library(ISLR)
## Warning: package 'ISLR' was built under R version 4.4.2
library(tree)
## Warning: package 'tree' was built under R version 4.4.3
library(rpart)
library(caret)
## Warning: package 'caret' was built under R version 4.4.2
## Loading required package: ggplot2
## Loading required package: lattice
library(randomForest)
## Warning: package 'randomForest' was built under R version 4.4.3
## 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
#PROBLEM 3
p = seq(0, 1, 0.001)
gini.index = 2 * p * (1 - p)
class.error = 1 - pmax(p, 1 - p)
cross.entropy = - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), ylab = "gini.index, class.error, cross.entropy", col = c("red", "blue", "green"))
#PROBLEM 8 PART A)
set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats)/2)
strain = Carseats[train, ]
stest = Carseats[-train, ]
#PROBLEM 8 PART B)
tree.seats = tree(Sales ~ ., data = strain)
summary(tree.seats)
##
## Regression tree:
## tree(formula = Sales ~ ., data = strain)
## 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.seats)
text(tree.seats, pretty = 0)
treeseat.pred = predict(tree.seats, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)
## [1] 4.922039
#The MSE obtained from the regression tree is 4.15.
#PROBLEM 8 PART c)
set.seed(1)
cv.seats = cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b")
prune.car = prune.tree(tree.seats, best = 10)
plot(prune.car)
text(prune.car,pretty=0)
treeseat.pred = predict(prune.car, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)
## [1] 4.918134
#The new MSE is 4.82. This means that pruning the tree does not improve the test MSE because it increased compared to the prior value of 4.15. Even if pruning minimizes cross-validated error, it can still have higher actual test MSE because there might be variance between cross-validation and test set.
#PROBLEM 8 PART D)
set.seed(1)
bag.seats = randomForest(Sales~., data = strain, mtry = 10, ntree = 551, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = stest)
mean((bagseat.pred - stest$Sales)^2)
## [1] 2.599099
importance(bag.seats)
## %IncMSE IncNodePurity
## CompPrice 26.18616309 170.781666
## Income 5.25063979 90.717958
## Advertising 13.25673204 97.498810
## Population -2.14346969 58.289311
## Price 60.58241525 503.478806
## ShelveLoc 50.77308639 380.258594
## Age 19.03720001 158.282846
## Education 1.24264920 44.834257
## Urban -0.08461165 9.883299
## US 4.71515903 17.907727
varImpPlot(bag.seats)
#The tope three variables that seem to be the most important are: Price, ShelveLoc and Age.
#PROBLEM 8 PART E)
set.seed(1)
rando.seats = randomForest(Sales~., data = strain, mtry = 10, importance = TRUE)
randseat.pred = predict(rando.seats, newdata = stest)
mean((randseat.pred - stest$Sales)^2)
## [1] 2.605253
#The test MSE is 2.61. This is higher value than 2.60, meaning that the random forests do not result in improved resultws over bagging.
importance(rando.seats)
## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863
#PROBLEM 9 PART A)
library(ISLR)
set.seed(1)
train = sample(dim(OJ)[1],800)
OJ_train = OJ[train,]
OJ_test = OJ[-train,]
#PROBLEM 9 PART B)
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 training error rate is 0.165, and the tree has 8 termminal nodes.
#PROBLEM 9 PART C)
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 < 0.196196 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 ) *
#The terminal nodes chosen for the interpretation is 9. The split cliteriro is LoyalCH <0.036, and the number of observations in the branch is 109. It has a deviance of 100.90 and overall prediction of MM. There are less than 2% of the observation sin the branch that tkae the value of CH, which leaves that remaining 98% to take the value of MM.
#PROBLEM 9 PART D)
plot(OJ_tree)
text(OJ_tree,pretty=TRUE)
#The most important variable seems to be LoyalCH for "Purchase" as the first branch shows the strength of customer loyalty to CH.
#PROBLEM 9 PART E)
tree.pred = predict(OJ_tree, newdata = OJ_test, type = "class")
table(tree.pred,OJ_test$Purchase)
##
## tree.pred CH MM
## CH 160 38
## MM 8 64
(147+62)/270
## [1] 0.7740741
#Above shows that 77% of test observations are correctly classified. This means that the error rate is abotu 23%.
#PROBLEM 9 PART F)
cv_OJ = cv.tree(OJ_tree, FUN = prune.misclass)
cv_OJ
## $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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"
#PROBLEM 9 PART G)
plot(cv_OJ$size,cv_OJ$dev,type='b', xlab = "Tree size", ylab = "Deviance")
#PROBLEM 9 PART H)
#From the plot above, we can se that the tree with 5 node correspondes to the lowest cross-validated classification error rate on the y-axis.
#PROBLEM 9 PART I)
prune_OJ = prune.misclass(OJ_tree, best=5)
plot(prune_OJ)
text(prune_OJ,pretty=0)
#PROBLEM 9 PART J)
summary(OJ_tree)
##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ_train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800
summary(prune_OJ)
##
## Classification tree:
## snip.tree(tree = OJ_tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800
#The trainig error rates between the pruned and unpruned trees seems to be the same. However, pruning helped create a more interpretable tree.
#PROBLEM 9 PART K)
tree.pred = predict(OJ_tree, newdata = OJ_test, type = "class")
table(tree.pred,OJ_test$Purchase)
##
## tree.pred CH MM
## CH 160 38
## MM 8 64
pruneOJ.pred = predict(prune_OJ, newdata = OJ_test, type = "class")
table(pruneOJ.pred, OJ_test$Purchase)
##
## pruneOJ.pred CH MM
## CH 160 36
## MM 8 66
(147+62)/270
## [1] 0.7740741
#The test error rates between the pruned and unpruned trees seem to be the same with aproximately 0.77.