assignment 7

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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

p = seq(0, 1, 0.01)
gini = p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
class.err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini, entropy, class.err), col = c("pink", "red", "purple"))

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.

library(ISLR)
attach(Carseats)
set.seed(1)

train = sample(dim(Carseats)[1], dim(Carseats)[1]/2)
Carseats.train = Carseats[train, ]
Carseats.test = Carseats[-train, ]

Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

library(tree)

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"         "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
plot(tree.carseats)
text(tree.carseats, pretty = 0)
mean((Carseats.test$Sales - pred.carseats)^2)

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, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
pruned.carseats = prune.tree(tree.carseats, best = 9)
par(mfrow = c(1, 1))
plot(pruned.carseats)
mean((Carseats.test$Sales - pred.pruned)^2)

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.

library(randomForest)
bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500, 
    importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
importance(bag.carseats)
##                %IncMSE IncNodePurity
## CompPrice   16.9874366    126.852848
## Income       3.8985402     78.314126
## Advertising 16.5698586    123.702901
## Population   0.6487058     62.328851
## Price       55.3976775    514.654890

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.

importance(rf.carseats)
rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500, 
    importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)
## Population   1.131119      82.24483
## Price       46.600559     451.70021
## ShelveLoc   37.352447     278.79756
## Age         19.992113     194.99430
## Education    1.945616      51.70741

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.

library(ISLR)
attach(OJ)
set.seed(1013)
OJ.test = OJ[-train

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?

library(tree)
oj.tree = tree(Purchase ~ ., data = OJ.train)
summary(oj.tree)
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH"   "PriceDiff"
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7517 = 596.1 / 793

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.

oj.tree
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
##  1) root 800 1075.00 CH ( 0.60250 0.39750 )  
##    2) LoyalCH < 0.5036 359  422.80 MM ( 0.27577 0.72423 ) 
##    3) LoyalCH > 0.5036 441  343.30 CH ( 0.86848 0.13152 )  
##      6) LoyalCH < 0.764572 186  210.30 CH ( 0.74731 0.25269 )  
##       12) PriceDiff < -0.165 29   34.16 MM ( 0.27586 0.72414 ) *
##       13) PriceDiff > -0.165 157  140.90 CH ( 0.83439 0.16561 )  
##         26) PriceDiff < 0.265 82   95.37 CH ( 0.73171 0.26829 ) *

Create a plot of the tree, and interpret the results.

plot(oj.tree)
text(oj.tree, 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?

oj.pred = predict(oj.tree, OJ.test, type = "class")
table(OJ.test$Purchase, oj.pred)

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.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? size 6
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.OJ,best=6)

Compare the training error rates between the pruned and unpruned trees. Which is higher?

summary(tree.OJ)
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## Residual mean deviance:  0.7432 = 587.8 / 791 
## Misclassification error rate: 0.1588 = 127 / 800
summary(prune.OJ)
## 
## Classification tree:
## snip.tree(tree = tree.OJ, nodes = c(10L, 4L))
## Number of terminal nodes:  6
## Residual mean deviance:  0.7748 = 614.4 / 793

Compare the test error rates between the pruned and unpruned trees. Which is higher?

treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)
pruneOJ.pred = predict(prune.OJ, newdata = OJtest, type = "class")
prunedOJvalerr = (36 + 8) / 270
prunedOJvalerr

assignment 7

2023-04-21

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