library(ISLR)
library(tree)
## Warning: package 'tree' was built under R version 4.0.2
library(randomForest)
## Warning: package 'randomForest' was built under R version 4.0.2
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.

Conceptual Exercise

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 xaxis 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

p=seq(0,1,0.01)

gini <- 2*p*(1-p) # Gini Index
classerror <- 1-pmax(p,1-p) # Missclass error rate
crossentropy <- -(p*log(p)+(1-p)*log(1-p)) # The Cross-entropy

plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='ˆpm1',ylab='f')

lines(p,gini,type='l')
lines(p,classerror,col='blue')
lines(p,crossentropy,col='red')

legend(x='top',legend=c('gini','classification error','cross entropy'),
       col=c('red','blue','black'),lty=1,text.width = 0.25)

Applied Exercise

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.

(a) Split the data set into a training set and a test set.

data("Carseats")
set.seed(12)

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?

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"       "Population"  "Age"         "Advertising"
## [6] "CompPrice"   "Income"     
## Number of terminal nodes:  15 
## Residual mean deviance:  2.109 = 390.2 / 185 
## Distribution of residuals:
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## -3.0380 -1.0150 -0.1154  0.0000  1.1230  3.5600
plot(tree.carseats)
text(tree.carseats, pretty = 0)

We can quickly interpret some information about our tree from the outputs above, like the number of terminal nodes = 15.

pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)
## [1] 5.08059

The test MSE is about 5.08

(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)
plot(cv.carseats$size, cv.carseats$dev, type = "b")

prune.carseats <- prune.tree(tree.carseats, best = 11) # k = 11 seems to be the smallest tree with the lowest error

plot(prune.carseats)
text(prune.carseats, pretty = 0)

pred.pruned = predict(prune.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)
## [1] 5.227138

We see that the MSE is a bit higher

(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 = Carseats.train, mtry = 10, ntree = 500, 
    importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)
## [1] 2.881548

We now see a much lower MSE using the bagging approach

VIB <- varImpPlot(bag.carseats) # this gives us the importance 

print(VIB[order(VIB[, 1]), ]) 
##                %IncMSE IncNodePurity
## Education   -1.3594319     30.649131
## Income      -0.9862782     54.515838
## Urban        0.4086038      4.280356
## US           2.7983404      7.347901
## Population   5.2019837     87.831061
## Advertising 15.3111508    120.266789
## CompPrice   18.4564304    120.468131
## Age         18.6626506    129.856836
## Price       53.8316620    384.388771
## ShelveLoc   65.6664567    655.125601

We see that the variables Price, and ShelveLoc were the most important predictors, the mean decrease in accuracy coming from Price and ShlevLoc are very eye catching , as is the mean decreases from those variables in node impurity.

(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] 2.875741
irf <- varImpPlot(rf.carseats)

print(irf[order(irf[, 1]), ])
##                %IncMSE IncNodePurity
## Education   -1.2662650     42.813568
## Income      -1.0441090     70.583814
## Urban        0.5468306      5.607485
## Population   2.8650507     94.439292
## US           4.5695307     16.895064
## CompPrice   11.0180736    132.420448
## Advertising 14.5338574    150.944568
## Age         14.6554169    146.690470
## Price       40.3582866    385.529299
## ShelveLoc   51.9604957    546.059806

It is interseting to see that the MSE is higher, but now the variable Price is in the top two important variables. Changing m raises the MSE by about .022

Question 9

This problem involves the OJ data set which is part of the ISLR package.

(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(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]

(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 / 800

The error rate .16588, and there are 9 terminal nodes with 5 variables being used.

(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 < 0.196197 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 ) *

Node 8 denoted by the * tells us that this is a terminal node. The output above explicitly tells us that the split criterion is LoyalCH < .0356415, and that the number of observations in that branch is 59 with a deviance of 10.14. We also see that 98 % of the observations have Minute Maide as a value of sales.

(d) Create a plot of the tree, and interpret the results.

plot(oj.tree)
text(oj.tree, pretty = 0)

We can interpret that the left branch tells us when there is lower loyalty to Citrus Hill or CH, that PriceDiff and SpecialCH could result in a sale of Citrus Hill. The only thing that the tree shows may reuslt in Citrus Hill day ones is if there is a percentage discount for Minute Maid or the variable MM.

(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$Purchase, oj.pred)
##     oj.pred
##       CH  MM
##   CH 160   8
##   MM  38  64
unpruned.test.err <- mean(oj.pred != OJ.test$Purchase)
print(unpruned.test.err)# this is the test error rate
## [1] 0.1703704
1-(160+64)/270 # so is this
## [1] 0.1703704

test error is about 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 = "Deviance")

(h) Which tree size corresponds to the lowest cross-validated classification error rate?

7 seems to be the lowest tree size

(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 = 7)
plot(prune.oj)
text(prune.oj, pretty = 0)

(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?

summary(prune.oj)
## 
## 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 / 800

We see that the pruned tree produces a lower misclass rate.

(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$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)
## [1] 0.1703704
pred.pruned = predict(prune.oj, OJ.test, type = "class")
misclass.pruned <- sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)
## [1] 0.162963

The pruned tree has a lower test error.