HW 7 STA 4143
Linda Davila
4/13/2023
library(ISLR)
library(tree)
library(rpart)
library(caret)## Loading required package: ggplot2## Loading required package: latticeProblem 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.
pm1 = seq(0.0, 1.0, .01)
pm2 = 1 - pm1
classe = 1- pmax(pm1, 1 - pm1)
entropy = - pm1 * log(pm1) - pm2* log(pm2)
gini = pm1 * (1 - pm1) + pm2 *(1- pm2)
plot(pm1, entropy, typ = "l", xlab = "pm1", ylab = "Value")
lines(pm1, gini, col = "green")
lines(pm1, classe, col = "blue")
## 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)
sample = sample(1:nrow(Carseats), nrow(Carseats)/2)
train = Carseats[sample,]
test = Carseats[-sample,](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. Only 6 variables were used in constructing the tree with a total of 18 nodes.
set.seed(1)
tree.c = tree(Sales ~ ., data = train)
plot(tree.c)
text(tree.c, pretty = 0)
pred = predict(tree.c, test)
mean((test$Sales - pred) ^ 2)## [1] 4.922039summary(tree.c)##
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Advertising" "CompPrice"
## [6] "US"
## Number of terminal nodes: 18
## Residual mean deviance: 2.167 = 394.3 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.88200 -0.88200 -0.08712 0.00000 0.89590 4.09900(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE? The optimal level of tree complexity is 17. It improved the MSE a little bit, bringing it down to 4.827162.
set.seed(1)
cv.tree = cv.tree(tree.c)
plot(cv.tree$size, cv.tree$dev, type = "b")
summary(cv.tree)## Length Class Mode
## size 17 -none- numeric
## dev 17 -none- numeric
## k 17 -none- numeric
## method 1 -none- characterprune.c = prune.tree(tree.c, best = 17)
plot(prune.c)
text(prune.c, pretty = 0)
prune.pred = predict(prune.c, test)
mean((test$Sales - prune.pred) ^ 2)## [1] 4.827162summary(prune.c)##
## Regression tree:
## snip.tree(tree = tree.c, nodes = 21L)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Advertising" "CompPrice"
## [6] "US"
## Number of terminal nodes: 17
## Residual mean deviance: 2.248 = 411.3 / 183
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.29900 -0.88200 -0.06379 0.00000 0.89590 4.09900(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. The test MSE is 2.605253. The most important variables are Price and Shelveloc, the have the highest %IncMSE and IncNodePurity
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':
##
## marginset.seed(1)
bag.c = randomForest(Sales ~., data = train , mtry = 10, importance = TRUE )
bag.c##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 10, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.889221
## % Var explained: 63.26importance(bag.c)## %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 18.073863bag.cp = predict(bag.c, newdata = test)
bag.mse = mean((bag.cp - test$Sales)^2)
bag.mse## [1] 2.605253varImpPlot(bag.c)
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained. The test MSE is 2.90282. The most important variables are again Price and Shelveloc.
set.seed(1)
random.c = randomForest(Sales ~., data = train , mtry = 5, ntree = 30, importance = TRUE )
random.c##
## Call:
## randomForest(formula = Sales ~ ., data = train, mtry = 5, ntree = 30, importance = TRUE)
## Type of random forest: regression
## Number of trees: 30
## No. of variables tried at each split: 5
##
## Mean of squared residuals: 3.550771
## % Var explained: 54.84random.f = predict(random.c, newdata = test)
random.mse = mean((random.f - test$Sales)^2)
random.mse## [1] 2.90282importance(random.c)## %IncMSE IncNodePurity
## CompPrice 4.2908277 143.26934
## Income 1.2067338 106.71853
## Advertising 2.8235763 85.25177
## Population -1.7883987 90.00200
## Price 11.0694587 446.45794
## ShelveLoc 9.5648921 326.38912
## Age 4.5656248 196.56374
## Education 0.3828607 54.48205
## Urban -1.0498078 13.97890
## US 1.6095756 28.77391(f) Now analyze the data using BART, and report your results. The MSE using BART is 0.1603478. The lowest MSE obtained
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':
##
## clusterset.seed(1)
bart.c = gbart(train, train$Sales, x.test = test)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 15, 200
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 10.360000, 1.000000
## xp1,xp[np*p]: 11.220000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 100 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,4.01382e-30,7.57815
## *****sigma: 0.000000
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,15,0
## *****printevery: 100
##
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 4s
## trcnt,tecnt: 1000,1000bart.cp = bart.c$yhat.test.mean
mean((test$Sales - bart.cp)^2)## [1] 0.1603478Problem 9
This problem involves the OJ data set which is part of the ISLR2 package.
library(ISLR2)##
## Attaching package: 'ISLR2'## The following objects are masked from 'package:ISLR':
##
## Auto, Creditattach(OJ)(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train.ojs=sample(nrow(OJ), 800)
train.oj=OJ[train.ojs,]
test.oj=OJ[-train.ojs, ](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? The error rate is 0.1588 and there are 9 nodes in the tree.
set.seed(1)
tree.oj = tree(Purchase ~ ., data = train.oj)
plot(tree.oj)
text(tree.oj, pretty = 0)
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(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. Number 8 customer brand loyalty for Citrus hill is split at less than .035 with 140.5 observations and a yprob of ( 0.01695 0.98305 )
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 ) *(d) Create a plot of the tree, and interpret the results. There are 16 terminal nodes
plot(tree.oj)
text(tree.oj, 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? The test error rate is
0.1703704.
set.seed(1)
oj.pred = predict(tree.oj, test.oj, type = "class")
table(test.oj$Purchase, oj.pred)## oj.pred
## CH MM
## CH 160 8
## MM 38 64mean(oj.pred!=test.oj$Purchase)## [1] 0.1703704confusionMatrix(test.oj$Purchase, oj.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
## (f) 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)
summary(cv.oj)## Length Class Mode
## size 6 -none- numeric
## dev 6 -none- numeric
## k 6 -none- numeric
## method 1 -none- characternames(cv.oj)## [1] "size" "dev" "k" "method"(g) 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)
(h) Which tree size corresponds to the lowest cross-validated
classification error rate? It looks like tree size four has the
lowest but 8 is very close (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.
set.seed(1)
p.oj=prune.tree(tree.oj, best=4)
summary(p.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = 4:6)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 4
## Residual mean deviance: 0.8678 = 690.7 / 796
## Misclassification error rate: 0.205 = 164 / 800set.seed(1)
p.oj=prune.tree(tree.oj, best=5)
summary(p.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(4L, 12L, 5L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "ListPriceDiff"
## Number of terminal nodes: 5
## Residual mean deviance: 0.8239 = 655 / 795
## Misclassification error rate: 0.205 = 164 / 800(j) Compare the training error rates between the pruned and unpruned trees. Which is higher? The training error rate for the pruned trees is higher than the unpruned trees.
set.seed(1)
unpruned.p = predict(tree.oj, newdata = train.oj, type="class")
table(train.oj$Purchase, unpruned.p)## unpruned.p
## CH MM
## CH 450 35
## MM 92 223mean(train.oj$Purchase!=unpruned.p)## [1] 0.15875pruned.p = predict(p.oj, newdata = train.oj, type="class")
table(train.oj$Purchase, pruned.p)## pruned.p
## CH MM
## CH 356 129
## MM 40 275mean(train.oj$Purchase!=pruned.p)## [1] 0.21125(k) Compare the test error rates between the pruned and unpruned trees. Which is higher? The test error rate for the pruned tree is 0.1925926 wich is higher than the unpruned tree at 0.1703704.
te.pred.prune = predict(p.oj, test.oj, type='class')
table(test.oj$Purchase, te.pred.prune)## te.pred.prune
## CH MM
## CH 135 33
## MM 19 83mean(test.oj$Purchase!=te.pred.prune)## [1] 0.1925926