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.

p1 <- seq(0, 1, 0.001)
p2 <- 1 - p1


classification_error <- 1 - pmax(p1, p2)

gini <- p1*(1-p1) + p2*(1-p2)

entropy <- -p1*log(p1) - p2*log(p2)


data.frame(p1, p2, classification_error, gini, entropy) %>%
  pivot_longer(cols = c(classification_error, gini, entropy), names_to = "metric") %>%
  ggplot(aes(x = p1, y = value, col = factor(metric))) + 
  geom_line(size = 1) + 
  scale_y_continuous(breaks = seq(0, 1, 0.1), minor_breaks = NULL) + 
  scale_color_hue(labels = c("Classification Error", "Entropy", "Gini")) +
  labs(col = "Metric", 
       y = "Value", 
       x = "Proportion (of class '1')")

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.

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

set.seed(1, sample.kind = "Rounding") 

train_index <- sample(1:nrow(Carseats), nrow(Carseats) / 2)

train <- Carseats[train_index, ]
test <- Carseats[-train_index, ]
  1. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree_model <- tree(Sales ~ ., train)

plot(tree_model)
text(tree_model, pretty = 0, cex = 0.7)

summary(tree_model)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Advertising" "Income"     
## [6] "CompPrice"  
## Number of terminal nodes:  18 
## Residual mean deviance:  2.36 = 429.5 / 182 
## Distribution of residuals:
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## -4.2570 -1.0360  0.1024  0.0000  0.9301  3.9130
test_pred <- predict(tree_model, test)
mean((test_pred - test$Sales)^2)
## [1] 4.148897

Shelf location is the first variable to split followed by Price. I would think that a third node would split at age, after that splitting doesn’t makes too much sense. Test MSE of 4.148897

  1. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(2)

cv_tree_model <- cv.tree(tree_model, 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")

I would go with 9 or 10 nodes. After 9 the RSS slightly increases. Going futher than 9 nodes the diagram would become too complex to be useful.

  1. 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(3)

bagged_trees_model <- randomForest(y = train$Sales, 
                                   x = train[ ,-1], 
                                   mtry = ncol(train) - 1, # 10
                                   importance = T) 
test_pred <- predict(bagged_trees_model, test)
mean((test_pred - test$Sales)^2)
## [1] 2.55751

The test MSE was lowered to 2.55751 using random forest.

importance(bagged_trees_model) %>%
  as.data.frame() %>%
  arrange(desc(IncNodePurity))
##                %IncMSE IncNodePurity
## Price       61.0815445    510.222759
## ShelveLoc   46.5308006    320.758343
## Age         20.9071380    196.309827
## CompPrice   14.3725974    133.408852
## Advertising 14.2546456    124.572583
## Income       5.9039670     78.035000
## Population   0.5521805     62.833496
## Education    1.3675478     42.260461
## US           6.8176330     15.273874
## Urban       -2.7169457      7.955904

The importance () function says to go with 10 variables.

  1. 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(3)
  
  rf_temp <- randomForest(y = train$Sales, 
                          x = train[ ,-1], 
                          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")

### Problem 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(4)

train_index <- sample(1:nrow(OJ), 800)

train <- OJ[train_index, ]
test <- OJ[-train_index, ]
  1. 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_model <- tree(Purchase ~ ., train)

summary(tree_model)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = train)
## Variables actually used in tree construction:
## [1] "LoyalCH"     "SalePriceMM" "PriceDiff"  
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7677 = 608.8 / 793 
## Misclassification error rate: 0.1788 = 143 / 800

The training error rate is 17.88%. The number of terminal nodes is 7. The variables used in the split were: LoyalCH, SalePriceMM, and PriceDiff.

  1. Create a plot of the tree, and interpret the results.
plot(tree_model)
text(tree_model, pretty = 0, cex = 0.7)

This tree shows people’s preference for Citrus Hill or Minute Maid drinks. The first terminal node shows loyalty, while the bottom terminal nodes show how price might sway someone’s answer.

  1. 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?
test_pred <- predict(tree_model, test, type = "class")
table(test_pred, test_actual = test$Purchase)
##          test_actual
## test_pred  CH  MM
##        CH 153  38
##        MM  15  64
1 - mean(test_pred == test$Purchase)
## [1] 0.1962963
1 - mean(test$Purchase == "CH")
## [1] 0.3777778
  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(5)

cv_tree_model <- cv.tree(tree_model, K = 10, FUN = prune.misclass)
cv_tree_model
## $size
## [1] 7 4 2 1
## 
## $dev
## [1] 166 166 166 315
## 
## $k
## [1] -Inf    0    4  164
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
data.frame(size = cv_tree_model$size, CV_Error = cv_tree_model$dev / nrow(train)) %>%
  mutate(min_CV_Error = as.numeric(min(CV_Error) == CV_Error)) %>%
  ggplot(aes(x = size, y = CV_Error)) +
  geom_line(col = "grey55") +
  geom_point(size = 2, aes(col = factor(min_CV_Error))) +
  scale_x_continuous(breaks = seq(1, 7), minor_breaks = NULL) +
  scale_y_continuous(labels = scales::percent_format()) +
  scale_color_manual(values = c("deepskyblue3", "green")) +
  theme(legend.position = "none") +
  labs(title = "OJ Dataset - Classification Tree",
       subtitle = "Selecting tree 'size' (# of terminal nodes) using cross-validation",
       x = "Tree Size",
       y = "CV Error")

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

The plot shows tree sizes of 2, 4, and 7. I will choose tree size 2.

  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_tree_model <- prune.tree(tree_model, best = 4)
summary(pruned_tree_model)
## 
## Classification tree:
## snip.tree(tree = tree_model, nodes = 4:6)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes:  4 
## Residual mean deviance:  0.8413 = 669.7 / 796 
## Misclassification error rate: 0.1888 = 151 / 800