Assignment 7
Simon Miller
2023-04-19
Data
data("Carseats")
data("OJ")Question 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.
Hint: In a setting with two classes, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.
x1=seq(0, 1, .01)
x2=1-x1
c=1-pmax(x1, x2)
g=x1*x2*2
e=-(x1*log(x1))-(x2*log(x2))
plot(x1, c, type="l", col="Blue", ylab="Value", xlab="pm1", ylim=c(min(0), max(1)))
lines(x1, g, col="Red")
lines(x1, e, col="Green")
text(.5, .35, "Classification", col="Blue")
text(.5, .55, "Gini", col="Red")
text(.5, .75, "Entropy", col="Green")
grid()
Question 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.
set.seed(1)
train=sample(1:nrow(Carseats),200)
test=Carseats[-train,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
regtree=tree(Sales~., data=Carseats, subset=train)
plot(regtree)
text(regtree, pretty=0, cex=.5)
summary(regtree)##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats, subset = 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 4.09900pred=predict(regtree, test)
error=mean((test$Sales-pred)^2)
error## [1] 4.922039From the results we can see that we only use 6 variables and the test MSE is 4.922039.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv=cv.tree(regtree, FUN=prune.tree)
plot(cv$size, cv$dev, type="b")
p=prune.tree(regtree, best=12)
plot(p)
text(p, pretty=0, cex=.5)
pred=predict(p, test)
error=mean((test$Sales-pred)^2)
error## [1] 4.966929We can see that using 12 as the optimal level and pruning the model actually raises the MSE to 4.966929.
- 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.
set.seed (1)
bag=randomForest(Sales~., data=Carseats, subset=train, mtry=10, importance=TRUE)
pred=predict(bag, test)
error=mean((test$Sales-pred)^2)
error## [1] 2.605253importance(bag)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863The test MSE for bagging is 2.605253.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
bag=randomForest(Sales~., data=Carseats, subset=train, mtry=3, importance=TRUE)
pred=predict(bag, test)
error=mean((test$Sales-pred)^2)
error## [1] 3.054306importance(bag)## %IncMSE IncNodePurity
## CompPrice 12.9540442 157.53376
## Income 2.1683293 129.18612
## Advertising 8.7289900 111.38250
## Population -2.5290493 102.78681
## Price 33.9482500 393.61313
## ShelveLoc 34.1358807 289.28756
## Age 12.0804387 172.03776
## Education 0.2213600 72.02479
## Urban 0.9793293 14.73763
## US 4.1072742 33.91622bag=randomForest(Sales~., data=Carseats, subset=train, mtry=5, importance=TRUE)
pred=predict(bag, test)
error=mean((test$Sales-pred)^2)
error## [1] 2.736102importance(bag)## %IncMSE IncNodePurity
## CompPrice 18.6728671 158.17550
## Income 3.1491322 109.09946
## Advertising 10.8214206 108.75115
## Population -1.5217588 76.78543
## Price 48.0253398 445.75915
## ShelveLoc 41.7681844 348.48696
## Age 14.9304083 166.65241
## Education 0.6073698 58.68779
## Urban 0.6843160 11.95040
## US 8.2025379 27.20123bag=randomForest(Sales~., data=Carseats, subset=train, mtry=7, importance=TRUE)
pred=predict(bag, test)
error=mean((test$Sales-pred)^2)
error## [1] 2.670094importance(bag)## %IncMSE IncNodePurity
## CompPrice 22.2364796 163.634007
## Income 4.4330080 100.551727
## Advertising 11.0280661 107.823416
## Population -0.2689341 67.196288
## Price 52.2754197 484.547908
## ShelveLoc 44.1715023 360.811941
## Age 13.6084631 162.451138
## Education 2.0785295 48.640094
## Urban 0.1480040 9.686486
## US 4.7421244 22.120670We can see that with a lower mtry level that it raises the test MSE.
- Now analyze the data using BART, and report your results.
Question 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.
set.seed(1)
train=sample(1:nrow(OJ),800)
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?
regtree=tree(Purchase~., data=OJ, subset=train)
summary(regtree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ, subset = 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 model uses 5 variables and has 9 nodes. The training error rate is .1588.
- 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.
regtree## 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 ) *From node 7, the splitting value is LoyalCH, we can see that there are 261 observations from the node with the splits being 95.79% CH and 4.21% MM and the deviance at 91.20
- Create a plot of the tree, and interpret the results.
plot(regtree)
text(regtree, pretty=0, cex=.5)
We can see that LoyalCH is the most important variable.
- 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=predict(regtree, test, type="class")
table(test$Purchase, pred)## pred
## CH MM
## CH 160 8
## MM 38 641-mean(pred==test$Purchase)## [1] 0.1703704The test error rate is .1703704.
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv=cv.tree(regtree, 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$size, cv$dev, xlab="Tree Size", ylab="Cross-validated Classification")
- Which tree size corresponds to the lowest cross-validated classification error rate?
The size 7 corresponds to the lowest cross-validation classification.
- 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.
p=prune.tree(regtree, best=7)- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(p)##
## Classification tree:
## snip.tree(tree = regtree, nodes = c(10L, 4L))
## 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 for the pruned model is higher at .1625
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
predu=predict(regtree, test, type="class")
tpredu=sum(test$Purchase != predu)
tpredu/length(predu)## [1] 0.1703704predp=predict(p, test, type="class")
tpredp=sum(test$Purchase != predp)
tpredp/length(predp)## [1] 0.162963The test error for the unpruned model is higher at a .1703704.