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Chapter 8 - Tree-Based Methods
Lab Solution
1 Problem 9
(a). 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 (1)
train = sample ( dim (OJ)[1], 800) OJ.train = OJ[train, ] OJ.test = OJ[-train, ]
(b). Fit a tree to the training data, with Purchase as the response and the other variables as predic- tors. 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)
Classification tree:
tree(formula = Purchase ~ ., data = OJ.train)
Variables actually used in tree construction:
[1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
Number of terminal nodes: 8
Residual mean deviance: 0.7305 = 578.6 / 792
Misclassification error rate: 0.165 = 132 / 800
As shown in the output, the tree has 8 terminal nodes.
(c). 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 1064.00 CH ( 0.61750 0.38250 )
2) LoyalCH < 0.508643 350 409.30 MM ( 0.27143 0.72857 )
4) LoyalCH < 0.264232 166 122.10 MM ( 0.12048 0.87952 )
8) LoyalCH < 0.0356415 57 10.07 MM ( 0.01754 0.98246 ) *
9) LoyalCH > 0.0356415 109 100.90 MM ( 0.17431 0.82569 ) *
5) LoyalCH > 0.264232 184 248.80 MM ( 0.40761 0.59239 )
10) PriceDiff < 0.195 83 91.66 MM ( 0.24096 0.75904 )
20) SpecialCH < 0.5 70 60.89 MM ( 0.15714 0.84286 ) *
21) SpecialCH > 0.5 13 16.05 CH ( 0.69231 0.30769 ) *
11) PriceDiff > 0.195 101 139.20 CH ( 0.54455 0.45545 ) *
3) LoyalCH > 0.508643 450 318.10 CH ( 0.88667 0.11333 )
6) LoyalCH < 0.764572 172 188.90 CH ( 0.76163 0.23837 )
12) ListPriceDiff < 0.235 70 95.61 CH ( 0.57143 0.42857 ) *
13) ListPriceDiff > 0.235 102 69.76 CH ( 0.89216 0.10784 ) *
7) LoyalCH > 0.764572 278 86.14 CH ( 0.96403 0.03597 ) *
The * symbols denote terminal nodes. If we select the one labeled “7)", we can see that there are 278 observations with a value for LoyalCH > 0.764572, and they are classified as “CH"
(d). Create a plot of the tree, and interpret the results.
plot (oj.tree) text (oj.tree, pretty = 0)
cv.oj = cv.tree (oj.tree, FUN = prune.tree)
(g). 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")
Tree Size
Deviance
(h). Which tree size corresponds to the lowest cross-validated classification error rate?
(i). 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.
oj.pruned = prune.tree (oj.tree, best = cv.oj$size[ which.min (cv.oj$dev)])
(j). Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary (oj.pruned)
Classification tree:
snip.tree(tree = oj.tree, nodes = 4:5)
Variables actually used in tree construction:
[1] "LoyalCH" "ListPriceDiff"
Number of terminal nodes: 5
Residual mean deviance: 0.7829 = 622.4 / 795
Misclassification error rate: 0.1825 = 146 / 800
The pruned tree is higher
(k). Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict (oj.tree, OJ.test, type = "class") misclass.unpruned = sum (OJ.test$Purchase != pred.unpruned) misclass.unpruned/ length (pred.unpruned)
## [1] 0.
pred.pruned = predict (oj.pruned, OJ.test, type = "class") misclass.pruned = sum (OJ.test$Purchase != pred.pruned) misclass.pruned/ length (pred.pruned)
## [1] 0.
The pruned tree has a higher test error
