Assignment 7

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 ˆpm1. The x-axis should

FIGURE 8.14. Left: A partition of the predictor space corresponding to Exercise 4a. Right: A tree corresponding to Exercise 4b. display ˆpm1, 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, ˆpm1 = 1 − ˆpm2. 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", "Missclassification","Gini"),lty=c(1,1,1),lwd=c(2.5,2.5,2.5),col=c('red','green','blue'))

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.

1. Split the data set into a training set and a test set.

library(ISLR2)

## Warning: package 'ISLR2' was built under R version 4.4.3

library(tree)

## Warning: package 'tree' was built under R version 4.4.3

library(caret)

## Warning: package 'caret' was built under R version 4.4.3

## Loading required package: ggplot2

## Warning: package 'ggplot2' was built under R version 4.4.3

## Loading required package: lattice

set.seed(1)
train_index<-sample(1:nrow(Carseats),nrow(Carseats)/2)
train_data<-Carseats[train_index,]
test_data<-Carseats[-train_index,]

2. Fit a regression tree to the training set. Plot the tree, and interpret the results.

reg_tree <- tree(Sales ~ ., data = train_data)
summary(reg_tree)

## 
## Regression tree:
## tree(formula = Sales ~ ., data = train_data)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Advertising" "CompPrice"  
## [6] "US"         
## Number of terminal nodes:  18 
## Residual mean deviance:  2.167 = 394.3 / 182 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -3.88200 -0.88200 -0.08712  0.00000  0.89590  4.09900

plot(reg_tree,compress=TRUE,margin=0.1)

## Warning in text.default(x[1L], y[1L], "|", ...): "compress" is not a graphical
## parameter

## Warning in text.default(x[1L], y[1L], "|", ...): "margin" is not a graphical
## parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "compress" is not a
## graphical parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "margin" is not a
## graphical parameter

text(reg_tree, pretty = 0,cex=0.4)

pred_tree <- predict(reg_tree, newdata = test_data)
mse_tree <- mean((pred_tree - test_data$Sales)^2)
mse_tree

## [1] 4.922039

What test MSE do you obtain?

The test MSE is 4.92

3. Use cross-validation in order to determine the optimal level of tree complexity.

cv_tree <- cv.tree(reg_tree)
plot(cv_tree$size, cv_tree$dev, type = "b")

best_size <- cv_tree$size[which.min(cv_tree$dev)]
pruned_tree <- prune.tree(reg_tree, best = best_size)
plot(pruned_tree,compress=TRUE,margin=0.1)

## Warning in text.default(x[1L], y[1L], "|", ...): "compress" is not a graphical
## parameter

## Warning in text.default(x[1L], y[1L], "|", ...): "margin" is not a graphical
## parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "compress" is not a
## graphical parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "margin" is not a
## graphical parameter

text(pruned_tree, pretty = 0,cex=0.5)

pred_pruned <- predict(pruned_tree, newdata = test_data)
mse_pruned <- mean((pred_pruned - test_data$Sales)^2)
mse_pruned

## [1] 4.922039

Does pruning the tree improve the test MSE?

No, pruning the tree generated a test MSE of 4.922, thus indicating minimal impact.

4. Use the bagging approach in order to analyze this data.

library(randomForest)

## Warning: package 'randomForest' was built under R version 4.4.3

## randomForest 4.7-1.2

## Type rfNews() to see new features/changes/bug fixes.

## 
## Attaching package: 'randomForest'

## The following object is masked from 'package:ggplot2':
## 
##     margin

set.seed(1)
bag_model <- randomForest(Sales ~ ., data = train_data, mtry = ncol(train_data) - 1, importance = TRUE)
pred_bag <- predict(bag_model, newdata = test_data)
mse_bag <- mean((pred_bag - test_data$Sales)^2)
mse_bag

## [1] 2.605253

importance(bag_model)

##                %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     18.073863

What test MSE do you obtain? Use the importance() function to determine which variables are most important.

Test MSE decreased to 2.605 using bagging. Price, ShelveLoc, and Age are the most influential predictors of sales.

5. Use random forests to analyze this data.

set.seed(1)
rf_model <- randomForest(Sales ~ ., data = train_data, mtry = 4, importance = TRUE)
pred_rf <- predict(rf_model, newdata = test_data)
mse_rf <- mean((pred_rf - test_data$Sales)^2)
mse_rf

## [1] 2.787584

importance(rf_model)

##                %IncMSE IncNodePurity
## CompPrice   15.7891655     160.57944
## Income       4.1275509     121.12953
## Advertising  9.6425758     111.54581
## Population  -1.3596645      85.92575
## Price       43.4055391     423.06225
## ShelveLoc   37.8850232     311.97119
## Age         13.8924424     174.18229
## Education    0.1960888      62.77782
## Urban        0.1393816      12.92952
## US           6.3532441      30.42255

varImpPlot(rf_model)

m_values <- 1:(ncol(train_data) - 1)
rf_errors <- sapply(m_values, function(m) {
  model <- randomForest(Sales ~ ., data = train_data, mtry = m)
  mean((predict(model, newdata = test_data) - test_data$Sales)^2)
})
plot(m_values, rf_errors, type = "b", xlab = "mtry", ylab = "Test MSE")

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.

Utilizing random forests with mtry=4, test MSE was 2.788 indicating a slightly worse MSE than bagging, and significantly worse than the basic regression tree. Price, ShelveLock, and Age are most important variables.

6. Now analyze the data using BART, and report your results.

library(BART)

## Warning: package 'BART' was built under R version 4.4.3

## Loading required package: nlme

## Loading required package: survival

## 
## Attaching package: 'survival'

## The following object is masked from 'package:caret':
## 
##     cluster

x_train <- train_data[, -which(names(train_data) == "Sales")]
y_train <- train_data$Sales
x_test <- test_data[, -which(names(test_data) == "Sales")]

set.seed(1)
bart_fit <- 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: 2.781850, 1.091850
## x1,x[n*p]: 107.000000, 1.000000
## xp1,xp[np*p]: 111.000000, 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: 2,0.95,0.273474,3,0.23074,7.57815
## *****sigma: 1.088371
## *****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: 4s
## trcnt,tecnt: 1000,1000

mean((bart_fit$yhat.test.mean - test_data$Sales)^2)

## [1] 1.450842

BART model returned MSE 1.451, substantially lower than regression, pruned, bagging, and random forests, suggesting BART is particularly effective at capturing complex, nonlinear relationships in the Carseats dataset

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

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

set.seed(1)
train_index <- sample(1:nrow(OJ), 800)
train_data <- OJ[train_index, ]
test_data <- OJ[-train_index, ]

2. 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..

oj_tree <- tree(Purchase ~ ., data = train_data)
summary(oj_tree)

## 
## Classification tree:
## tree(formula = Purchase ~ ., data = train_data)
## 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

..describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?

The tree used LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, and PctDiscMM variables to make splits. These variables suggest that customer brand loyalty, price differences, and promotional activity play key roles in distinguishing between Citris Hill and Minute Maid purchases. The tree has 9 terminal nodes.

3. Type in the name of the tree object in order to get a detailed text output.

oj_tree

## 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 ) *

Pick one of the terminal nodes, and interpret the information displayed.

Node 8: terminal node where LoyalCH<0.0356, with 59 obversations - Predicted class has very high class probability: 98.3%, indicating almost all customers with extremely low brand loyalty chose Minute Maid. Node deviance is 10.14, indicating high purity (confident and consistent).

4. Create a plot of the tree, and..

plot(oj_tree,compress=TRUE,margin=0.1)

## Warning in text.default(x[1L], y[1L], "|", ...): "compress" is not a graphical
## parameter

## Warning in text.default(x[1L], y[1L], "|", ...): "margin" is not a graphical
## parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "compress" is not a
## graphical parameter

## Warning in plot.xy(xy.coords(x, y), type = type, ...): "margin" is not a
## graphical parameter

text(oj_tree, pretty = 0,cex=0.7)

..interpret the results.

When LoyalCH<0.5036, customers are more likely to purchase Minute Maid, while higher loyalty scores are classified as choosing Citrus Hill. The tree shows that brand loyalty is the strongest predictor, with price and promotion effects modifying the decision in specific loyalty segments.

5. Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels.

oj_pred <- predict(oj_tree, test_data, type = "class")
confusion <- table(Predicted = oj_pred, Actual = test_data$Purchase)
confusion

##          Actual
## Predicted  CH  MM
##        CH 160  38
##        MM   8  64

test_error <- 1 - sum(diag(confusion)) / sum(confusion)
test_error

## [1] 0.1703704

What is the test error rate?

Test error rate: 17.04%

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

cv_oj <- cv.tree(oj_tree, FUN = prune.misclass)

7. 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 = "CV Classification Error")

8. Which tree size corresponds to the lowest cross-validated classification error rate?

best_size <- cv_oj$size[which.min(cv_oj$dev)]
best_size

## [1] 7

Tree size corresponding to lowest cross-validated classification error rate: 7

1. 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.

pruned_oj <- prune.misclass(oj_tree, best = best_size)
plot(pruned_oj)
text(pruned_oj, pretty = 0)

10. Compare the training error rates between the pruned and unpruned trees.

summary(oj_tree)$misclass

## [1] 127 800

summary(pruned_oj)$misclass

## [1] 130 800

Which is higher?

Pruned tree with 7 terminal nodes: 16.25%

11. Compare the test error rates between the pruned and unpruned trees.

oj_pred_pruned <- predict(pruned_oj, test_data, type = "class")
conf_pruned <- table(Predicted = oj_pred_pruned, Actual = test_data$Purchase)
test_error_pruned <- 1 - sum(diag(conf_pruned)) / sum(conf_pruned)
test_error_pruned

## [1] 0.162963

Which is higher?

Unpruned tree: 17.04%