1. 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.

ANSWER 3a:

require(ISLR)
## Loading required package: ISLR
require(dplyr)
## Loading required package: dplyr
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
require(caret)
## Loading required package: caret
## Loading required package: lattice
## Loading required package: ggplot2
require(ggplot2)
require(tidyr)
## Loading required package: tidyr
prop_class_1 <- seq(0, 1, 0.001)
prop_class_2 <- 1 - prop_class_1


classification_error <- 1 - pmax(prop_class_1, prop_class_2)

gini <- prop_class_1*(1-prop_class_1) + prop_class_2*(1-prop_class_2)

entropy <- -prop_class_1*log(prop_class_1) - prop_class_2*log(prop_class_2)


data.frame(prop_class_1, prop_class_2, classification_error, gini, entropy) %>%
  pivot_longer(cols = c(classification_error, gini, entropy), names_to = "comb") %>%
  ggplot(aes(x = prop_class_1, y = value, col = factor(comb)))+
  geom_line(size = 1) + 
  scale_y_continuous(breaks = seq(0, 1, 0.1), minor_breaks = NULL) + 
  scale_color_hue(labels = c("Classification Error", "Entropy", "Gini")) +
  labs(col = "comb", 
       y = "Value", 
       x = "Proportion (of class '1')") 

  #+ 
  #scale_colour_manual(values = c("Classification Error" = "orange",  "Entropy" = "blue", "Gini" = "red"))
  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.

ANSWER 8a:

Data is split as shown below for train and test sets:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

set.seed(2, sample.kind = "Rounding") 

train_index <- sample(1:nrow(Carseats), nrow(Carseats) / 2)

train <- Carseats[train_index, ]
test <- Carseats[-train_index, ]
  1. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

ANSWER 8b:

ShelveLoc and Price appear to be the most important factors in predicting car seat sales as they appear at the top of the tree. There is a total of 17 terminal nodes and the test MSE result = 4.844991 as shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tree)
## Loading required package: tree
tree_model <- tree(Sales ~ ., train)

plot(tree_model)
text(tree_model, cex = .50)

summary(tree_model)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Income"      "CompPrice"  
## [6] "Population"  "Advertising"
## Number of terminal nodes:  17 
## Residual mean deviance:  2.341 = 428.4 / 183 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -3.76700 -1.00900 -0.01558  0.00000  0.94900  3.58600
test_pred <- predict(tree_model, test)
mean((test_pred - test$Sales)^2)
## [1] 4.844991
  1. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?

ANSWER 8c:

The model selected with the lowest cross-validation error is the one with 17 Terminal Nodes as shown below. The optimal tree is the original fully grown tree without pruning, therefore, the test MSE is the same result from part 8b.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

set.seed(3)

cv_tree_model <- cv.tree(tree_model, K = 10)

data.frame(n_leaves = cv_tree_model$size,
           CV_RSS = cv_tree_model$dev) %>%
  mutate(min_CV_RSS = as.numeric(min(CV_RSS) == CV_RSS)) %>%
  ggplot(aes(x = n_leaves, y = CV_RSS)) +
  geom_line(col = "blue") +
  geom_point(size = 2, aes(col = factor(min_CV_RSS))) +
  scale_x_continuous(breaks = seq(1, 17, 2)) +
  scale_y_continuous(labels = scales::comma_format()) +
  scale_color_manual(values = c("deepskyblue3", "orange")) +
  theme(legend.position = "none") +
  labs(title = "Carseats Dataset - Regression Tree",
       subtitle = "Selecting the complexity parameter with cross-validation",
       x = "Terminal Nodes",
       y = "CV RSS")

pruned_tree_model <- prune.tree(tree_model, best = 17)

test_pred <- predict(pruned_tree_model, test)
mean((test_pred - test$Sales)^2)
## [1] 4.844991
  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.

ANSWER 8d:

By using the randomForest() function we can implement bagged trees as shown below. The test MSE obtained = 2.368674. By using the importance() function we find that the important variables are Price and Shelveloc for bagged trees.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(randomForest)
## Loading required package: randomForest
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
## 
## Attaching package: 'randomForest'
## The following object is masked from 'package:ggplot2':
## 
##     margin
## The following object is masked from 'package:dplyr':
## 
##     combine
require(tibble)
## Loading required package: tibble
set.seed(1)

bagged_trees_mdl <- randomForest(y = train$Sales, x = train[ ,-1],  mtry = ncol(train) - 1, importance = T) 

test_pred <- predict(bagged_trees_mdl, test)
mean((test_pred - test$Sales)^2)
## [1] 2.368674
importance(bagged_trees_mdl) %>% as.data.frame() %>% rownames_to_column("varname") %>% arrange(desc(IncNodePurity))
##        varname     %IncMSE IncNodePurity
## 1        Price 55.76266629    493.804969
## 2    ShelveLoc 53.87451311    446.816951
## 3    CompPrice 25.47984338    173.982449
## 4          Age 12.07366106    117.502364
## 5  Advertising 13.97464644     96.929928
## 6       Income  1.72616791     71.465227
## 7   Population  1.01449985     68.297498
## 8    Education  0.08382003     37.513944
## 9        Urban -3.06299457      6.909530
## 10          US  0.14346468      5.985091
  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.

ANSWER 8e:

Reducing the mtry increases the test MSE and increasing the mtry rapidly reduces the test error as shown in the plot below. The best random forest model uses an mtry = 10. The test MSE = 2.368674. By using the importance() function we find that the important variables are Price and Shelveloc for bagged trees as shown below.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

test_MSE <- c()

j <- 1
for (Mtry in 1:10) {
  set.seed(1)
  
  rf_temp <- randomForest(y = train$Sales, 
                          x = train[ ,-1], 
                          mtry = Mtry, 
                          importance = T)
  
  test_pred <- predict(rf_temp, test)
  
  test_MSE[j] <- mean((test_pred - test$Sales)^2)
  
  j <- j + 1
}


data.frame(mtry = 1:10, test_MSE = test_MSE) %>%
  mutate(min_test_MSE = as.numeric(min(test_MSE) == test_MSE)) %>%
  ggplot(aes(x = mtry, y = test_MSE)) +
  geom_line(col = "blue") +
  geom_point(size = 2, aes(col = factor(min_test_MSE))) +
  scale_x_continuous(breaks = seq(1, 10), minor_breaks = NULL) +
  scale_color_manual(values = c("deepskyblue3", "orange")) +
  theme(legend.position = "none") +
  labs(title = "Carseats Dataset - Random Forests",
       subtitle = "Selecting 'mtry' using the test MSE",
       x = "mtry",
       y = "Test MSE")

tail(test_MSE, 1)
## [1] 2.368674
importance(rf_temp) %>%as.data.frame() %>%rownames_to_column("varname") %>% arrange(desc(IncNodePurity))
##        varname     %IncMSE IncNodePurity
## 1        Price 55.76266629    493.804969
## 2    ShelveLoc 53.87451311    446.816951
## 3    CompPrice 25.47984338    173.982449
## 4          Age 12.07366106    117.502364
## 5  Advertising 13.97464644     96.929928
## 6       Income  1.72616791     71.465227
## 7   Population  1.01449985     68.297498
## 8    Education  0.08382003     37.513944
## 9        Urban -3.06299457      6.909530
## 10          US  0.14346468      5.985091
  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.

ANSWER 9a:

train and test sets are split below as shown:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

#glimpse(OJ) # OJ dataset shows 1,070 rows

set.seed(5)

train_index <- sample(1:nrow(OJ), 800)

train <- OJ[train_index, ]
test <- OJ[-train_index, ]
  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?

ANSWER 9b:

The training error rate = 18.38%. There is 7 terminal nodes in the classification tree as shown below. There are only 3 predictors used in the tree construction(splits): LoyalCH, PriceDiff and DiscCH.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

tree_mdl <- tree(Purchase ~ ., train)
summary(tree_mdl)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = train)
## Variables actually used in tree construction:
## [1] "LoyalCH"   "PriceDiff" "DiscCH"   
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7786 = 617.4 / 793 
## Misclassification error rate: 0.1838 = 147 / 800
  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.

ANSWER 9c:

Choosing Node #12: which is also a terminal node. From the root node, the following splits produce the terminal Node 12:

Node #12 is the subset of purchases where 0.5036 < LoyalCH < 0.705699 and PriceDiff < 0.25. The overall prediction isCH and the node shows 52.94% CH vs 47.05%MM.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

tree_mdl
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 800 1064.00 CH ( 0.61750 0.38250 )  
##    2) LoyalCH < 0.5036 354  435.50 MM ( 0.30508 0.69492 )  
##      4) LoyalCH < 0.142213 100   45.39 MM ( 0.06000 0.94000 ) *
##      5) LoyalCH > 0.142213 254  342.20 MM ( 0.40157 0.59843 )  
##       10) PriceDiff < 0.235 136  153.00 MM ( 0.25000 0.75000 ) *
##       11) PriceDiff > 0.235 118  160.80 CH ( 0.57627 0.42373 ) *
##    3) LoyalCH > 0.5036 446  352.30 CH ( 0.86547 0.13453 )  
##      6) LoyalCH < 0.705699 154  189.50 CH ( 0.69481 0.30519 )  
##       12) PriceDiff < 0.25 85  117.50 CH ( 0.52941 0.47059 )  
##         24) DiscCH < 0.15 77  106.60 MM ( 0.48052 0.51948 ) *
##         25) DiscCH > 0.15 8    0.00 CH ( 1.00000 0.00000 ) *
##       13) PriceDiff > 0.25 69   45.30 CH ( 0.89855 0.10145 ) *
##      7) LoyalCH > 0.705699 292  106.30 CH ( 0.95548 0.04452 ) *

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.

  1. Create a plot of the tree, and interpret the results.

ANSWER 9d:

LoyalCH is the most important variable, followed by PriceDiff and DiscCH. We can see Node #12is the third terminal node from right to left (its shown where PriceDiff< 0.25). LoyalCH ranges from 0 to 1, the first split takes the less loyal to CH to the left and those more loyal to CH to the right.

Those that scored the lowest (LoyalCH < 0.142213) have been predicted to buy MM.Those that were a bit more loyal to CH (0.142213 < LoyalCH < 0.5036) will buy MMif the price is not to high (Price < 0.235), but if the price is high enough they will end up purchasing CH

Those on the right side of the tree are the most loyal to CH(LoyalCH > 0.705699) and this is also their predicted purchase. Those with lower brand loyalty (0.5036 < LoyalCH < 0.705699) will purchase CHif its cheaper (Price > 0.25) or at a discount(PriceDiff < 0.25 & DiscCH > 0.15). For cases where CH were not cheap enough(PriceDiff < 0.25) and at a discount(DiscCH < 0.15) the predicted purchase is MM.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(tree)

plot(tree_mdl)
text(tree_mdl, cex = 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?

ANSWER 9e:

The test error rate is 0.2444444 as shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

test_pred <- predict(tree_mdl, test, type = "class")
table(test_pred, test_actual = test$Purchase)
##          test_actual
## test_pred  CH  MM
##        CH 125  32
##        MM  34  79
1 - mean(test_pred == test$Purchase)
## [1] 0.2444444
  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.

ANSWER 9f:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

set.seed(2)

cv_tree_model <- cv.tree(tree_mdl, K = 10, FUN = prune.misclass)
cv_tree_model
## $size
## [1] 7 4 2 1
## 
## $dev
## [1] 164 164 177 298
## 
## $k
## [1] -Inf    1    9  138
## 
## $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.

ANSWER 9g:

The cross-validated classification error rate and tree size plot is shown below:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

data.frame(size = cv_tree_model$size, CV_Error = cv_tree_model$dev / nrow(train)) %>% mutate(min_CV_Error = as.numeric(min(CV_Error) == CV_Error)) %>%
  ggplot(aes(x = size, y = CV_Error)) +
  geom_line(col = "blue") +
  geom_point(size = 2, aes(col = factor(min_CV_Error))) +
  scale_x_continuous(breaks = seq(1, 7), minor_breaks = NULL) +
  scale_y_continuous(labels = scales::percent_format()) +
  scale_color_manual(values = c("deepskyblue3", "orange")) +
  theme(legend.position = "none") +
  labs(title = "OJ Dataset - Classification Tree",
       subtitle = "Selecting tree 'size' (# of terminal nodes) using cross-validation",
       x = "Tree Size",
       y = "CV Error")

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

ANSWER 9h:

Using the previous plot from part 9g we can see that trees of sizes 4 and 7 have the lowest (and same) cross validation error.

  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.

ANSWER 9i:

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

pruned_tree_model <- prune.tree(tree_mdl, best = 4)
pruned_tree_model
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
## 1) root 800 1064.00 CH ( 0.61750 0.38250 )  
##   2) LoyalCH < 0.5036 354  435.50 MM ( 0.30508 0.69492 )  
##     4) LoyalCH < 0.142213 100   45.39 MM ( 0.06000 0.94000 ) *
##     5) LoyalCH > 0.142213 254  342.20 MM ( 0.40157 0.59843 ) *
##   3) LoyalCH > 0.5036 446  352.30 CH ( 0.86547 0.13453 )  
##     6) LoyalCH < 0.705699 154  189.50 CH ( 0.69481 0.30519 ) *
##     7) LoyalCH > 0.705699 292  106.30 CH ( 0.95548 0.04452 ) *
  1. Compare the training error rates between the pruned and unpruned trees. Which is higher?

ANSWER 9j:

The training error = 0.21 for the pruned tree and its also higher than the unpruned tree training error as shown below. We expect the training error of a tree to decrease as the number of splits increase.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)
require(gam)
## Loading required package: gam
## Loading required package: splines
## Loading required package: foreach
## Loaded gam 1.20
mean(predict(pruned_tree_model, type = "class") != train$Purchase) 
## [1] 0.21
#training error for pruned tree - 4 terminal nodes

mean(predict(tree_mdl, type = "class") != train$Purchase)  
## [1] 0.18375
#training error for unpruned tree - 7 terminal nodes
  1. Compare the test error rates between the pruned and unpruned trees. Which is higher?

ANSWER 9k:

The Test Error = 0.24 for the unpruned tree is higher than the pruned tree Test Error as shown below.

require(ISLR)
require(dplyr)
require(caret)
require(ggplot2)

mean(predict(pruned_tree_model, type = "class", newdata = test) != test$Purchase)
## [1] 0.1703704
#Test Error for the pruned tree

mean(predict(tree_mdl, type = "class", newdata = test) != test$Purchase)
## [1] 0.2444444
#Test Error for the unpruned tree