Assignment 7
Drew Hinck
2023-04-20
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 p̂ m1. The xaxis should display p̂ m1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
Hint: In a setting with two classes, p̂ m1 = 1 - p̂ m2. You could make this plot by hand, but it will be much easier to make in R.
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("grey", "blue", "green"))
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)
library(tree)
library(randomForest)
attach(Carseats)
set.seed(1)
train_ind <- sample(dim(Carseats)[1], dim(Carseats)[1]/2)
train <- Carseats[train_ind, ]
test <- Carseats[-train_ind, ](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.carseats = tree(Sales ~ ., data = train)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = 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)
tree.pred = predict(tree.carseats, test)
obs.sales = test$Sales
mean((tree.pred-obs.sales)^2)## [1] 4.922039The test MSE obtained for the regression tree is 4.922039.
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
carseats.cv <- cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(carseats.cv$size, carseats.cv$dev, type = "b")
plot(carseats.cv$k, carseats.cv$dev, type = "b")
pruned.carseats = prune.tree(tree.carseats, best = 13)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
prune.pred <- predict(pruned.carseats, test)
prune.mse <- mean((test$Sales - prune.pred)^2)
prune.mse## [1] 4.96547Using cross-validation and pruning the tree actually increased the MSE from 4.922039 to 4.96547.
(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.
bag.car <- randomForest(Sales ~ ., data = train, mtry = 10, importance = TRUE)
bag.car##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 10, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.931324
## % Var explained: 62.72bag.pred <- predict(bag.car, test)
bag.err <- mean((test$Sales - bag.pred)^2)
bag.err## [1] 2.657296importance(bag.car)## %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 15.157608The model that uses the bagging approach produces an MSE of 2.657296.
The ouput obtained using the importance function shows that the most
important variables in the model are Price,
ShelveLoc, CompPrice, Age, and
then Advertising.
(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.car5=randomForest(Sales~.,data=train,mtry=5,importance=T)
rf.car3=randomForest(Sales~.,data=train,mtry=3,importance=T)
rf.car5##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 5, importance = T)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 5
##
## Mean of squared residuals: 2.979207
## % Var explained: 62.11rf.car3##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 3, importance = T)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 3
##
## Mean of squared residuals: 3.320908
## % Var explained: 57.77importance(rf.car5)## %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 25.59797importance(rf.car3)## %IncMSE IncNodePurity
## CompPrice 13.7896822 157.75688
## Income 3.8824616 122.65512
## Advertising 7.4859338 115.76957
## Population -0.6597586 99.08349
## Price 38.6048179 391.96348
## ShelveLoc 34.1547272 297.41187
## Age 10.5834571 168.84092
## Education -0.2100206 74.83891
## Urban -1.2733873 15.02380
## US 4.6450025 31.11326rf.5=predict(rf.car5,newdata=test)
rf.3=predict(rf.car3,newdata=test)
mean((test$Sales - rf.5)^2)## [1] 2.701665mean((test$Sales - rf.3)^2)## [1] 2.973152By changing the number of variables considered at each split, the
error rate went from 2.70696 using 5 variables to 3.067615 using 3
variables. The most important variables in both models are once again
Price, ShelveLoc, CompPrice,
Age, and then Advertising.
(f) Now analyze the data using BART, and report your results.
library(BART)## Warning: package 'BART' was built under R version 4.2.3## Loading required package: nlme## Loading required package: nnet## Loading required package: survival## Warning: package 'survival' was built under R version 4.2.3set.seed(1)
x=Carseats[,1:10]
y=Carseats[,"Sales"]
xtrain=x[train_ind, ]
ytrain=y[train_ind]
xtest=x[-train_ind, ]
ytest=y[-train_ind]
bart.fit=gbart(xtrain,ytrain,x.test=xtest)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 13, 200
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 10.360000, 1.000000
## xp1,xp[np*p]: 11.220000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 100 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,3.99291e-30,7.57815
## *****sigma: 0.000000
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,13,0
## *****printevery: 100
##
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 4s
## trcnt,tecnt: 1000,1000yhat.bart=bart.fit$yhat.test.mean
mean((ytest-yhat.bart)^2)## [1] 0.184202The BART model produced a much lower error rate at 0.184202.
Problem 9
This problem involves the OJ data set which is part of the ISLR2 package.
detach(Carseats)
attach(OJ)(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train2=sample(1:nrow(OJ),800)
oj.train=OJ[train2, ]
oj.test=OJ[-train2, ](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~.,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 the tree has 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 ) *Looking at Node 5 LoyalCH is a terminal node with a
split of 0.280875. There are 188 observations and the final prediction
for this is MM. At this node there is a y probability of ( 0.44149
0.55851 ). This means that about 44% of the observations in this branch
have a CH value and the other 56% have an MM value.
(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree)
The plot shows that LoyalCH is the most important
predictor in the tree as it is the most important variable in each of
the top three nodes.
(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$Purchase, oj.pred)## oj.pred
## CH MM
## CH 160 8
## MM 38 641-mean(oj.pred==oj.test$Purchase)## [1] 0.1703704The test error rate for this model is 17%.
(f) 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)
cv.oj## $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
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Cross-Validation Classification Error")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
A tree size of 9 appears to have the lowest cross-validated classification error rate, but this would not prune the tree.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.prune <- prune.misclass(oj.tree, best = 5)
plot(oj.prune)
text(oj.prune, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.prune)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The pruned model produced a training error rate of 0.1625, which is lower than the unpruned test error rate of 0.1703704.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, oj.test, type = "class")
misclass.unpruned = sum(oj.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1703704pred.pruned = predict(oj.prune, oj.test, type = "class")
misclass.pruned = sum(oj.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.162963The pruned model has a test error rate of 16.3% and the unpruned model has a test error rate of 17%.