Exercise 7 - Chapter 8: 3, 8, 9
Dom Rodriguez
8/15/2021
Q3. Consider the Gini index, classification error, and cross-entropy in a simple 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.
p <- seq(0, 1, 0.01)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), col = c("red", "green", "blue"))
Q8. 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)## Warning: package 'ISLR' was built under R version 4.0.5set.seed(1)
train <- sample(1:nrow(Carseats), nrow(Carseats) / 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 error rate do you obtain?
library(tree)## Warning: package 'tree' was built under R version 4.0.5tree.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" "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 4.09900plot(tree.carseats)
text(tree.carseats, pretty = 0)
yhat <- predict(tree.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 4.922039We may conclude that the Test MSE is about 4.15.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test error rate ?
cv.carseats <- cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
tree.min <- which.min(cv.carseats$dev)
points(tree.min, cv.carseats$dev[tree.min], col = "red", cex = 2, pch = 20)
In this case, the tree of size 8 is selected by cross-validation. We now prune the tree to obtain the 8-node tree.
prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 5.113254We may see that pruning the tree increases the Test MSE to 5.1.
- 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.
bag.carseats <- randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500, importance = TRUE)
yhat.bag <- predict(bag.carseats, newdata = Carseats.test)
mean((yhat.bag - Carseats.test$Sales)^2)## [1] 2.657296We may see that bagging decreases the Test MSE to 2.6.
importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 23.07909904 171.185734
## Income 2.82081527 94.079825
## Advertising 11.43295625 99.098941
## Population -3.92119532 59.818905
## Price 54.24314632 505.887016
## ShelveLoc 46.26912996 361.962753
## Age 14.24992212 159.740422
## Education -0.07662320 46.738585
## Urban 0.08530119 8.453749
## US 4.34349223 15.157608We may conclude that “Price” and “ShelveLoc” are the two most important variables.
- Use random forests to analyze this data. What test error rate do you obtain? Use the “importance()” function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats <- randomForest(Sales ~ ., data = Carseats.train, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(rf.carseats, newdata = Carseats.test)
mean((yhat.rf - Carseats.test$Sales)^2)## [1] 3.049406In this case, with \(m=p–√\), we have a Test MSE of 3.3.
importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 12.9489323 158.48521
## Income 2.2754686 129.59400
## Advertising 8.9977589 111.94374
## Population -2.2513981 102.84599
## Price 33.4226950 391.60804
## ShelveLoc 34.0233545 290.56502
## Age 12.2185108 171.83302
## Education 0.2592124 71.65413
## Urban 1.1382113 14.76798
## US 4.1925335 33.75554We may conclude that, in this case also, “Price” and “ShelveLoc” are the two most important variables.
Q9. This problem involves the “OJ” data set which is part of the “ISLR” package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train <- sample(1:nrow(OJ), 800)
OJ.train <- OJ[train, ]
OJ.test <- OJ[-train, ]- Fit a tree to the training data, with “Purchase” as the response and the other variables except for “Buy” 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.oj <- tree(Purchase ~ ., data = OJ.train)
summary(tree.oj)##
## 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 / 800The fitted tree has 8 terminal nodes and a training error rate of 0.165.
- 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.
tree.oj## 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 ) *We pick the node labelled 8, which is a terminal node because of the asterisk. The split criterion is LoyalCH < 0.035, the number of observations in that branch is 57 with a deviance of 10.07 and an overall prediction for the branch of MM. Less than 2% of the observations in that branch take the value of CH, and the remaining 98% take the value of MM.
- Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)
We may see that the most important indicator of “Purchase” appears to be “LoyalCH”, since the first branch differentiates the intensity of customer brand loyalty to CH. In fact, the top three nodes contain “LoyalCH”.
- 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 ?
tree.pred <- predict(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 641 - (147 + 62) / 270## [1] 0.2259259We may conclude that the test error rate is about 22%.
- Apply the “cv.tree()” function to the training set in order to determine the optimal size tree.
cv.oj <- cv.tree(tree.oj, FUN = prune.misclass)
cv.oj## $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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
h. Which tree size corresponds to the lowest cross-validated classification error rate?
We may see that the 2-node tree is the smallest tree with the lowest classification error rate.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.misclass(tree.oj, best = 2)
plot(prune.oj)
text(prune.oj, pretty = 0)
j. Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = 3:2)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 2
## Residual mean deviance: 0.9768 = 779.5 / 798
## Misclassification error rate: 0.205 = 164 / 800The misclassification error rate is slightly higher for the pruned tree (0.1825 vs 0.165).
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, OJ.test, type = "class")
table(prune.pred, OJ.test$Purchase)##
## prune.pred CH MM
## CH 142 24
## MM 26 781 - (119 + 81) / 270## [1] 0.2592593In this case, the pruning process increased the test error rate to about 26%, but it produced a way more interpretable tree.