##Q3 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 ˆ pm1. Thexaxis should display ˆ pm1, 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, ˆ pm1 =1− ˆ pm2. You could make this plot by hand, but it will be much easier to make in R.

p <- seq(0, 1, 0.01)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
par(bg = "maroon")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("darkolivegreen4" , "wheat", "tomato2"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("darkolivegreen4" , "wheat", "tomato2"), pch=c(15,17,19))

##Q8 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.

  1. Split the data set into a training set and a test set.
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
library(tree)
## Warning: package 'tree' was built under R version 4.0.5
library(dplyr)
## Warning: package 'dplyr' was built under R version 4.0.5
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
set.seed(24)
c<-Carseats
train<-sample(1:nrow(c), nrow(c)/2)

c.tr<-c[train,]
c.te<-c[-train,]
  1. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.c=tree(Sales~., data=c.tr)
plot(tree.c)
text(tree.c, pretty=0)

yhat.tree=predict(tree.c, c.te)
obs.sales<-c.te$Sales
mean((yhat.tree-obs.sales)^2)
## [1] 4.541017
  1. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.c=cv.tree(tree.c)
#Let's find the lowest $dev value here
cv.c
## $size
##  [1] 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2  1
## 
## $dev
##  [1] 1012.989 1041.573 1020.262 1043.493 1057.997 1064.050 1131.822 1130.429
##  [9] 1211.419 1182.640 1250.044 1239.531 1352.795 1438.036 1382.154 1686.631
## 
## $k
##  [1]      -Inf  17.46067  18.21007  25.43324  29.77121  32.67272  44.19972
##  [8]  46.09869  49.55016  50.87585  60.04473  63.14420  99.94859 132.15272
## [15] 143.22869 395.57046
## 
## $method
## [1] "deviance"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
plot(cv.c$size,cv.c$dev, xlab = 'size', ylab = 'cv error ')
abline(h = min(cv.c$dev) + .2*sd(cv.c$dev))

We can see in the chart that the size that holds the lowest cv error is size 16. However, we can also see there are a couple other smaller variable sizes that are also very close. Let’s see if we can find which size provides the lowest MSE.

prune.c<-list()
yhat.prune<-list()
mse.prune<-list()
for (i in 2:16) {
    prune.c[[i]]=prune.tree(tree.c, best = i)
    yhat.prune[[i]]=predict(prune.c[[i]], c.te)
    obs.sales<-c.te$Sales
    mse.prune[[i]]<-mean((yhat.prune[[i]]-obs.sales)^2)  
}

mse.mat<-as.matrix(mse.prune)
mse.df<-as.data.frame(mse.mat)

par(bg = "pink" )
matplot(2:16, mse.df[-1,1], xlab = 'Size', ylab = 'MSE', main = "lowest MSE by size")
lines(2:16, mse.df[-1,1], type = "o")

mse.df <- data.frame(size=2:16,lapply(mse.df, unlist) )
mse.df$mse.rank<-rank(mse.df$V1)
names(mse.df)[2]<-'mse'
mse.df[mse.df$mse.rank %in% 1:3,]
##    size      mse mse.rank
## 13   14 4.727879        3
## 14   15 4.499964        1
## 15   16 4.541017        2

We can see in the rank that 15 nodes gives the lowest MSE with 4.499964

  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.
library(randomForest)
## Warning: package 'randomForest' was built under R version 4.0.5
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
## 
## Attaching package: 'randomForest'
## The following object is masked from 'package:dplyr':
## 
##     combine
set.seed(24)
bag.c<-randomForest(Sales~., data = c.tr, mtry = 10, importance = TRUE)

yhat.bag<-predict(bag.c, newdata = c.te)
mean((yhat.bag - c.te$Sales)^2)
## [1] 2.692938
importance(bag.c)
##                %IncMSE IncNodePurity
## CompPrice   22.8321261    176.578526
## Income       4.1064747     77.183648
## Advertising 12.8743758     99.953925
## Population   0.3779757     76.359264
## Price       51.4435675    465.155918
## ShelveLoc   51.6714645    438.596460
## Age         21.2202209    201.116544
## Education    1.3558349     36.305982
## Urban       -2.0566741      6.577210
## US           0.6451619      5.857904

lets try mtry = 1:10

rf.c<- list()
yhat.rf<-list()
mse.rf<-list()
for ( i in 1:10 ) {
    set.seed(24)
    rf.c[[i]]<-randomForest(Sales ~ ., data = c.tr, mtry = i, importance = TRUE)
    yhat.rf[[i]]<-predict(rf.c[[i]], newdata = c.te)
    mse.rf[[i]]<- mean((yhat.rf[[i]] - c.te$Sales)^2)
}
par(bg = "green" )
matplot(1:10, mse.rf, xlab = 'mtry', ylab = 'MSE', main = "Lowest MSE by Predictors")
lines(1:10, mse.rf, type = "o")

mseRF<-t(as.data.frame(lapply(mse.rf, unlist)))
mse.rf<-data.frame(size=1:10,mseRF)
mse.rf.rownames<-paste0("mtry", 1:10)
rownames(mse.rf)<-mse.rf.rownames
mse.rf$mse.rank<-rank(mse.rf$mseRF)
mse.rf[order(mse.rf$mse.rank),]
##        size    mseRF mse.rank
## mtry10   10 2.692938        1
## mtry8     8 2.704865        2
## mtry9     9 2.722630        3
## mtry7     7 2.759607        4
## mtry6     6 2.843481        5
## mtry5     5 2.897282        6
## mtry4     4 2.970406        7
## mtry3     3 3.204131        8
## mtry2     2 3.573238        9
## mtry1     1 4.770207       10

In the table above we can see that mtry = 10 holds the lowest MSE value of 2.692938

##Q9 This problem involves the OJ data set which is part of the ISLR package.

  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(10)
library(ISLR)
library(tree)
sam<-sample(1:nrow(OJ), 800)

oj.tr<-OJ[sam,]
oj.te<-OJ[-sam,]
  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?
tree.oj<-tree(Purchase~., data = oj.tr)
summary(tree.oj)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.tr)
## Variables actually used in tree construction:
## [1] "LoyalCH"   "DiscMM"    "PriceDiff"
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7983 = 633 / 793 
## Misclassification error rate: 0.1775 = 142 / 800

Misclassification error rate: \(17.75%\) Number of terminal nodes: 7 Uses only 3 of the 18 variables: “LoyalCH” “DiscMM” “PriceDiff”

  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 1067.000 CH ( 0.61375 0.38625 )  
##    2) LoyalCH < 0.48285 290  315.900 MM ( 0.23448 0.76552 )  
##      4) LoyalCH < 0.035047 51    9.844 MM ( 0.01961 0.98039 ) *
##      5) LoyalCH > 0.035047 239  283.600 MM ( 0.28033 0.71967 )  
##       10) DiscMM < 0.47 220  270.500 MM ( 0.30455 0.69545 ) *
##       11) DiscMM > 0.47 19    0.000 MM ( 0.00000 1.00000 ) *
##    3) LoyalCH > 0.48285 510  466.000 CH ( 0.82941 0.17059 )  
##      6) LoyalCH < 0.764572 245  300.200 CH ( 0.69796 0.30204 )  
##       12) PriceDiff < 0.145 99  137.000 MM ( 0.47475 0.52525 )  
##         24) DiscMM < 0.47 82  112.900 CH ( 0.54878 0.45122 ) *
##         25) DiscMM > 0.47 17   12.320 MM ( 0.11765 0.88235 ) *
##       13) PriceDiff > 0.145 146  123.800 CH ( 0.84932 0.15068 ) *
##      7) LoyalCH > 0.764572 265  103.700 CH ( 0.95094 0.04906 ) *
  1. LoyalCH < 0.035047 51 9.844 MM ( 0.01961 0.98039 ) * If LoyalCH is less than 0.48285 AND less than 0.035047, then the Purchase is MM
  1. Create a plot of the tree, and interpret the results
par(bg = "violet")
plot(tree.oj)
text(tree.oj, pretty = 0)

  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.preds<-predict(tree.oj, oj.te, type = 'class')
obs.purch<-oj.te$Purchase
caret::confusionMatrix(tree.preds, obs.purch)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 135  20
##         MM  27  88
##                                           
##                Accuracy : 0.8259          
##                  95% CI : (0.7753, 0.8692)
##     No Information Rate : 0.6             
##     P-Value [Acc > NIR] : 9.992e-16       
##                                           
##                   Kappa : 0.6412          
##                                           
##  Mcnemar's Test P-Value : 0.3815          
##                                           
##             Sensitivity : 0.8333          
##             Specificity : 0.8148          
##          Pos Pred Value : 0.8710          
##          Neg Pred Value : 0.7652          
##              Prevalence : 0.6000          
##          Detection Rate : 0.5000          
##    Detection Prevalence : 0.5741          
##       Balanced Accuracy : 0.8241          
##                                           
##        'Positive' Class : CH              
##

We get an error rate of 17.41%

  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj<-cv.tree(tree.oj, FUN = prune.misclass)
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
par(bg = "yellow")
plot(cv.oj$size,cv.oj$dev/800, type ="b", xlab = "Size", ylab = 'CV Error Rate', main = 'Lowest CV Error Rate by Size')

  1. Which tree size corresponds to the lowest cross-validated classification error rate?
best.trees<-data.frame(tree_size = cv.oj$size, CvErrors = cv.oj$dev, Rate = paste0(cv.oj$dev/8,"%"))
best.trees[order(best.trees$Rate),]
##   tree_size CvErrors    Rate
## 1         7      157 19.625%
## 2         5      157 19.625%
## 3         2      161 20.125%
## 4         1      309 38.625%

We can see in the table above that the tree size of 7 and 5 give us the lowest CV error rate of 19.625%

  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)
  1. Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.predz<-predict(prune.oj, oj.te, type = "class")
caret::confusionMatrix(prune.predz, obs.purch)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 135  20
##         MM  27  88
##                                           
##                Accuracy : 0.8259          
##                  95% CI : (0.7753, 0.8692)
##     No Information Rate : 0.6             
##     P-Value [Acc > NIR] : 9.992e-16       
##                                           
##                   Kappa : 0.6412          
##                                           
##  Mcnemar's Test P-Value : 0.3815          
##                                           
##             Sensitivity : 0.8333          
##             Specificity : 0.8148          
##          Pos Pred Value : 0.8710          
##          Neg Pred Value : 0.7652          
##              Prevalence : 0.6000          
##          Detection Rate : 0.5000          
##    Detection Prevalence : 0.5741          
##       Balanced Accuracy : 0.8241          
##                                           
##        'Positive' Class : CH              
##

best = 5 17.41% best = 7 17.41% Both of the trees with 5 and 7 terminal nodes have the same error rate which is the lowest compared to the others.