home work 7
Karim I
2022-11-07
R Markdown
- 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, 0.001)
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("red", "black", "navy"))
- 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.
library(ISLR2)
library(tree)
attach(Carseats)- Split the data set into a training set and a test set.
set.seed(1)
train=sample(nrow(Carseats),nrow(Carseats)/2)
test=Carseats[-train,"Sales"]- 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=Carseats,subset=train)
summary.carseats=summary(tree.carseats)
summary.carseats##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats, subset = 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 4.09900plot(tree.carseats)
text(tree.carseats,
pretty = 0,
cex = 0.5)
tree.pred=predict(tree.carseats,newdata=Carseats[-train,])
mean((tree.pred-test)^2)## [1] 4.922039MSE Test 4.922039
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats=cv.tree(tree.carseats,FUN=prune.tree)
tree.min=which.min(cv.carseats$dev)
best.node.no=cv.carseats$size[tree.min]prune.carseats=prune.tree(tree.carseats,best=best.node.no)
prune.pred=predict(prune.carseats, Carseats[-train,])
mean((prune.pred-test)^2)## [1] 4.922039plot(cv.carseats$size, cv.carseats$dev, type = "b")
points(best.node.no, cv.carseats$dev[tree.min], col = "navy", cex = 2, pch = 20)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
Using cross-validation to select the appropriate amount of tree complexity and trimming the tree did not appear to increase the Test MSE and generated the same result.
- 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)## randomForest 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
bag.carseats=randomForest(Sales~.,data=Carseats,mtry=10,subset=train,importance=T)
bag.carseats##
## Call:
## randomForest(formula = Sales ~ ., data = Carseats, mtry = 10, importance = T, subset = train)
## 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: 63.26yhat.bag=predict(bag.carseats,newdata=Carseats[-train,])
mean((yhat.bag-test)^2)## [1] 2.605253importance(bag.carseats)## %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 18.073863varImpPlot(bag.carseats)
The result produced after conducting Test MSE of 2.605 which is nearly
half of what the Regression Trees presented. The most important
variables being Price and ShelveLoc.
- 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(1)
rf5.carseats=randomForest(Sales~.,data=Carseats,mtry=5,subset=train,importance=T)
rf3.carseats=randomForest(Sales~.,data=Carseats,mtry=3,subset=train,importance=T)
rf5.carseats##
## Call:
## randomForest(formula = Sales ~ ., data = Carseats, mtry = 5, importance = T, subset = train)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 5
##
## Mean of squared residuals: 3.060785
## % Var explained: 61.08rf3.carseats##
## Call:
## randomForest(formula = Sales ~ ., data = Carseats, mtry = 3, importance = T, subset = train)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 3
##
## Mean of squared residuals: 3.387281
## % Var explained: 56.92yhat5.rf=predict(rf5.carseats,newdata=Carseats[-train,])
yhat3.rf=predict(rf3.carseats,newdata=Carseats[-train,])
mean((yhat5.rf-test)^2)## [1] 2.714168mean((yhat3.rf-test)^2)## [1] 3.009147par(mfrow=c(2,2))
varImpPlot(rf5.carseats)
varImpPlot(rf3.carseats)
Test MSE for the Random Forest method creates 2.71 for yhat 5 and 3.00
for yhat 3.009
- Now analyze the data using BART, and report your results.
library(BART)## Loading required package: nlme## Loading required package: nnet## Loading required package: survivalTest MSE was 1.469
x = Carseats[, 2:11]
y = Carseats[, "Sales"]
xtrain = x[train,]
ytrain = y[train]
xtest = x[test, ]
ytest = y[test]
set.seed(1)
bartfit = gbart(xtrain,
ytrain,
x.test = xtest)## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 196
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 107.000000, 1.000000
## xp1,xp[np*p]: 121.000000, 0.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 63 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,0.23074,7.57815
## *****sigma: 1.088371
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****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: 1000,1000yhat.bart = bartfit$yhat.test.mean
mean((ytest - yhat.bart)^2)## [1] 1.46948- This problem involves the OJ data set which is part of the ISLR2 package.
oj = OJ- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train=sample(1:nrow(OJ),800)
oj.train=OJ[train, ]
oj.test=OJ[-train, ]- 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?
oj.tree=tree(Purchase~.,OJ,subset=train)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ, subset = 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- 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.
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 ) *- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree,
pretty = 0,
cex = 0.5)
- 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?
oj.pred=predict(oj.tree,oj.test,type='class')
1-mean(oj.pred==oj.test$Purchase)## [1] 0.1703704Test MSE is .17 or 17%
- 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] 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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size,
oj.cv$dev,
type = "b",
xlab = "Tree Size",
ylab = "CV Error Rate")
- Which tree size corresponds to the lowest cross-validated classification error rate?
tree size 7 has lowest error rate.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned=prune.misclass(oj.tree,best=2)
oj.pruned## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1073.0 CH ( 0.6062 0.3937 )
## 2) LoyalCH < 0.5036 365 441.6 MM ( 0.2932 0.7068 ) *
## 3) LoyalCH > 0.5036 435 337.9 CH ( 0.8690 0.1310 ) *plot(oj.pruned)
text(oj.pruned)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 3:2)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 2
## Residual mean deviance: 0.9768 = 779.5 / 798
## Misclassification error rate: 0.205 = 164 / 800Pruned tree is 0.205 and Tree is .17 so pruned tree is higher
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree,
oj.test,
type="class")
misclass.unpruned = sum(oj.test$Purchase != pred.unpruned)
misclass.unpruned / length(pred.unpruned)## [1] 0.1703704pred.pruned = predict(oj.pruned,
oj.test,
type="class")
misclass.pruned = sum(oj.test$Purchase != pred.pruned)
misclass.pruned / length(pred.pruned)## [1] 0.1851852pruned is higher by .015