Problem 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 p^m1. The x-axis should display 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, p^m1 = 1 − p^m2. You could make this plot by hand, but it will be much easier to make in R.
p <- seq(0, 1, 0.01)
gini = 2*p*(1-p)
classerror = 1 - pmax(p, 1-p)
crossentropy = -(p*log(p)+(1-p)*log(1-p))
plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='f')
lines(p,gini,type='l', col='#4daf4a')
lines(p,classerror,col='#377eb8')
lines(p,crossentropy,col='#e41a1c')
legend(x='top',legend=c('Gini','Classification Error','Entropy'),
col=c('#4daf4a','#377eb8','#e41a1c'),lty=1,text.width = 0.25)
Problem 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.
- Split the data set into a training set and a test set.
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str(Carseats)## 'data.frame': 400 obs. of 11 variables:
## $ Sales : num 9.5 11.22 10.06 7.4 4.15 ...
## $ CompPrice : num 138 111 113 117 141 124 115 136 132 132 ...
## $ Income : num 73 48 35 100 64 113 105 81 110 113 ...
## $ Advertising: num 11 16 10 4 3 13 0 15 0 0 ...
## $ Population : num 276 260 269 466 340 501 45 425 108 131 ...
## $ Price : num 120 83 80 97 128 72 108 120 124 124 ...
## $ ShelveLoc : Factor w/ 3 levels "Bad","Good","Medium": 1 2 3 3 1 1 3 2 3 3 ...
## $ Age : num 42 65 59 55 38 78 71 67 76 76 ...
## $ Education : num 17 10 12 14 13 16 15 10 10 17 ...
## $ Urban : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 1 2 2 1 1 ...
## $ US : Factor w/ 2 levels "No","Yes": 2 2 2 2 1 2 1 2 1 2 ...set.seed(2025)
train.Index <- createDataPartition(Sales, p=0.8, list = FALSE)
train <- Carseats[train.Index,]
test <- Carseats[-train.Index,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.fit <- tree(Sales ~ ., data = train)
summary(tree.fit)##
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Income" "CompPrice"
## [6] "Advertising"
## Number of terminal nodes: 19
## Residual mean deviance: 2.643 = 798.1 / 302
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.3790 -1.1670 0.0775 0.0000 1.2050 4.3010plot(tree.fit)
text(tree.fit, pretty = 0, cex = 0.55)
tree.pred <- predict(tree.fit, newdata = test)
(mse <- mean((test$Sales - tree.pred) ^2))## [1] 3.841789The most important variables in the tree are ShelveLoc and Price. The test error rate is around 3.84.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(2025)
cv_tree_model <- cv.tree(tree.fit, K = 10)
data.frame(n_leaves = cv_tree_model$size, CV_RSS = cv_tree_model$dev) %>%
mutate(min_CV_RSS = as.numeric(min(CV_RSS) == CV_RSS)) %>%
ggplot(aes(x = n_leaves, y = CV_RSS)) +
geom_line(col = "grey55") +
geom_point(size = 2, aes(col = factor(min_CV_RSS))) +
scale_x_continuous(breaks = seq(1, 17, 2)) +
scale_y_continuous(labels = scales::comma_format()) +
scale_color_manual(values = c("deepskyblue3", "green")) +
theme(legend.position = "none") +
labs(title = "Carseats Dataset - Regression Tree",
subtitle = "Selecting the complexity parameter with cross-validation",
x = "Terminal Nodes",
y = "CV RSS")
From the plot, we can see that the optimal tree is the fully grown tree without pruning, since the best number of terminal nodes is 13. We verify that below.
which.min(cv_tree_model$dev)## [1] 6cv_tree_model$size[6]## [1] 13Now we check how the MSE differs by specifying best=13.
prune.model = prune.tree(tree.fit, best = 13)
prune.pred <- predict(prune.model, test)
mean((prune.pred - test$Sales)^2)## [1] 3.841629There is no difference in the test MSE between unpruned and pruned trees. The fully grown tree is the optimal tree in this case.
- 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(2025)
rf.model <- randomForest(Sales ~ ., data = train, mtry = 10, ntree = 500, importance = T)
rf.model##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 10, ntree = 500, importance = T)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.530131
## % Var explained: 69.35rf.pred <- predict(rf.model, test)
(rf.mse = mean((test$Sales - rf.pred)^2))## [1] 1.644367importance(rf.model) |>
as.data.frame() %>%
rownames_to_column("varname") %>%
arrange(desc(IncNodePurity))## varname %IncMSE IncNodePurity
## 1 ShelveLoc 79.7926818 816.425095
## 2 Price 69.9106181 737.900601
## 3 CompPrice 28.2415076 258.895130
## 4 Age 24.0131573 255.103750
## 5 Income 13.5697669 164.714906
## 6 Advertising 17.8783436 163.631355
## 7 Population -1.8123672 84.026687
## 8 Education -0.1807842 73.647902
## 9 US 6.4733304 24.361600
## 10 Urban 0.3599708 9.510284The test error obtained after bagging is 1.644 which is less than what we obtained from the trees. The most important variables are ShelveLoc, Price, and CompPrice.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
test_MSE <- c()
i <- 1
for (Mtry in 1:10) {
set.seed(2025)
rf_temp <- randomForest(Sales ~ ., data = train, mtry = Mtry, importance = T)
test_pred <- predict(rf_temp, test)
test_MSE[i] <- mean((test_pred - test$Sales)^2)
i <- i + 1
}
data.frame(mtry = 1:10, test_MSE = test_MSE) %>%
mutate(min_test_MSE = as.numeric(min(test_MSE) == test_MSE)) %>%
ggplot(aes(x = mtry, y = test_MSE)) +
geom_line(col = "grey55") +
geom_point(size = 2, aes(col = factor(min_test_MSE))) +
scale_x_continuous(breaks = seq(1, 10), minor_breaks = NULL) +
scale_color_manual(values = c("deepskyblue3", "green")) +
theme(legend.position = "none") +
labs(title = "Carseats Dataset - Random Forests",
subtitle = "Selecting 'mtry' using the test MSE",
x = "mtry",
y = "Test MSE")
By tuning the mtry parameters obtained the best random forest model with an mtry of 9
test_MSE[9]## [1] 1.592166and the test_MSE is 1.592
rf.model.1 <- randomForest(Sales ~ ., data = train, mtry = 9, importance = T)
importance(rf.model.1) |>
as.data.frame() %>%
rownames_to_column("varname") %>%
arrange(desc(IncNodePurity))## varname %IncMSE IncNodePurity
## 1 ShelveLoc 81.6223752 819.459494
## 2 Price 66.1708476 738.717960
## 3 CompPrice 29.7620325 257.958231
## 4 Age 22.7082569 254.026790
## 5 Advertising 19.3180729 171.403524
## 6 Income 12.8022062 167.781751
## 7 Population 1.1874768 92.229784
## 8 Education -1.0015192 71.743042
## 9 US 4.9008990 24.928679
## 10 Urban -0.1782073 9.674885The test error decreased after tuning the mtry parameter of the random forest model, with the best performance achieved at mtry = 9. The most important predictors were ShelveLoc and Price.
Problem 9
- This problem involves the OJ data set which is part of the ISLR package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
detach(Carseats)
attach(OJ)
str(OJ)## 'data.frame': 1070 obs. of 18 variables:
## $ Purchase : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
## $ WeekofPurchase: num 237 239 245 227 228 230 232 234 235 238 ...
## $ StoreID : num 1 1 1 1 7 7 7 7 7 7 ...
## $ PriceCH : num 1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
## $ PriceMM : num 1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
## $ DiscCH : num 0 0 0.17 0 0 0 0 0 0 0 ...
## $ DiscMM : num 0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
## $ SpecialCH : num 0 0 0 0 0 0 1 1 0 0 ...
## $ SpecialMM : num 0 1 0 0 0 1 1 0 0 0 ...
## $ LoyalCH : num 0.5 0.6 0.68 0.4 0.957 ...
## $ SalePriceMM : num 1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
## $ SalePriceCH : num 1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
## $ PriceDiff : num 0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
## $ Store7 : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
## $ PctDiscMM : num 0 0.151 0 0 0 ...
## $ PctDiscCH : num 0 0 0.0914 0 0 ...
## $ ListPriceDiff : num 0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
## $ STORE : num 1 1 1 1 0 0 0 0 0 0 ...set.seed(2025)
train.Index <- sample(nrow(OJ), 800)
train.OJ <- OJ[train.Index,]
test.OJ <- OJ[-train.Index,]- 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.OJ <- tree(Purchase ~ ., data = train.OJ)
summary(tree.fit.OJ)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train.OJ)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "PriceMM" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7291 = 577.4 / 792
## Misclassification error rate: 0.1688 = 135 / 800The tree uses only five variables LoyalCH, PriceDiff, PriceMM, ListPriceDiff and PctDiscMM. The training error rate is 135 / 800 = 0.1688 and the tree has 8 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.OJ## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1062.000 CH ( 0.62125 0.37875 )
## 2) LoyalCH < 0.48285 296 328.400 MM ( 0.24324 0.75676 )
## 4) LoyalCH < 0.136698 94 33.080 MM ( 0.04255 0.95745 ) *
## 5) LoyalCH > 0.136698 202 258.100 MM ( 0.33663 0.66337 )
## 10) PriceDiff < 0.05 88 80.360 MM ( 0.17045 0.82955 )
## 20) PriceMM < 2.11 63 69.160 MM ( 0.23810 0.76190 ) *
## 21) PriceMM > 2.11 25 0.000 MM ( 0.00000 1.00000 ) *
## 11) PriceDiff > 0.05 114 157.500 MM ( 0.46491 0.53509 ) *
## 3) LoyalCH > 0.48285 504 437.700 CH ( 0.84325 0.15675 )
## 6) LoyalCH < 0.764572 243 288.100 CH ( 0.72016 0.27984 )
## 12) ListPriceDiff < 0.235 96 133.000 MM ( 0.48958 0.51042 )
## 24) PctDiscMM < 0.196196 78 105.600 CH ( 0.58974 0.41026 ) *
## 25) PctDiscMM > 0.196196 18 7.724 MM ( 0.05556 0.94444 ) *
## 13) ListPriceDiff > 0.235 147 113.200 CH ( 0.87075 0.12925 ) *
## 7) LoyalCH > 0.764572 261 91.200 CH ( 0.95785 0.04215 ) *We will interpret the terminal node at 11) PriceDiff > 0.05 114 157.500 MM ( 0.46491 0.53509 )*
The splitting variable is PriceDiff and the splitting value is 0.05. There are 114 points in the subtree below this node. The deviance for all points contained in the region below this node is 157.5 . The * in this line denotes that this is infact a terminal node. The prediction at this node is Sales = CH. About 46.4% of points in this node have MM as value of Sales. The remaining 53.5% has CH has value of Sales.
- Create a plot of the tree, and interpret the results.
plot(tree.fit.OJ)
text(tree.fit.OJ, pretty = 0, cex = 0.55)
LoyalCH is the most important variable. The top 3 nodes contain LoyalCH. If LoyalCH < 0.1367, the tree predicts MM. If LoyalCH > 0.76, the tree predicts CH. For intermediate values of LoyalCH, the decision also depends on the value of the four additional variables.
- 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.fit.OJ, test.OJ, type = 'class')
conf_mat = confusionMatrix(test.OJ$Purchase, pred.oj)test_error_rate <- 1 - conf_mat$overall['Accuracy']
print(test_error_rate)## Accuracy
## 0.1851852The test error rate is 0.185.
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv_oj = cv.tree(tree.fit.OJ, FUN = prune.tree)- 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 classification error rate?
which.min(cv_oj$dev)## [1] 2cv_oj$size[2]## [1] 7The tree size of 7 gives the lowest cross-validation error.
- 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.OJ, best = 7)- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(prune_oj)##
## Classification tree:
## snip.tree(tree = tree.fit.OJ, nodes = 10L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7423 = 588.6 / 793
## Misclassification error rate: 0.1688 = 135 / 800There is no difference in the training error rates between the pruned and unpruned trees.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
unpruned_pred = predict(tree.fit.OJ, test.OJ, type = "class")
unpruned_error = sum(test.OJ$Purchase != unpruned_pred)
unpruned_error/length(unpruned_pred)## [1] 0.1851852pruned_pred = predict(prune_oj, test.OJ, type = "class")
pruned_error = sum(test.OJ$Purchase != pruned_pred)
pruned_error/length(pruned_pred)## [1] 0.1851852There are no changes in the error rates since we used the same 9 nodes in the pruned tree as well.