library(ISLR2)
library(tree)
library(rpart)
library(caret)

## Loading required package: ggplot2

## Loading required package: lattice

library(randomForest)

## randomForest 4.7-1.1

## 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: nnet

## Loading required package: survival

## 
## Attaching package: 'survival'

## The following object is masked from 'package:caret':
## 
##     cluster

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 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, ˆpm1 = 1 − ˆpm2. 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("blue", "red", "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.

attach(Carseats)
set.seed (1)
train <- sample (1: nrow(Carseats), nrow(Carseats) / 2)
inTrain = Carseats[train, ]
inTest = 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  = inTrain)
summary(tree.carseats)

## 
## Regression tree:
## tree(formula = Sales ~ ., data = inTrain)
## 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  4.09900

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

carseats.pred <- predict(tree.carseats , inTest)
mean((carseats.pred - inTest$Sales)^2)

## [1] 4.922039

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?

set.seed(2)
cv.carseats <- cv.tree(tree.carseats , FUN = prune.tree)

par(mfrow = c(1, 2))
plot(cv.carseats$size , cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")

prune.carseats <- prune.tree(tree.carseats , best = 9)
plot(prune.carseats)
text(prune.carseats , pretty = 0)

prune.pred <- predict(prune.carseats , inTest)
mean((prune.pred - inTest$Sales)^2)

## [1] 4.918134

Test MSE is 4.918134. Pruning the tree improves the test MSE, but not a lots of changing.

(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.

set.seed(3)
bag.seats = randomForest(Sales~., data = inTrain, mtry = 10, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = inTest)
mean((bagseat.pred - inTest$Sales)^2)

## [1] 2.630702

importance(bag.seats)

##                %IncMSE IncNodePurity
## CompPrice   25.6368749    169.596380
## Income       5.5092406     93.989153
## Advertising 13.0320872     99.531672
## Population  -1.3427420     56.478915
## Price       52.4411414    498.870909
## ShelveLoc   47.4964524    384.768230
## Age         16.7659587    153.617412
## Education   -0.4141856     44.838301
## Urban        0.2914200      9.449681
## US           5.1460185     15.033204

varImpPlot(bag.seats)

Test MSE is 2.630702. Price, ShelveLoc and CompPrice are three most important predictors of Sale

(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.

set.seed(4)
rf.seats = randomForest(Sales~., data = inTrain, mtry = 7, importance = TRUE)
rfseat.pred = predict(rf.seats, newdata = inTest)
mean((rfseat.pred - inTest$Sales)^2)

## [1] 2.598644

importance(rf.seats)

##                %IncMSE IncNodePurity
## CompPrice   21.5102063     168.65006
## Income       4.6811374      97.73172
## Advertising  9.7272392     102.75612
## Population  -0.4470593      69.81935
## Price       51.8398697     481.01558
## ShelveLoc   44.0746429     351.57487
## Age         14.1101538     166.63692
## Education   -0.5184763      49.87776
## Urban        2.2814303       9.24646
## US           4.7481416      21.50501

varImpPlot(rf.seats)

Test MSE is 2.598644, random forests improves the test MSE. Price, ShelveLoc and CompPrice still are three most important predictors of Sale

(f) Now analyze the data using BART, and report your results.

set.seed(5)

x <- Carseats[, 1:10]
y <- Carseats[, "Sales"]
xtrain <- x[train, ]
ytrain <- y[train]

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

bartfit <- 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: 3s
## trcnt,tecnt: 1000,1000

seats.bart <- bartfit$yhat.test.mean
mean (( ytest - seats.bart)^2)

## [1] 0.1547725

Test MSE is 0.1547725. The test MSE of BART is the lowest test error.

Problem 9

This problem involves the `OJ` data set which is part of the `ISLR2` 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(nrow(OJ), 800)
OJtrain = OJ[train, ]
OJtest = 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?

tree.OJ = tree(Purchase ~ ., data = OJtrain)
summary(tree.OJ)

## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## 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

5 variables were actually used in the tree construction. The training error rate is 0.1588. There are 9 terminal nodes on the tree.

(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 < 0.196196 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 5. The splitting variable at this node is LoyalCH. It also shows that the splitting value of this node is 0.280875. There are 188 points in the sub-tree below this node. 55.9% of the observations take the value of MM and 44.1% of the observations take the value of CH.

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

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

LoyalCH, PriceDiff, SpecialCH, PctDiscMMand ListPriceDiff are the most important variables.

(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, OJtest, type = "class")
confusionMatrix(OJtest$Purchase, pred.OJ)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  38  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.7333          
##     P-Value [Acc > NIR] : 0.0001259       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.8081          
##             Specificity : 0.8889          
##          Pos Pred Value : 0.9524          
##          Neg Pred Value : 0.6275          
##              Prevalence : 0.7333          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.8485          
##                                           
##        'Positive' Class : CH              
##

Test error rate is 0.1704

(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.

OJ.cv = cv.tree(tree.OJ, FUN = prune.misclass)
OJ.cv

## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 150 150 149 158 172 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## 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(OJ.cv$size, OJ.cv$dev, type = "b", xlab = "Tree Size", ylab = "cv classification error rate")

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

Size of 7 has lowest cross-validation error.

(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.tree(tree.OJ,best=7)

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

summary(prune.OJ)

## 
## Classification tree:
## snip.tree(tree = tree.OJ, 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 / 800

Training error rate is 0.1625, which is higher than unpruned trees.

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

unpruned_pred = predict(tree.OJ, OJtest, type = "class")
confusionMatrix(OJtest$Purchase, unpruned_pred)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  38  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.7333          
##     P-Value [Acc > NIR] : 0.0001259       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.8081          
##             Specificity : 0.8889          
##          Pos Pred Value : 0.9524          
##          Neg Pred Value : 0.6275          
##              Prevalence : 0.7333          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.8485          
##                                           
##        'Positive' Class : CH              
##

pruned_pred<-predict(prune.OJ,OJtest,type = "class")
confusionMatrix(OJtest$Purchase, pruned_pred)

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  36  66
##                                          
##                Accuracy : 0.837          
##                  95% CI : (0.7875, 0.879)
##     No Information Rate : 0.7259         
##     P-Value [Acc > NIR] : 1.185e-05      
##                                          
##                   Kappa : 0.6336         
##                                          
##  Mcnemar's Test P-Value : 4.693e-05      
##                                          
##             Sensitivity : 0.8163         
##             Specificity : 0.8919         
##          Pos Pred Value : 0.9524         
##          Neg Pred Value : 0.6471         
##              Prevalence : 0.7259         
##          Detection Rate : 0.5926         
##    Detection Prevalence : 0.6222         
##       Balanced Accuracy : 0.8541         
##                                          
##        'Positive' Class : CH             
##

Pruned tree has a lower test error rate compared to the unpruned tree.

Assignment #7 - Chapter 8

Tien Vo

2023-04-26

Problem 3

Hint: In a setting with two classes, ˆpm1 = 1 − ˆpm2. You could make this plot by hand, but it will be much easier to make in R.

Problem 8

Problem 9

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

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

(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?

(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.

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

(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?

(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.

(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

(h) Which tree size 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.

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

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

Assignment #7 - Chapter 8

Tien Vo

2023-04-26

Problem 3

Hint: In a setting with two classes, ˆpm1 = 1 − ˆpm2. You could make this plot by hand, but it will be much easier to make in R.

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.

(b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

(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.

(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.

(f) Now analyze the data using BART, and report your results.

Problem 9

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

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

(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?

(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.

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

(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?

(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.

(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

(h) Which tree size 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.

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

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

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.

(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.

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

(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?