title: “assigment 7” author: “oscar cancino” date: “4/22/2022” output: html_document
- 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 xaxis 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, pˆm1 = 1 − pˆm2. 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 = "white")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("green" , "yellow", "red"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("green" , "yellow", "red"), pch=c(15,17,19))
- 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.
- Split the data set into a training set and a test set.
library(ISLR)
set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats) / 2)
Car.train = Carseats[train, ]
Car.test = Carseats[-train,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)## Warning: package 'tree' was built under R version 4.1.3reg.tree = tree(Sales~.,data = Carseats, subset=train)
reg.tree = tree(Sales~.,data = Car.train)
summary(reg.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = Car.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(reg.tree)
text(reg.tree ,pretty =0)
yhat = predict(reg.tree,newdata = Car.test)
mean((yhat - Car.test$Sales)^2)## [1] 4.922039the test mse comes out to 4.92
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.car = cv.tree(reg.tree)
plot(cv.car$size, cv.car$dev, type = "b")
prune.car = prune.tree(reg.tree, best = 8)
plot(prune.car)
text(prune.car,pretty=0)
yhat=predict(prune.car, newdata= Car.test)
mean((yhat-Car.test$Sales)^2)## [1] 5.113254- 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 4.1.3## randomForest 4.7-1## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
bag.car = randomForest(Sales~.,data=Car.train,mtry = 10, importance = TRUE)
yhat.bag = predict(bag.car,newdata=Car.test)
mean((yhat.bag-Car.test$Sales)^2)## [1] 2.605253importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863varImpPlot(bag.car)
the most important varianles is the price at each site and the quality of the shelving location. the rest mse comes out to 2.60, which is less than above
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
library(randomForest)
set.seed(1)
rf.car = randomForest(Sales~.,data=Car.train,mtry = 3, importance = TRUE)
yhat.rf = predict(rf.car,newdata=Car.test)
mean((yhat.rf-Car.test$Sales)^2)## [1] 2.960559the MSe comes out to 2.960559 which indicates that random forest does not provide an improvemnt over bagging
- This problem involves the OJ data set which is part of the ISLR package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
set.seed(1)
train = sample(dim(OJ)[1],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~., 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 / 800the fitted tree has 9 terminal nodes with a training error rate of 0.1588
- 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 < 0.196197 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196197 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 ) *i picked node 9 which is a terminal node. the split criterion is loyalch >.0356415 the number of observations is 118 with a deviance of 116.40 amd an overall prediction for the branch mm.
- Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree,pretty=TRUE)
the most important indicator of “purchase” appears to be “loyalCH” since the first branch diffrent the intesity of customer brand loyalty to CH. in fact, the top three nodes contain “loyalCH”
- 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.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] 0.774074177% of the test observation are correctly classified so the test error rate is 23%
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.OJ$size,cv.OJ$dev,type='b', xlab = "Tree size", ylab = "Deviance")
- Which tree size corresponds to the lowest cross-validated classification error rate?
we might see that the 4-node tree is the smallest tree with the lowest classification error rate
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ = prune.misclass(OJ.tree, best=5)
plot(prune.OJ)
text(prune.OJ,pretty=0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
tree.pred = predict(prune.OJ, newdata = OJ.test, type = "class")
table(tree.pred,OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 36
## MM 8 66(147+62)/270## [1] 0.7740741in this case the prunning has the same test error, but it produced a way more interpretable tree