HW7
Gregory Kresge EOW378
7/29/2020
Chapter 08 (page 332): 3, 8, 9
#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 \(\hat{P}_{m1}\). The x-axis should display \(\hat{P}_{m1}\), 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, \(\hat{P}_{m1}\)=1ā\(\hat{P}_{m2}\). You could make this plot by hand, but it will be much easier to make in R.
p=seq(0,1,0.01)
giniIndex= 2*p*(1-p)
classificationError= 1-pmax(p,1-p)
entropy= -(p*log(p)+(1-p)*log(1-p))
plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='f')
lines(p,giniIndex,type='b', col='red')
lines(p,classificationError, type = 'b', col='green')
lines(p,entropy,type = 'b', col='blue')
legend(x='bottom',legend=c('gini','class error','cross entropy'),
col=c('red','green','blue'),lty=1,text.width = 0.22)
#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.
attach(Carseats)(a) Split the data set into a training set and a test set.
set.seed(1)
train=sample(1:nrow(Carseats), nrow(Carseats)*.70)
train.carseats=Carseats[train,]
test=(-train)
test.carseats=Carseats[test,]
Sales.test=Carseats$Sales[test](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.carseats=tree(Salesā¼., data=train.carseats)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = train.carseats)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Income" "CompPrice"
## [6] "Advertising"
## Number of terminal nodes: 18
## Residual mean deviance: 2.409 = 631.1 / 262
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.77800 -0.96100 -0.08865 0.00000 1.01800 4.14100plot(tree.carseats)
text(tree.carseats, pretty = 0)
pred.carseats.sales=predict(tree.carseats, newdata = test.carseats)
MSE.Carseats = mean((pred.carseats.sales - Sales.test)^2)
MSE.Carseats## [1] 4.208383Based on the above regression tree, the test MSE is 4.208383.
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.carseats = cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "p")
tree.min = which.min(cv.carseats$dev)
points(tree.min, cv.carseats$dev[tree.min], col = "green", cex = 2, pch = 20)
tree.carseats.prune = prune.tree(tree.carseats, best = tree.min)
plot(tree.carseats.prune)
text(tree.carseats.prune, pretty = 0)
pred.carseats.sales.prune=predict(tree.carseats.prune, newdata = test.carseats)
MSE.Carseats.prune = mean((pred.carseats.sales.prune - Sales.test)^2)
MSE.Carseats.prune## [1] 4.691032MSE.Carseats## [1] 4.208383The pruned regression tree has a test MSE of 4.691032. The un-pruned tree has a lower MSE than the pruned tree.
(d) Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
bag.carseats = randomForest(Sales ~ ., data = train.carseats, mtry = 10, ntree = 500, importance = TRUE)
pred.carseats.bag.sales = predict(bag.carseats, newdata = test.carseats)
mean((pred.carseats.bag.sales - Sales.test)^2)## [1] 2.571169importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 34.6322504 233.60705
## Income 5.3645204 116.93827
## Advertising 18.8175105 153.05938
## Population -3.0858810 64.26621
## Price 70.9386948 698.15948
## ShelveLoc 74.2328945 645.49148
## Age 20.8561302 224.71954
## Education 1.3723565 61.47839
## Urban -1.9986734 10.51832
## US 0.8095402 10.19895(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
carseats.rf = train(Sales~., data=train.carseats, method ='rf',trControl = trainControl("cv", number = 10),importance = TRUE)
carseats.rf$bestTune## mtry
## 3 11carseats.rf$finalModel##
## Call:
## randomForest(x = x, y = y, mtry = param$mtry, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 11
##
## Mean of squared residuals: 2.538977
## % Var explained: 68.48plot(varImp(carseats.rf))
The final random forest model uses 11 variables and has a test MSE of 2.618235.
#9
This problem involves the OJ data set which is part of the ISLR package.
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)
train=sample(1:nrow(OJ), 800)
train.OJ=OJ[train,]
test=(-train)
test.OJ=OJ[test,]
purchase.test=OJ$Purchase[test](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?
set.seed(1)
tree.oj=tree(Purchaseā¼., data=train.OJ)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train.OJ)
## 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 / 800There are 9 terminal nodes. The training error rate is 0.1588.
(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.
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 ) *Looking at terminal node 21, we can see this node is for if there is a special on Citrus Hill orange juice. There are 15 observations when there is a special on Citrus Hill, and the prediction for the node is a purchase of Citrus Hill. 60% of the observations for that node are Citrus Hill purchases.
(d) Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)
Looking at the plot of the tree, it is aparent that main factors that determine which brand is purcahsed is customer loyalty (in this case to Citrus Hill brand), and price (either price difference between the brands or if there is a special or discount on one brand or the other.) For insantance, looking at the left branch of the tree, this is customers with low brand loyalty to Citrus Hill. However, given a significantly lower prices for Citrus Hill, or a special on Citrus Hill, the customer is likely to buy that product instead of Minute Maid.
(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(tree.oj, newdata = test.OJ, type = 'class')
table(pred.oj, purchase.test)## purchase.test
## pred.oj CH MM
## CH 160 38
## MM 8 64oj.test.err = (8+38)/(160+38+8+64)
oj.test.err## [1] 0.1703704The test error rate is .1703704.
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
cv.oj = cv.tree(tree.oj)
which.min(cv.oj$dev)## [1] 5The optimal tree size is 5.
(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 = "p")
tree.min = which.min(cv.oj$dev)
points(tree.min, cv.oj$dev[tree.min], col = "green", cex = 2, pch = 20)
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
which.min(cv.oj$dev)## [1] 5Tree size = 5 corresponds 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.
tree.oj.prune = prune.tree(tree.oj, best = tree.min)
plot(tree.oj.prune)
text(tree.oj.prune, pretty = 0)
(j) Compare the training error rates between the pruned and un-pruned trees. Which is higher?
summary(tree.oj.prune)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(4L, 12L, 5L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "ListPriceDiff"
## Number of terminal nodes: 5
## Residual mean deviance: 0.8239 = 655 / 795
## Misclassification error rate: 0.205 = 164 / 800Pruned training error rate is 0.205. This is higher than the un-pruned training error rate of 0.1588.
(k) Compare the test error rates between the pruned and unprunedtrees. Which is higher?
set.seed(1)
pred.oj.purchase.prune=predict(tree.oj.prune, newdata = test.OJ, type = 'class')
table(pred.oj.purchase.prune, purchase.test)## purchase.test
## pred.oj.purchase.prune CH MM
## CH 136 21
## MM 32 81oj.test.err = (32+21)/(136+21+32+81)
oj.test.err## [1] 0.1962963Pruned test error rate is 0.1962963. This is larger than the un-pruned test error rate of .1703704.