8.3 Lab: Decision Trees: Fitting Classification Trees
8.3.1 Fitting Classification Trees
We will use classification trees to analyze the Carseats
data set. In these data, Sales is a continuous variable, and so
we begin by recording it as a binary variable. We use the
if else function to create a variable, called
High, which takes on a value of Yes if the
Sales variable exceeds 8, and takes on a value of No
otherwise.
library(tree)## Warning: package 'tree' was built under R version 4.3.3library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.3.3attach(Carseats)
High <- factor(ifelse(Sales<=8, "No","Yes"))Finally, we use the data.frame()function to fit a
classification tree in order to predict High using all
variables but Sales. The syntax of the
tree()function is quite similar to that of the
lm()function.
tree.carseats <- tree(High~. - Sales, Carseats)
summary(tree.carseats)##
## Classification tree:
## tree(formula = High ~ . - Sales, data = Carseats)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Income" "CompPrice" "Population"
## [6] "Advertising" "Age" "US"
## Number of terminal nodes: 27
## Residual mean deviance: 0.4575 = 170.7 / 373
## Misclassification error rate: 0.09 = 36 / 400- We see that the training error rate is 9%.
- A small deviance indicates a tree that provides a good fit to the (training) data.
- The residual mean deviance reported is simply the deviance divided by \(n − |T_0|\), which in this case is 400 − 27 = 373.
We use the plot() function to display the tree
structure, and the text() function to display the node
labels. The argument pretty = 0 instructs
R to include the category names for any qualitative
predictors, rather than simply displaying a letter for each
category.
#plot(tree.carseats)
#text(tree.carseats)
tree.carseats## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 400 541.500 No ( 0.59000 0.41000 )
## 2) ShelveLoc: Bad,Medium 315 390.600 No ( 0.68889 0.31111 )
## 4) Price < 92.5 46 56.530 Yes ( 0.30435 0.69565 )
## 8) Income < 57 10 12.220 No ( 0.70000 0.30000 )
## 16) CompPrice < 110.5 5 0.000 No ( 1.00000 0.00000 ) *
## 17) CompPrice > 110.5 5 6.730 Yes ( 0.40000 0.60000 ) *
## 9) Income > 57 36 35.470 Yes ( 0.19444 0.80556 )
## 18) Population < 207.5 16 21.170 Yes ( 0.37500 0.62500 ) *
## 19) Population > 207.5 20 7.941 Yes ( 0.05000 0.95000 ) *
## 5) Price > 92.5 269 299.800 No ( 0.75465 0.24535 )
## 10) Advertising < 13.5 224 213.200 No ( 0.81696 0.18304 )
## 20) CompPrice < 124.5 96 44.890 No ( 0.93750 0.06250 )
## 40) Price < 106.5 38 33.150 No ( 0.84211 0.15789 )
## 80) Population < 177 12 16.300 No ( 0.58333 0.41667 )
## 160) Income < 60.5 6 0.000 No ( 1.00000 0.00000 ) *
## 161) Income > 60.5 6 5.407 Yes ( 0.16667 0.83333 ) *
## 81) Population > 177 26 8.477 No ( 0.96154 0.03846 ) *
## 41) Price > 106.5 58 0.000 No ( 1.00000 0.00000 ) *
## 21) CompPrice > 124.5 128 150.200 No ( 0.72656 0.27344 )
## 42) Price < 122.5 51 70.680 Yes ( 0.49020 0.50980 )
## 84) ShelveLoc: Bad 11 6.702 No ( 0.90909 0.09091 ) *
## 85) ShelveLoc: Medium 40 52.930 Yes ( 0.37500 0.62500 )
## 170) Price < 109.5 16 7.481 Yes ( 0.06250 0.93750 ) *
## 171) Price > 109.5 24 32.600 No ( 0.58333 0.41667 )
## 342) Age < 49.5 13 16.050 Yes ( 0.30769 0.69231 ) *
## 343) Age > 49.5 11 6.702 No ( 0.90909 0.09091 ) *
## 43) Price > 122.5 77 55.540 No ( 0.88312 0.11688 )
## 86) CompPrice < 147.5 58 17.400 No ( 0.96552 0.03448 ) *
## 87) CompPrice > 147.5 19 25.010 No ( 0.63158 0.36842 )
## 174) Price < 147 12 16.300 Yes ( 0.41667 0.58333 )
## 348) CompPrice < 152.5 7 5.742 Yes ( 0.14286 0.85714 ) *
## 349) CompPrice > 152.5 5 5.004 No ( 0.80000 0.20000 ) *
## 175) Price > 147 7 0.000 No ( 1.00000 0.00000 ) *
## 11) Advertising > 13.5 45 61.830 Yes ( 0.44444 0.55556 )
## 22) Age < 54.5 25 25.020 Yes ( 0.20000 0.80000 )
## 44) CompPrice < 130.5 14 18.250 Yes ( 0.35714 0.64286 )
## 88) Income < 100 9 12.370 No ( 0.55556 0.44444 ) *
## 89) Income > 100 5 0.000 Yes ( 0.00000 1.00000 ) *
## 45) CompPrice > 130.5 11 0.000 Yes ( 0.00000 1.00000 ) *
## 23) Age > 54.5 20 22.490 No ( 0.75000 0.25000 )
## 46) CompPrice < 122.5 10 0.000 No ( 1.00000 0.00000 ) *
## 47) CompPrice > 122.5 10 13.860 No ( 0.50000 0.50000 )
## 94) Price < 125 5 0.000 Yes ( 0.00000 1.00000 ) *
## 95) Price > 125 5 0.000 No ( 1.00000 0.00000 ) *
## 3) ShelveLoc: Good 85 90.330 Yes ( 0.22353 0.77647 )
## 6) Price < 135 68 49.260 Yes ( 0.11765 0.88235 )
## 12) US: No 17 22.070 Yes ( 0.35294 0.64706 )
## 24) Price < 109 8 0.000 Yes ( 0.00000 1.00000 ) *
## 25) Price > 109 9 11.460 No ( 0.66667 0.33333 ) *
## 13) US: Yes 51 16.880 Yes ( 0.03922 0.96078 ) *
## 7) Price > 135 17 22.070 No ( 0.64706 0.35294 )
## 14) Income < 46 6 0.000 No ( 1.00000 0.00000 ) *
## 15) Income > 46 11 15.160 Yes ( 0.45455 0.54545 ) *In order to properly evaluate the performance of a classification
tree on these data, we must estimate the test error rather than simply
computing the training error. We split the observations into a training
set and a test set, build the tree using the training set, and evaluate
its performance on the test data. The predict() function
can be used for this purpose. In the case of a classification tree, the
argument type = "class" instructs R to
return the actual class prediction. This approach leads to correct
predictions for around 77% of the locations in the test data set.
set.seed(2)
train <- sample(1:nrow(Carseats), 200)
Carseats.test <- Carseats[-train, ]
High.test <- High[-train]
tree.carseats <- tree(High ~ . - Sales, Carseats,subset = train)
tree.pred <- predict(tree.carseats , Carseats.test , type = "class")
table(tree.pred , High.test)## High.test
## tree.pred No Yes
## No 104 33
## Yes 13 50(104 + 50) / 200## [1] 0.77Next, we consider whether pruning the tree might lead to improved
results. The function cv.tree() performs cross-validation
in order to determine the optimal level of tree complexity; cost
complexity pruning cv.tree() is used in order to select a
sequence of trees for consideration.
set.seed(7)
cv.carseats <- cv.tree(tree.carseats , FUN = prune.misclass)
names(cv.carseats)## [1] "size" "dev" "k" "method"cv.carseats## $size
## [1] 21 19 14 9 8 5 3 2 1
##
## $dev
## [1] 75 75 75 74 82 83 83 85 82
##
## $k
## [1] -Inf 0.0 1.0 1.4 2.0 3.0 4.0 9.0 18.0
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"Despite its name, dev corresponds to the number of cross-validation errors. The tree with 9 terminal nodes results in only 74 cross-validation errors. We plot the error rate as a function of both size and k.
par(mfrow = c(1, 2))
plot(cv.carseats$size , cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
We now apply the prune.misclass() function in order to
prune the tree to obtain the nine-node tree.
prune.carseats <- prune.misclass(tree.carseats , best = 9)
plot(prune.carseats)
text(prune.carseats , pretty = 0)
How well does this pruned tree perform on the test data set? Once again,
we apply the predict() function.
tree.pred <- predict(prune.carseats , Carseats.test , type = "class")
table(tree.pred , High.test)## High.test
## tree.pred No Yes
## No 97 25
## Yes 20 58(97 + 58) / 200## [1] 0.775Now 77.5% of the test observations are correctly classified, so not only has the pruning process produced a more interpretable tree, but it has also slightly improved the classification accuracy.
If we increase the value of best, we obtain a larger pruned tree with lower classification accuracy:
prune.carseats <- prune.misclass(tree.carseats , best = 14)
plot(prune.carseats)
text(prune.carseats , pretty = 0)
tree.pred <- predict(prune.carseats , Carseats.test ,type = "class")
table(tree.pred , High.test)## High.test
## tree.pred No Yes
## No 102 31
## Yes 15 52(102 + 52) / 200## [1] 0.77