Decision Trees
Rahab Lawshea
4/7/2022
3. Consider the Gini index, classification error, and cross-entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of pˆm1. The x- axis should display pˆm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
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)
plot(p, entropy, type = "l", col = "purple", xlab = "p", ylab = "Values of Error")
lines(p, gini, col = "red")
lines(p, class.err, col = "blue")
legend(0.3, 0.25, c("Classification error", "Gini index", "Cross entropy"), col = c("blue", "red", "purple"), lty = c(1, 1))
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.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5attach(Carseats)
set.seed(1)
train = sample(dim(Carseats)[1], dim(Carseats)[1]/2)
carseats.train = Carseats[train, ]
carseats.test = Carseats[-train, ](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test error rate do you obtain?
library(tree)## Warning: package 'tree' was built under R version 4.0.5carseats.tree = tree(Sales~., data = carseats.train)
summary(carseats.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Advertising" "CompPrice"
## [6] "US"
## Number of terminal nodes: 18
## Residual mean deviance: 2.167 = 394.3 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.88200 -0.88200 -0.08712 0.00000 0.89590 4.09900There are 6 variables used: ShelveLoc, Price, Age, Advertising, CompPrice, and US. The tree has 18 terminal nodes and the residual mean deviance is 2.167
plot(carseats.tree)
text(carseats.tree, pretty=0)
carseats.pred = predict(carseats.tree, carseats.test)
mean((carseats.test$Sales - carseats.pred)^2)## [1] 4.922039The test MSE is ~4.922
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test error rate?
carseats.cv = cv.tree(carseats.tree, FUN = prune.tree)
plot(carseats.cv$size, carseats.cv$dev, type = 'b')
carseats.cv$size[which.min(carseats.cv$dev)]## [1] 18The optimal level of tree complexity is 18, however, since 18 and 16 are close in value based on the plot, let’s take a look at the MSE value to see if we notice a reduction.
carseats.prune = prune.tree(carseats.tree, best = 16)
plot(carseats.prune)
text(carseats.prune, pretty = 0)
prune.pred = predict(carseats.prune, carseats.test)
mean((carseats.test$Sales - prune.pred)^2)## [1] 4.903443carseats.prune = prune.tree(carseats.tree, best = 15)
plot(carseats.prune)
text(carseats.prune, pretty = 0)
prune.pred = predict(carseats.prune, carseats.test)
mean((carseats.test$Sales - prune.pred)^2)## [1] 4.924799Taking a look at the MSE produced by a value of 16 we can see that pruning the tree down to 16 did reduce the MSE value from 4.922 to 4.903. However, pruning down to 15 increased the MSE to 4.925. Using the optimal complexity of 18 will keep the MSE value the same but pruning the tree down to 16 did reduce the MSE.
(d) Use the bagging approach in order to analyze this data. What test error rate do you obtain? Use the importance() function to determine which variables are most important.
library(randomForest)
set.seed(1)
carseats.bag = randomForest(Sales~., data = carseats.train, mtry = 10, ntree = 500, importance = T)
bag.pred = predict(carseats.bag, carseats.test)
mean((carseats.test$Sales - bag.pred)^2)## [1] 2.644557Using the bagging approach decreased the MSE to 2.64.
importance(carseats.bag)## %IncMSE IncNodePurity
## CompPrice 10.8264762 85268.731
## Income 3.9884983 46974.308
## Advertising 9.5963049 48385.617
## Population -1.9578708 29030.815
## Price 42.4824814 252448.246
## ShelveLoc 42.1190535 188285.403
## Age 14.0949144 76789.024
## Education -2.6177621 22007.010
## Urban -0.6679903 4044.419
## US 3.9078747 9167.896Using the importance function we can see that ‘Price’ and ‘Shelveloc’ are the most important variables.
(e) Use random forests to analyze this data. What test error rate 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.
set.seed(1)
carseats.rf = randomForest(Sales~., data = carseats.train, mtry = 5, ntree = 500, importance = T)
pred.rf = predict(carseats.rf, carseats.test)
mean((carseats.test$Sales - pred.rf)^2)## [1] 2.847225set.seed(1)
carseats.rf = randomForest(Sales~., data = carseats.train, mtry = 7, ntree = 500, importance = T)
pred.rf = predict(carseats.rf, carseats.test)
mean((carseats.test$Sales - pred.rf)^2)## [1] 2.680464set.seed(1)
carseats.rf = randomForest(Sales~., data = carseats.train, mtry = 10, ntree = 500, importance = T)
pred.rf = predict(carseats.rf, carseats.test)
mean((carseats.test$Sales - pred.rf)^2)## [1] 2.644557Looking at the values above we can see that decreasing m seems to increase the MSE and increasing m decreases the MSE.
importance(carseats.rf)## %IncMSE IncNodePurity
## CompPrice 10.8264762 85268.731
## Income 3.9884983 46974.308
## Advertising 9.5963049 48385.617
## Population -1.9578708 29030.815
## Price 42.4824814 252448.246
## ShelveLoc 42.1190535 188285.403
## Age 14.0949144 76789.024
## Education -2.6177621 22007.010
## Urban -0.6679903 4044.419
## US 3.9078747 9167.896“Price” and “ShelveLoc” are still the most important variables to the model.
9. This problem involves the OJ data set which is part of the ISLR package.
(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
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 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?
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"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800Variables used: LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, PctDiscMM. There are 9 terminal nodes and the training error rate is 0.1588.
(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 1073.00 CH ( 0.60625 0.39375 )
## 2) LoyalCH < 0.5036 365 441.60 MM ( 0.29315 0.70685 )
## 4) LoyalCH < 0.280875 177 140.50 MM ( 0.13559 0.86441 )
## 8) LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ) *
## 9) LoyalCH > 0.0356415 118 116.40 MM ( 0.19492 0.80508 ) *
## 5) LoyalCH > 0.280875 188 258.00 MM ( 0.44149 0.55851 )
## 10) PriceDiff < 0.05 79 84.79 MM ( 0.22785 0.77215 )
## 20) SpecialCH < 0.5 64 51.98 MM ( 0.14062 0.85938 ) *
## 21) SpecialCH > 0.5 15 20.19 CH ( 0.60000 0.40000 ) *
## 11) PriceDiff > 0.05 109 147.00 CH ( 0.59633 0.40367 ) *
## 3) LoyalCH > 0.5036 435 337.90 CH ( 0.86897 0.13103 )
## 6) LoyalCH < 0.764572 174 201.00 CH ( 0.73563 0.26437 )
## 12) ListPriceDiff < 0.235 72 99.81 MM ( 0.50000 0.50000 )
## 24) PctDiscMM < 0.196197 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196197 17 12.32 MM ( 0.11765 0.88235 ) *
## 13) ListPriceDiff > 0.235 102 65.43 CH ( 0.90196 0.09804 ) *
## 7) LoyalCH > 0.764572 261 91.20 CH ( 0.95785 0.04215 ) *Choosing terminal node 8(we know this because of the *), the splitting variable at this node is LoyalCH, and the splitting value is 0.0356. There are 59 points(observations) in the subtree below this node, and deviance for this region is 10.14. An overall prediction for this node is MM(Minute Maid). About 1.7% of nodes have the value CH(Citrus Hill), while 98.3% of the nodes have the value MM.
(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
LoyalCH is the most important variable, which can be seen by how it is the top 3 nodes of this tree. If LoyalCH is less than 0.5036 and < 0.280875 the tree predicts MM will be purchased. If LoyalCH is > 0.5036 and < 0.76 the tree predicts CH will be purchased.
(e) 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)## oj.pred
## CH MM
## CH 160 8
## MM 38 64mean(oj.test$Purchase != oj.pred)## [1] 0.1703704The test error rate is 17.04%
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
oj.cv = cv.tree(oj.tree, FUN = prune.misclass)
oj.cv## $size
## [1] 9 8 7 4 2 1
##
## $dev
## [1] 150 150 149 158 172 315
##
## $k
## [1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size, oj.cv$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
oj.cv$size[which.min(oj.cv$dev)]## [1] 7(h) Which tree size corresponds to the lowest cross-validated classification error rate?
Tree size 9 has the lowest cross-validation error.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned = prune.misclass(oj.tree, best = 5)
plot(oj.pruned)
text(oj.pruned,pretty=0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The training error rate of the pruned tree is higher than the unpruned tree with a value of 0.1625 vs 0.1588
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, oj.test, type = "class")
misclass.unpruned = sum(oj.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1703704pred.pruned = predict(oj.pruned, oj.test, type = "class")
misclass.pruned = sum(oj.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.162963The test error rate at 0.17 for the unpruned tree is higher than the test error at 0.16 for the pruned tree.