Assignment 7
John Shunk
2022-11-09
3.) 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 display ˆpm1, 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)
matplot(p, cbind(gini, entropy, class.err), col = c("orange", "blue", "green"))
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.
- Split the data set into a training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.1.3attach(Carseats)
set.seed(96)
train = sample(dim(Carseats)[1], dim(Carseats)[1]/3)
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)## Warning: package 'tree' was built under R version 4.1.3tree.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" "Income" "CompPrice"
## [6] "Education" "Advertising"
## Number of terminal nodes: 19
## Residual mean deviance: 2.408 = 597.1 / 248
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.87000 -1.10000 0.06133 0.00000 1.05900 3.93500plot(tree.carseats)
text(tree.carseats, pretty = 0)
pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)## [1] 4.106416MSE is roughly 5.25.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
Best size = 13.
pruned.carseats = prune.tree(tree.carseats, best = 13)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)## [1] 4.316175MSE is roughly 5.44.
- 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)## Warning: package 'randomForest' was built under R version 4.1.3## randomForest 4.7-1## Type rfNews() to see new features/changes/bug fixes.bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500,
importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)## [1] 2.387076importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 27.2327728 254.929288
## Income 8.1227939 111.716806
## Advertising 19.0661236 157.406322
## Population -0.8751688 65.173074
## Price 57.8265312 519.423753
## ShelveLoc 73.3318600 756.953261
## Age 17.8904175 190.879169
## Education 5.5626592 76.472595
## Urban -0.1915783 11.364312
## US 0.9419029 7.527342MSE is now 2.54. Most important variables are Price, ShelveLoc, and Age.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500,
importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)## [1] 2.292494importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 23.4940193 228.09041
## Income 5.2477228 129.95889
## Advertising 17.9395611 196.25554
## Population -2.0898898 93.32153
## Price 49.7200070 506.39863
## ShelveLoc 59.9138565 624.47180
## Age 16.7206769 219.21207
## Education 2.4953601 83.05996
## Urban -1.6208138 14.78184
## US 0.8603896 17.17162MSE is now roughly 2.50. Price, ShelveLoc, and Age are important again, along with Advertising, Income, and CompPrice.
9.) 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(12)
train = sample(dim(OJ)[1], 802)
OJ.train = OJ[train, ]
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)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "SalePriceMM" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 7
## Residual mean deviance: 0.765 = 608.2 / 795
## Misclassification error rate: 0.1584 = 127 / 802The tree only uses two variables: LoyalCH and PriceDiff. It has 8 terminal nodes. Training error rate (misclassification error) for the tree is 0.1484.
- 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 802 1072.00 CH ( 0.61097 0.38903 )
## 2) LoyalCH < 0.482935 305 331.00 MM ( 0.23279 0.76721 )
## 4) LoyalCH < 0.0589885 69 18.11 MM ( 0.02899 0.97101 ) *
## 5) LoyalCH > 0.0589885 236 285.20 MM ( 0.29237 0.70763 )
## 10) SalePriceMM < 2.04 125 125.10 MM ( 0.20000 0.80000 ) *
## 11) SalePriceMM > 2.04 111 149.10 MM ( 0.39640 0.60360 ) *
## 3) LoyalCH > 0.482935 497 432.00 CH ( 0.84306 0.15694 )
## 6) LoyalCH < 0.764572 222 270.20 CH ( 0.70270 0.29730 )
## 12) PriceDiff < 0.085 73 99.54 MM ( 0.42466 0.57534 )
## 24) ListPriceDiff < 0.235 48 56.07 MM ( 0.27083 0.72917 ) *
## 25) ListPriceDiff > 0.235 25 29.65 CH ( 0.72000 0.28000 ) *
## 13) PriceDiff > 0.085 149 131.60 CH ( 0.83893 0.16107 ) *
## 7) LoyalCH > 0.764572 275 98.63 CH ( 0.95636 0.04364 ) *Let’s pick terminal node labeled “8)”. The splitting variable at this node is LoyalCH. The splitting value of this node is 0.036. There are 60 points in the subtree below this node. The deviance for all points contained in region below this node is 61. A * in the line denotes that this is in fact a terminal node. The prediction at this node is Sales = MM. About 1.7% points in this node have CH as value of Sales. Remaining 98.3% points have MM as value of Sales.
- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
LoyalCH is the most important variable of the tree, in fact top 3 nodes contain LoyalCH. If LoyalCH<0.276, the tree predicts MM. If LoyalCH>0.765, the tree predicts CH. For intermediate values of LoyalCH, the decision also depends on the value of PriceDiff.
- 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$Purchase, oj.pred)## oj.pred
## CH MM
## CH 127 36
## MM 15 901-((138+74)/(138+18+38+74))## [1] 0.2089552Test error rate is 20.9%.
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(oj.tree, FUN = prune.tree)- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
- Which tree size corresponds to the lowest cross-validated classif ication error rate?
Size of 8 gives lowest cross-validation error.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned = prune.tree(oj.tree, best = 8)## Warning in prune.tree(oj.tree, best = 8): best is bigger than tree size- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "SalePriceMM" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 7
## Residual mean deviance: 0.765 = 608.2 / 795
## Misclassification error rate: 0.1584 = 127 / 802Misclassification error of pruned tree is exactly same as that of original tree: 0.1484.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1902985pred.pruned = predict(oj.pruned, OJ.test, type = "class")
misclass.pruned = sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.1902985Pruned and unpruned trees have same test error rate of 0.209.