Assingment 7
Richard Rivas
2022-11-17
library(ISLR2)
library(tree)
library(rpart)
library(caret)## Loading required package: ggplot2## Loading required package: latticelibrary(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- Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that dis- plays each of these quantities as a function of ˆp m1 . The x-axis should display ˆp 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, ˆp m1 = 1 − ˆp m2 . You could make this plot by hand, but it will be much easier to make in R.
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", "green", "orange"))
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)
ctrain = Carseats[train, ]
ctest = Carseats[-train, ](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain? The test MSE is 4.922039
tree.seat = tree(Sales ~ ., data= ctrain)
plot(tree.seat)
text(tree.seat, pretty = 5)
summary(tree.seat)##
## Regression tree:
## tree(formula = Sales ~ ., data = ctrain)
## 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 4.09900cpred =predict(tree.seat, newdata = ctest)
mean((cpred - ctest$Sales)^2)## [1] 4.922039(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
Pruning the tree to 10 lowers the MSE to 4.918 from 4.922
set.seed(1)
cv.seats = cv.tree(tree.seat)
plot(cv.seats$size, cv.seats$dev, type = "b")
prune.seats = prune.tree(tree.seat, best = 10)
plot(prune.seats)
text(prune.seats, pretty=5)
prun.pred = predict(prune.seats, newdata = ctest)
mean((prun.pred - ctest$Sales)^2)## [1] 4.918134(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. MSE is 2.599099.
set.seed(1)
bag.seats = randomForest(Sales~., data = ctrain, mtry = 10, ntree = 551, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = ctest)
mean((bagseat.pred - ctest$Sales)^2)## [1] 2.599099(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained. The MSE is 2.605253
set.seed(1)
rando.seats = randomForest(Sales~., data = ctrain, mtry = 10, importance = TRUE)
rand.seat.predict = predict(rando.seats, newdata = ctest)
mean((rand.seat.predict - ctest$Sales)^2)## [1] 2.605253importance(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(f) Now analyze the data using BART, and report your results.
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.
attach(OJ)
train.OJ= sample(1:nrow(OJ),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? 7 variables were used int the tree construction. The training error rate is 0.095. There are 15 terminal nodes on the tree.
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" "StoreID" "DiscMM"
## [5] "PriceCH" "SalePriceMM" "WeekofPurchase"
## Number of terminal nodes: 15
## Residual mean deviance: 0.3951 = 73.09 / 185
## Misclassification error rate: 0.095 = 19 / 200plot(OJ.tree)
text(OJ.tree, pretty= 0)
(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. 13) LoyalCH > 0.70936 79 37.280 CH (
0.93671 0.06329 )
OJ.tree## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 200 242.600 CH ( 0.70500 0.29500 )
## 2) LoyalCH < 0.592 75 95.480 MM ( 0.33333 0.66667 )
## 4) LoyalCH < 0.060978 18 0.000 MM ( 0.00000 1.00000 ) *
## 5) LoyalCH > 0.060978 57 78.160 MM ( 0.43860 0.56140 )
## 10) PriceDiff < 0.065 25 27.550 MM ( 0.24000 0.76000 )
## 20) StoreID < 2.5 11 0.000 MM ( 0.00000 1.00000 ) *
## 21) StoreID > 2.5 14 19.120 MM ( 0.42857 0.57143 )
## 42) DiscMM < 0.3 6 5.407 MM ( 0.16667 0.83333 ) *
## 43) DiscMM > 0.3 8 10.590 CH ( 0.62500 0.37500 ) *
## 11) PriceDiff > 0.065 32 43.230 CH ( 0.59375 0.40625 )
## 22) StoreID < 1.5 5 0.000 CH ( 1.00000 0.00000 ) *
## 23) StoreID > 1.5 27 37.390 CH ( 0.51852 0.48148 )
## 46) PriceCH < 1.805 5 0.000 MM ( 0.00000 1.00000 ) *
## 47) PriceCH > 1.805 22 28.840 CH ( 0.63636 0.36364 )
## 94) SalePriceMM < 2.125 7 0.000 CH ( 1.00000 0.00000 ) *
## 95) SalePriceMM > 2.125 15 20.730 MM ( 0.46667 0.53333 ) *
## 3) LoyalCH > 0.592 125 64.700 CH ( 0.92800 0.07200 )
## 6) PriceDiff < 0.31 94 59.340 CH ( 0.90426 0.09574 )
## 12) LoyalCH < 0.70936 15 17.400 CH ( 0.73333 0.26667 ) *
## 13) LoyalCH > 0.70936 79 37.280 CH ( 0.93671 0.06329 )
## 26) WeekofPurchase < 248.5 24 21.630 CH ( 0.83333 0.16667 )
## 52) WeekofPurchase < 236.5 9 0.000 CH ( 1.00000 0.00000 ) *
## 53) WeekofPurchase > 236.5 15 17.400 CH ( 0.73333 0.26667 )
## 106) LoyalCH < 0.902367 10 6.502 CH ( 0.90000 0.10000 ) *
## 107) LoyalCH > 0.902367 5 6.730 MM ( 0.40000 0.60000 ) *
## 27) WeekofPurchase > 248.5 55 9.996 CH ( 0.98182 0.01818 )
## 54) DiscMM < 0.47 48 0.000 CH ( 1.00000 0.00000 ) *
## 55) DiscMM > 0.47 7 5.742 CH ( 0.85714 0.14286 ) *
## 7) PriceDiff > 0.31 31 0.000 CH ( 1.00000 0.00000 ) *(d) Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree, pretty = 0)
(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? Test error rate is 27.8%
OJ.tree.pred = predict(OJ.tree, newdata = OJ.test, type = "class")
table(OJ.tree.pred,OJ.test$Purchase)##
## OJ.tree.pred CH MM
## CH 404 134
## MM 108 224(108+134)/870## [1] 0.2781609(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
OJ.cv= cv.tree(OJ.tree,FUN = prune.misclass)
OJ.cv## $size
## [1] 15 14 9 6 4 2 1
##
## $dev
## [1] 43 41 37 41 44 43 62
##
## $k
## [1] -Inf 0.0 0.2 1.0 2.5 3.0 25.0
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJ.cv$size, OJ.cv$dev, type = "b", xlab = "Tree Size", ylab = "cv classification error rate")
(h) Which tree size corresponds to the lowest cross-validated
classi- fication error rate? Node six has the lowest
cross-validated classifcation error rate. (i) Produce a pruned
tree corresponding to the optimal tree size obtained using
cross-validation. If cross-validation does not lead to selection of a
pruned tree, then create a pruned tree with five terminal
nodes.
prune.OJ= prune.tree(OJ.tree,best = 7)(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(OJ.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "StoreID" "DiscMM"
## [5] "PriceCH" "SalePriceMM" "WeekofPurchase"
## Number of terminal nodes: 15
## Residual mean deviance: 0.3951 = 73.09 / 185
## Misclassification error rate: 0.095 = 19 / 200summary(prune.OJ)##
## Classification tree:
## snip.tree(tree = OJ.tree, nodes = c(21L, 3L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "StoreID" "PriceCH" "SalePriceMM"
## Number of terminal nodes: 8
## Residual mean deviance: 0.5445 = 104.5 / 192
## Misclassification error rate: 0.11 = 22 / 200(k) Compare the test error rates between the pruned and unpruned trees. Which is higher? The pruned tree has a higher test error rate of 0.4137255
OJ.tree.pred = predict(OJ.tree, newdata = OJ.test, type = "class")
table(OJ.tree.pred, OJ.test$Purchase)##
## OJ.tree.pred CH MM
## CH 404 134
## MM 108 224(108+134)/870## [1] 0.2781609prune.OJ.pred= predict(prune.OJ, newdata = OJ.test, type = "class")
table(prune.OJ.pred, OJ.test$Purchase)##
## prune.OJ.pred CH MM
## CH 404 103
## MM 108 255(108+103)/510## [1] 0.4137255