random sample of 800 obser- vations, and a test set containing the remaining observations.
set.seed(1)
train_idx <- sample(nrow(OJ), 800)
train_set <- OJ[train_idx, ]
test_set <- 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.fit <- tree(Purchase ~ ., data = train_set)
summary(tree.fit)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train_set)
## 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 / 800Answer: The training error rate is 15.88% (127 misclassified out of 800) and the tree has 9 terminal nodes.
- 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.fit## 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 ) *Answer: Looking at node 4 (split on LoyalCH < 0.280875): it has 177 observations, a deviance of 140.5 and 86.4% of the observations in this branch choose MM.
- Create a plot of the tree, and interpret the results.
plot(tree.fit)
text(tree.fit, 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?
tree.pred <- predict(tree.fit, test_set, type = "class")
table(tree.pred, test_set$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 64test_err <- mean(tree.pred != test_set$Purchase)
cat("Test Error Rate:", test_err, "\n")## Test Error Rate: 0.1703704- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
cv.oj <- cv.tree(tree.fit, FUN = prune.misclass)
cv.oj## $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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
- Which tree size corresponds to the lowest cross-validated classi- fication error rate?
Answer: Tree size 8 corresponds to the lowest deviance.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.tree(tree.fit, best = 6, method = "misclass")
plot(prune.oj)
text(prune.oj, pretty = 0)
- Compare the training error rates between the pruned and un- pruned trees. Which is higher?
unpruned_train_err <- summary(tree.fit)$misclass[1] / summary(tree.fit)$misclass[2]
pruned_train_err <- summary(prune.oj)$misclass[1] / summary(prune.oj)$misclass[2]
cat("Unpruned Training Error Rate:", unpruned_train_err, "\n")## Unpruned Training Error Rate: 0.15875cat("Pruned Training Error Rate:", pruned_train_err, "\n")## Pruned Training Error Rate: 0.1625- Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, test_set, type = "class")
prune_test_err <- mean(prune.pred != test_set$Purchase)
cat("Unpruned Test Error Rate:", test_err, "\n")## Unpruned Test Error Rate: 0.1703704cat("Pruned Test Error Rate:", prune_test_err, "\n")## Pruned Test Error Rate: 0.162963