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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 <- p * (1 - p) * 2
c.error <- 1 - pmax(p, 1 - p)
c.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
plot(c.error, ylim = c(0, 1), col = "#FF1493", ylab = "C.Error, C.Entropy, Gini.Index", xlab = "P")
lines(gini.index, lwd = 2, col = "#9932CC")
lines(c.entropy, lwd = 3, col = "#1E90FF")
legend('top', inset = .01, legend = c('Gini index', 'Classification error', 'Cross entropy'), col = c("#9932CC", "#FF1493", "#1E90FF"), pch = c(16,16,16))

  1. 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.
  1. Split the data set into a training set and a test set.
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
data <- Carseats
sample_size <- floor(0.70* nrow(Carseats))
set.seed(1)
train_index <- sample(seq_len(nrow(Carseats)), size = sample_size)
train <- Carseats[train_index, ]
test <- Carseats[-train_index, ]
  1. Fit a regression tree to the training set. Plot the tree, and interam pret the results. What test MSE do you obtain?
library(tree)
## Warning: package 'tree' was built under R version 4.0.5
set.seed(1)
tree.fit <- tree(Sales ~ ., data = train)
summary(tree.fit)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Income"      "CompPrice"  
## [6] "Advertising"
## Number of terminal nodes:  18 
## Residual mean deviance:  2.409 = 631.1 / 262 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -4.77800 -0.96100 -0.08865  0.00000  1.01800  4.14100
set.seed(1)
tree.pred <- predict(tree.fit, newdata = test)
mean((tree.pred - test$Sales)^2)
## [1] 4.208383
plot(tree.fit)
text(tree.fit, pretty = 0, cex=.8, font = 0.5)

  1. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
tree.cv = cv.tree(tree.fit)
plot(tree.cv$size, tree.cv$dev, type = "b")
tree.min <- which.min(tree.cv$dev)
points(tree.min, tree.cv$dev[tree.min], cex = 2, col = "red", pch = 20)

set.seed
## function (seed, kind = NULL, normal.kind = NULL, sample.kind = NULL) 
## {
##     kinds <- c("Wichmann-Hill", "Marsaglia-Multicarry", "Super-Duper", 
##         "Mersenne-Twister", "Knuth-TAOCP", "user-supplied", "Knuth-TAOCP-2002", 
##         "L'Ecuyer-CMRG", "default")
##     n.kinds <- c("Buggy Kinderman-Ramage", "Ahrens-Dieter", "Box-Muller", 
##         "user-supplied", "Inversion", "Kinderman-Ramage", "default")
##     s.kinds <- c("Rounding", "Rejection", "default")
##     if (length(kind)) {
##         if (!is.character(kind) || length(kind) > 1L) 
##             stop("'kind' must be a character string of length 1 (RNG to be used).")
##         if (is.na(i.knd <- pmatch(kind, kinds) - 1L)) 
##             stop(gettextf("'%s' is not a valid abbreviation of an RNG", 
##                 kind), domain = NA)
##         if (i.knd == length(kinds) - 1L) 
##             i.knd <- -1L
##     }
##     else i.knd <- NULL
##     if (!is.null(normal.kind)) {
##         if (!is.character(normal.kind) || length(normal.kind) != 
##             1L) 
##             stop("'normal.kind' must be a character string of length 1")
##         normal.kind <- pmatch(normal.kind, n.kinds) - 1L
##         if (is.na(normal.kind)) 
##             stop(gettextf("'%s' is not a valid choice", normal.kind), 
##                 domain = NA)
##         if (normal.kind == 0L) 
##             stop("buggy version of Kinderman-Ramage generator is not allowed", 
##                 domain = NA)
##         if (normal.kind == length(n.kinds) - 1L) 
##             normal.kind <- -1L
##     }
##     if (!is.null(sample.kind)) {
##         if (!is.character(sample.kind) || length(sample.kind) != 
##             1L) 
##             stop("'sample.kind' must be a character string of length 1")
##         sample.kind <- pmatch(sample.kind, s.kinds) - 1L
##         if (is.na(sample.kind)) 
##             stop(gettextf("'%s' is not a valid choice", sample.kind), 
##                 domain = NA)
##         if (sample.kind == 0L) 
##             warning("non-uniform 'Rounding' sampler used", domain = NA)
##         if (sample.kind == length(s.kinds) - 1L) 
##             sample.kind <- -1L
##     }
##     .Internal(set.seed(seed, i.knd, normal.kind, sample.kind))
## }
## <bytecode: 0x0000000013e680a0>
## <environment: namespace:base>
pruned.tree <- prune.tree(tree.fit, best = 11)
pruned.pred <- predict(pruned.tree, newdata = test)
mean((pruned.pred - test$Sales)^2)
## [1] 4.691032

The error rate did did not improved.

  1. 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 4.0.5
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
set.seed(1)
bag.car <- randomForest(Sales ~ ., data = train, mtry = 10, importance = TRUE )
bag.pred <- predict(bag.car, newdata = test)
mean((bag.pred - test$Sales)^2)
## [1] 2.573252
importance(bag.car)
##               %IncMSE IncNodePurity
## CompPrice   35.324343     230.55353
## Income       7.923790     118.79213
## Advertising 19.736816     155.03099
## Population  -3.479681      63.11975
## Price       70.613500     673.11982
## ShelveLoc   68.147204     637.69500
## Age         20.964215     228.90345
## Education    4.705263      63.13124
## Urban       -2.098091      11.66651
## US           1.389570      11.15329

Test error rate is 2.57, the variables of importance are = Price, ShelveLoc, and CompPrice.

  1. 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)
random.car5 = randomForest(Sales ~ ., data = train, mtry = 5, ntree = 500, importance = T)
random.pred.car5 = predict(random.car5, test)
mean((test$Sales - random.pred.car5)^2)
## [1] 2.60463
random.car6 = randomForest(Sales ~ ., data = train, mtry = 6, ntree = 500, importance = T)
random.pred.car6 = predict(random.car6, test)
mean((test$Sales - random.pred.car6)^2)
## [1] 2.551024
random.car7 = randomForest(Sales ~ ., data = train, mtry = 7, ntree = 500, importance = T)
random.pred.car7 = predict(random.car7, test)
mean((test$Sales - random.pred.car7)^2)
## [1] 2.565275
random.car8 = randomForest(Sales ~ ., data = train, mtry = 8, ntree = 500, importance = T)
random.pred.car8 = predict(random.car8, test)
mean((test$Sales - random.pred.car8)^2)
## [1] 2.567522
random.car9 = randomForest(Sales ~ ., data = train, mtry = 9, ntree = 500, importance = T)
random.pred.car9 = predict(random.car9, test)
mean((test$Sales - random.pred.car9)^2)
## [1] 2.609295
random.car10 = randomForest(Sales ~ ., data = train, mtry = 10, ntree = 500, importance = T)
random.pred.car10 = predict(random.car10, test)
mean((test$Sales - random.pred.car10)^2)
## [1] 2.598445
random.car1 = randomForest(Sales ~ ., data = train, mtry = 1, ntree = 500, importance = T)
random.pred.car1 = predict(random.car1, test)
mean((test$Sales - random.pred.car1)^2)
## [1] 4.557115
random.car2 = randomForest(Sales ~ ., data = train, mtry = 2, ntree = 500, importance = T)
random.pred.car2 = predict(random.car2, test)
mean((test$Sales - random.pred.car2)^2)
## [1] 3.315914
random.car3 = randomForest(Sales ~ ., data = train, mtry = 3, ntree = 500, importance = T)
random.pred.car3 = predict(random.car3, test)
mean((test$Sales - random.pred.car3)^2)
## [1] 2.894646
random.car4 = randomForest(Sales ~ ., data = train, mtry = 4, ntree = 500, importance = T)
random.pred.car4 = predict(random.car4, test)
mean((test$Sales - random.pred.car4)^2)
## [1] 2.672202

The mtry equal to 6 has the least error rate in compared to the others.

importance(random.car6)
##               %IncMSE IncNodePurity
## CompPrice   27.143517     219.92968
## Income       4.921334     130.21447
## Advertising 17.294698     155.03223
## Population  -1.667603      85.59735
## Price       60.661862     646.00362
## ShelveLoc   63.328288     590.63510
## Age         22.179583     252.02070
## Education    1.416713      74.92792
## Urban       -1.338920      13.06650
## US           4.552794      20.63057

The top 3 important variables are = Price, CompPrice and ShelveLoc. Which are the same as previous.

  1. This problem involves the OJ data set which is part of the ISLR package.
  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
data <- OJ
set.seed(1)
sample_oj <- sample(nrow(OJ), 800)
train.oj <- OJ[sample_oj, ]
test.oj <- OJ[-sample_oj, ]
  1. 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?
library(tree)
set.seed(1)
tree.fit.oj <- tree(Purchase ~ ., data = train.oj)
summary(tree.fit.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

Test error rate is 0.1588 and our tree has 9 terminal nodes. The variables used = [1] “LoyalCH” “PriceDiff” “SpecialCH” “ListPriceDiff” “PctDiscMM” .

  1. 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.
tree.fit.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 < 0.196197 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 ) *

We are selecting node 25 PctDiscMM, the node splits for then PctDiscMM > 0.196197. There are 17 observations in the leaf with the residual variance of 12.32. The prediction is MM with 0.11765 (11.765%) taking MM and 0.88235 (88.235%) taking CH.

  1. Create a plot of the tree, and interpret the results.
plot(tree.fit.oj)
text(tree.fit.oj, pretty=0, , cex=.8, font = 0.5)

  1. 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?
set.seed(1)
tree.pred.oj <- predict(tree.fit.oj, test.oj, type = 'class')
table(test.oj$Purchase, tree.pred.oj)
##     tree.pred.oj
##       CH  MM
##   CH 160   8
##   MM  38  64
caret::confusionMatrix(tree.pred.oj, test.oj$Purchase)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160  38
##         MM   8  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.6222          
##     P-Value [Acc > NIR] : 8.077e-14       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.9524          
##             Specificity : 0.6275          
##          Pos Pred Value : 0.8081          
##          Neg Pred Value : 0.8889          
##              Prevalence : 0.6222          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.7333          
##       Balanced Accuracy : 0.7899          
##                                           
##        'Positive' Class : CH              
##

The error rate is 17.037% = (8+38)/(160+64+8+38)

  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
oj.tree.cv = cv.tree(tree.fit.oj, FUN = prune.misclass)
oj.tree.cv
## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 145 145 146 146 167 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.tree.cv$size, oj.tree.cv$dev, type="b")

  1. Which tree size corresponds to the lowest cross-validated classification error rate?

8 and 9 have the lowest cross-validation error rate with < 150 cross validation errors.

  1. 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)
pruned.fit.oj = prune.misclass(tree.fit.oj, best=4)
summary(pruned.fit.oj)
## 
## Classification tree:
## snip.tree(tree = tree.fit.oj, nodes = c(4L, 10L, 3L))
## Variables actually used in tree construction:
## [1] "LoyalCH"   "PriceDiff"
## Number of terminal nodes:  4 
## Residual mean deviance:  0.8922 = 710.2 / 796 
## Misclassification error rate: 0.1788 = 143 / 800
  1. Compare the training error rates between the pruned and unpruned trees. Which is higher?

The pruned tree error rate is 17.88% which is higher to our 15.88% unpruned tree error rate.

  1. Compare the test error rates between the pruned and unpruned trees. Which is higher?
set.seed(1)
pruned.pred.oj <- predict(pruned.fit.oj, test.oj, type = 'class')
caret::confusionMatrix(pruned.pred.oj, test.oj$Purchase)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 161  41
##         MM   7  61
##                                           
##                Accuracy : 0.8222          
##                  95% CI : (0.7713, 0.8659)
##     No Information Rate : 0.6222          
##     P-Value [Acc > NIR] : 6.769e-13       
##                                           
##                   Kappa : 0.5954          
##                                           
##  Mcnemar's Test P-Value : 1.906e-06       
##                                           
##             Sensitivity : 0.9583          
##             Specificity : 0.5980          
##          Pos Pred Value : 0.7970          
##          Neg Pred Value : 0.8971          
##              Prevalence : 0.6222          
##          Detection Rate : 0.5963          
##    Detection Prevalence : 0.7481          
##       Balanced Accuracy : 0.7782          
##                                           
##        'Positive' Class : CH              
##

So the pruned tree has a test error rate of 17.78% (41+7)/(41+7+161+61) which is it higher than our un-pruned tree test error rate that is 17.037% = (8+38)/(160+64+8+38).