Conceptual

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 \(\hat{p}_{m1}\). 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, \(\hat{p}_{m1}\) = 1 - \(\hat{p}_{m2}\). You could make this plot by hand, but it will be much easier to make in R.

prob_1 <- seq(0, 1, 0.10)
prob_2 <- 1 - prob_1

gini_index <- 2 * prob_1 * prob_2
classification_error <- 1 - pmax(prob_1, prob_2)
entropy <- - (ifelse(prob_1 == 0, 0, prob_1 * log(prob_1)) +
              ifelse(prob_2 == 0, 0, prob_2 * log(prob_2)))

plot(prob_1,
     entropy,
     type = "l",
     ylim = c(0,1),
     col = "purple",
     )
lines(prob_1, gini_index, col = "blue")
lines(prob_1, classification_error, col = "darkgreen")

Applied

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.

# install.packages("BART")
library(ISLR2)
library(caret)
## Loading required package: ggplot2
## Loading required package: lattice
library(tree)
library(randomForest)
## randomForest 4.7-1.2
## Type rfNews() to see new features/changes/bug fixes.
## 
## Attaching package: 'randomForest'
## The following object is masked from 'package:ggplot2':
## 
##     margin
library(BART)
## Loading required package: nlme
## Loading required package: survival
## 
## Attaching package: 'survival'
## The following object is masked from 'package:caret':
## 
##     cluster
attach(Carseats)
  1. Split the data set into a training set and a test set.
set.seed(1)
trainindex <- createDataPartition(Carseats$Sales, p = 0.80, list = FALSE)

train_carseat <- Carseats[trainindex, ]
test_carseat <- Carseats[- trainindex, ]
  1. 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 ~ ., train_carseat)
summary(tree.carseats)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train_carseat)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "CompPrice"   "Age"         "Advertising"
## Number of terminal nodes:  18 
## Residual mean deviance:  2.39 = 724.3 / 303 
## Distribution of residuals:
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## -4.9380 -0.9822  0.1214  0.0000  0.9908  3.3760
plot(tree.carseats)
text(tree.carseats, pretty = 0)

tree_pred <- predict(tree.carseats, test_carseat)
mean((tree_pred - test_carseat$Sales) ^ 2)
## [1] 3.947858

From the decision tree, we can see that the test MSE using the training data is about 3.95. Out of the 12 variables, the decision tree made the predictions based on five variables: ShelveLoc, Price, CompPrice, Age, Advertising.

  1. 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)
cv.carseats
## $size
##  [1] 18 17 16 15 14 13 12 10  9  8  7  6  5  4  3  2  1
## 
## $dev
##  [1] 1518.114 1544.175 1532.682 1562.012 1564.222 1573.732 1575.487 1629.843
##  [9] 1613.281 1758.139 1792.464 1784.681 1773.340 1765.712 1783.284 1954.793
## [17] 2566.119
## 
## $k
##  [1]      -Inf  26.74673  27.28892  32.67834  40.44341  43.06716  43.24045
##  [8]  48.48233  51.91873  66.97883  68.81871  71.65623  80.16261 117.47087
## [15] 152.95942 284.39933 619.81164
## 
## $method
## [1] "deviance"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"

From the cross-validation, we can see that optimal level of tree complexity is at size 18. What this is saying is that pruning the decision tree does not improve the current decision tree. Since the cross-validation said the full model is most efficienct, the MSE has not changed and is also 3.95.

  1. 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.carseat <- randomForest(Sales ~ ., data = train_carseat, importance = TRUE)
bag.carseat
## 
## Call:
##  randomForest(formula = Sales ~ ., data = train_carseat, importance = TRUE) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 3
## 
##           Mean of squared residuals: 2.785152
##                     % Var explained: 64.92
set.seed(1)
yhat.bag <- predict(bag.carseat, newdata = test_carseat)
test_sales <- test_carseat$Sales
plot(yhat.bag, test_sales)
abline(0,1)

mean((yhat.bag - test_sales) ^ 2)
## [1] 2.878256
importance(bag.carseat)
##                   %IncMSE IncNodePurity
## CompPrice   14.7758597775     226.30590
## Income       7.0219098197     189.57453
## Advertising 20.1504739445     232.13827
## Population  -0.0006996264     149.40412
## Price       49.3932918406     622.92892
## ShelveLoc   53.3930058581     588.83879
## Age         16.4445354850     250.09345
## Education    3.1977026272     102.48771
## Urban        0.0509883165      22.39921
## US           4.1102636734      33.54182

Using bagging, we can see that the MSE error is about 2.88. And with importance, we can see that ShelveLoc and Price are the two most important variables.

  1. 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.carseat <- randomForest(Sales ~., data = train_carseat,
                            mtry = 5, importance = TRUE)
rf.carseat
## 
## Call:
##  randomForest(formula = Sales ~ ., data = train_carseat, mtry = 5,      importance = TRUE) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 5
## 
##           Mean of squared residuals: 2.460876
##                     % Var explained: 69.01
yhat.rf <- predict(rf.carseat, newdata = test_carseat)
mean((yhat.rf - test_sales) ^ 2)
## [1] 2.568048
importance(rf.carseat)
##               %IncMSE IncNodePurity
## CompPrice   20.184934     246.32343
## Income       7.372158     155.33685
## Advertising 23.544142     235.54125
## Population  -1.188342     113.81406
## Price       61.300857     680.92064
## ShelveLoc   67.591155     685.01236
## Age         19.124661     232.73986
## Education    3.395507      81.30260
## Urban       -2.697134      12.68769
## US           4.634857      21.45461

The test MSE of the random forest is 2.61 and we can see from importance that the variables ShelveLoc and Price are the most important. Since m is equal to 5 this will be the number of variables considered at the split of a random forest. If m is too small then it could lead to a random forest that does not provide significant information. While on the other hand if m is too large then all of the trees could start to be identical to one another.

  1. Now analyze the data using BART, and report your results.
x <- Carseats[, 2:11]
y <- Carseats[, "Sales"]

xtrain <- x[trainindex, ]
ytrain <- y[trainindex]

xtest <- x[-trainindex, ]
ytest <- y[-trainindex]

set.seed(1)

bartfit <- gbart(xtrain, ytrain, x.test = xtest)
## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 321, 14, 79
## y1,yn: 1.989065, -1.570935
## x1,x[n*p]: 138.000000, 1.000000
## xp1,xp[np*p]: 117.000000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 67 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.276302,3,0.202195,7.51093
## *****sigma: 1.018826
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****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: 1s
## trcnt,tecnt: 1000,1000
yhat.bart <- bartfit$yhat.test.mean
mean((yhat.bart - ytest) ^ 2)
## [1] 1.437575

With BART, the MSE is 1.46. This is good because the BART test is performing better then random forest model but is worse then the bagging model. Of the models test I would recommend to use either the bagging model or BART model.

Question 9

This problem involves the OJ data set which is part of the ISLR2 package.

attach(OJ)
  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)

index <- sample(nrow(OJ), size = 800)

train_oj <- OJ[index, ]
test_oj <- OJ[- index, ]
  1. 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(Purchase ~., 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 / 800

In the initial tree, it has decided to base it’s decisions on 5 variables: LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, PctDiscMM. Also, the initial test contained 9 terminal nodes. Lastly, the model has a training error rate of 15% almost 16%.

  1. 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 < 0.196196 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 8: LoyalCH < 0.0356415: - Split: Any LoyalCH value less then 0.0356415. This is how the model decides how the data will be split. - N: 59. This is the total number of points after the split - Deviance: 10.14. This is deviance of the 59 points in Node 8. - YVal: MM. This is the predicted value for any point that is placed within the node. So if LoyalCH < 0.0356415 then the model will predict the brand as MM (Minute Maid) - (YProb): ( 0.01695 0.98305 ). This is the probability of the predictions for the points in the node. In this example for node 8, there is a 1.695% of a chance the data point in Node 8 is Citrus Hill. While on the other hand there is a 98.305% of a chance the data point in Node 8 is Minute Maid.

  1. Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)

From the plot, it seems the most important indicator is LoyalCH. Also, it can be see that at the end of the branches majority of the end predictions are Citrus Hill.

  1. 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?
tree.oj.pred <- predict(tree.oj, test_oj, type = "class")
table(tree.oj.pred, test_oj$Purchase)
##             
## tree.oj.pred  CH  MM
##           CH 160  38
##           MM   8  64
(8 + 38) / 270
## [1] 0.1703704

The test error rate of the initial tree is 17.04%.

  1. 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, FUN = prune.misclass)
cv.oj
## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 145 145 146 146 167 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b")

  1. Which tree size corresponds to the lowest cross-validated classification error rate?

Based on the plot, the lowest cross-validated classification error rate is lowest when the tree size corresponds with 9 or 8. However, the inital tree had nine nodes, so the cross-validation is saying the initial tree performs the best.

  1. 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(tree.oj, best = 5)
summary(prune.oj)
## 
## Classification tree:
## snip.tree(tree = tree.oj, 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
  1. Compare the training error rates between the pruned and unpruned trees. Which is higher?
# pruned tree
prune.pred <- predict(prune.oj, train_oj, type = "class")
table(prune.pred, train_oj$Purchase)
##           
## prune.pred  CH  MM
##         CH 441  86
##         MM  44 229
(44 + 86) / 800
## [1] 0.1625

The training error of the pruned tree is: 16.25%

# unpruned tree

tree.oj.pred <- predict(tree.oj, train_oj, type = "class")
table(tree.oj.pred, train_oj$Purchase)
##             
## tree.oj.pred  CH  MM
##           CH 450  92
##           MM  35 223
(35 + 92) / 800
## [1] 0.15875

The training error of the unpruned tree is: 15.88%

The pruned tree has the higher training error rate of the two.

  1. Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, test_oj, type = "class")
table(prune.pred, test_oj$Purchase)
##           
## prune.pred  CH  MM
##         CH 160  36
##         MM   8  66
(8 + 36) / 270
## [1] 0.162963

The test error rate of the pruned tree is: 16.30%

tree.oj.pred <- predict(tree.oj, test_oj, type = "class")
table(tree.oj.pred, test_oj$Purchase)
##             
## tree.oj.pred  CH  MM
##           CH 160  38
##           MM   8  64
(8 + 38) / 270
## [1] 0.1703704

The test error rate of the unpruned tree is 17.04%

The unpruned tree has the higher test error rate of the two.