Assignment 7
ikq392
2022-11-13
Problem 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.index=2 * p * (1 - p)
class.error=1 - pmax(p, 1 - p)
cross.entropy=- (p * log(p) + (1 - p) * log(1 - p))
par(bg = "papayawhip")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("green" , "yellow", "red"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("green" , "yellow", "red"), pch=c(15,17,19))
Problem 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(ISLR2)
attach(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 MSE do you obtain?
library(tree)
tree.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)
pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)## [1] 4.922039The MSE is 4.92.
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
pruned.carseats = prune.tree(tree.carseats, best = 9)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)## [1] 4.918134The MSE stayed relatively the same.
(d) 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)
bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500,
importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)## [1] 2.657296importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 23.07909904 171.185734
## Income 2.82081527 94.079825
## Advertising 11.43295625 99.098941
## Population -3.92119532 59.818905
## Price 54.24314632 505.887016
## ShelveLoc 46.26912996 361.962753
## Age 14.24992212 159.740422
## Education -0.07662320 46.738585
## Urban 0.08530119 8.453749
## US 4.34349223 15.157608Bagging improves the MSE to 2.66. The most important variables are Price, ShelveLoc, and CompPrice.
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500,
importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)## [1] 2.701665importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 19.8160444 162.73603
## Income 2.8940268 106.96093
## Advertising 11.6799573 106.30923
## Population -1.6998805 79.04937
## Price 46.3454015 448.33554
## ShelveLoc 40.4412189 334.33610
## Age 12.5440659 169.06125
## Education 1.0762096 55.87510
## Urban 0.5703583 13.21963
## US 5.8799999 25.59797Random Forest increases the MSE to 2.70. The MSE changes dependent on changes in m. The most important factors are still Price, ShelveLoc, and CompPrice.
(f) Now analyze the data using BART, and report your results.
This errored out for me - I can’t determine the necessary adjustment.
library(BART)
x<-Carseats[,1:12]
y<-Carseats[, “medv”]
xtrain<-x[train, ]
ytrain<-y[train]
xtest<-x[-train, ]
ytest<-y[-train]
set.seed(1)
bartfit<-gbart(xtrain,ytrain,x.test=xtest
The test error would be computed with:
yhat.bart <- bartfit$yhat.test.mean
mean (( ytest - yhat.bart)^2)
Then, I could compare the MSE to boosting and random forests.
Problem 9.
This problem involves the OJ data set which is part of the ISLR2 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_sampleOJ=sample(dim(OJ)[1], 800)
oj.train=OJ[train_sampleOJ, ]
oj.test=OJ[-train_sampleOJ, ](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 / 800The training error rate is 0.1588 and there are 9 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 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 < 0.196196 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196196 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 ) *Node 24, PctDiscMM, is a terminal node with a 0.196196 split, 55 observations, and a yprob of (0.61818 0.38182).
(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
(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?
pred.oj = predict(oj.tree, oj.test, type = "class")
table(oj.test$Purchase, pred.oj)## pred.oj
## CH MM
## CH 160 8
## MM 38 64pred.unprune = predict(oj.tree, oj.test, type = "class")
misclass.unprune = sum(oj.test$Purchase !=pred.unprune)
misclass.unprune/length(pred.unprune)## [1] 0.1703704The MSE is 0.17.
(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.tree)
oj.cv## $size
## [1] 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 685.6493 698.8799 702.8083 702.8083 714.1093 725.4734 780.2099
## [8] 790.0301 1074.2062
##
## $k
## [1] -Inf 12.62207 13.94616 14.35384 26.21539 35.74964 43.07317
## [8] 45.67120 293.15784
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size, oj.cv$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
A tree size of 9 appears to correspond to the lowest cross-validated classification error rate.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj = prune.misclass(oj.tree, best = 9)
plot(prune.oj)
text(prune.oj, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800Both the pruned and unpruned trees have the same training MSE rates of 0.1588.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred = predict(prune.oj, oj.test, type = "class")
table(prune.pred, oj.test$Purchase)##
## prune.pred CH MM
## CH 160 38
## MM 8 641 - (160 + 64) / 270## [1] 0.1703704The pruned tree has a higher test MSE of .1704.