Practical Machine Learning - Quiz 3

Question 1

For this quiz we will be using several R packages. R package versions change over time, the right answers have been checked using the following versions of the packages.

AppliedPredictiveModeling: v1.1.6

caret: v6.0.47

ElemStatLearn: v2012.04-0

pgmm: v1.1

rpart: v4.1.8

If you aren’t using these versions of the packages, your answers may not exactly match the right answer, but hopefully should be close.

Load the cell segmentation data from the AppliedPredictiveModeling package using the commands:

library(AppliedPredictiveModeling)
data(segmentationOriginal)
library(caret)

## Loading required package: lattice

## Loading required package: ggplot2

library(rattle)

## Rattle: A free graphical interface for data science with R.
## Version 5.2.0 Copyright (c) 2006-2018 Togaware Pty Ltd.
## Type 'rattle()' to shake, rattle, and roll your data.

1. Subset the data to a training set and testing set based on the Case variable in the data set.

2. Set the seed to 125 and fit a CART model with the rpart method using all predictor variables and default caret settings.

3. In the final model what would be the final model prediction for cases with the following variable values:

1. TotalIntench2 = 23,000; FiberWidthCh1 = 10; PerimStatusCh1=2

2. TotalIntench2 = 50,000; FiberWidthCh1 = 10;VarIntenCh4 = 100

3. TotalIntench2 = 57,000; FiberWidthCh1 = 8;VarIntenCh4 = 100

4. FiberWidthCh1 = 8;VarIntenCh4 = 100; PerimStatusCh1=2

trainSet <- segmentationOriginal[segmentationOriginal$Case =="Train",]
testSet <- segmentationOriginal[segmentationOriginal$Case =="Test",]

set.seed(125)
model_rpart <- train(Class~.,data=trainSet,method="rpart")

fancyRpartPlot(model_rpart$finalModel)

Question 2

If K is small in a K-fold cross validation is the bias in the estimate of out-of-sample (test set) accuracy smaller or bigger? If K is small is the variance in the estimate of out-of-sample (test set) accuracy smaller or bigger. Is K large or small in leave one out cross validation?

Question 3

Load the olive oil data using the commands:

library(pgmm)
data(olive)
olive = olive[,-1]

(NOTE: If you have trouble installing the pgmm package, you can download the -code-olive-/code- dataset here: olive_data.zip. After unzipping the archive, you can load the file using the -code-load()-/code- function in R.)

These data contain information on 572 different Italian olive oils from multiple regions in Italy. Fit a classification tree where Area is the outcome variable. Then predict the value of area for the following data frame using the tree command with all defaults

newdata = as.data.frame(t(colMeans(olive)))

What is the resulting prediction? Is the resulting prediction strange? Why or why not?

str(olive)

## 'data.frame':    572 obs. of  9 variables:
##  $ Area       : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ Palmitic   : num  1075 1088 911 966 1051 ...
##  $ Palmitoleic: num  75 73 54 57 67 49 66 61 60 55 ...
##  $ Stearic    : num  226 224 246 240 259 268 264 235 239 213 ...
##  $ Oleic      : num  7823 7709 8113 7952 7771 ...
##  $ Linoleic   : num  672 781 549 619 672 678 618 734 709 633 ...
##  $ Linolenic  : num  36 31 31 50 50 51 49 39 46 26 ...
##  $ Arachidic  : num  60 61 63 78 80 70 56 64 83 52 ...
##  $ Eicosenoic : num  29 29 29 35 46 44 29 35 33 30 ...

table(olive$Area)

## 
##   1   2   3   4   5   6   7   8   9 
##  25  56 206  36  65  33  50  50  51

olive_rpart <- train(Area~.,data=olive,method="rpart")

## Warning in nominalTrainWorkflow(x = x, y = y, wts = weights, info =
## trainInfo, : There were missing values in resampled performance measures.

fancyRpartPlot(olive_rpart$finalModel)

predict(olive_rpart,newdata=newdata)

##        1 
## 2.783282

Question 4

Load the South Africa Heart Disease Data and create training and test sets with the following code:

library(ElemStatLearn)

data(SAheart)
set.seed(8484)
train = sample(1:dim(SAheart)[1],size=dim(SAheart)[1]/2,replace=F)
trainSA = SAheart[train,]
testSA = SAheart[-train,]

Then set the seed to 13234 and fit a logistic regression model (method=“glm”, be sure to specify family=“binomial”) with Coronary Heart Disease (chd) as the outcome and age at onset, current alcohol consumption, obesity levels, cumulative tabacco, type-A behavior, and low density lipoprotein cholesterol as predictors. Calculate the misclassification rate for your model using this function and a prediction on the “response” scale:

missClass = function(values,prediction){sum(((prediction > 0.5)*1) != values)/length(values)}

What is the misclassification rate on the training set? What is the misclassification rate on the test set?

set.seed(13234)

regModel <- train(chd~age+alcohol+obesity+tobacco+typea+ldl,data=trainSA,method="glm",family="binomial")

## Warning in train.default(x, y, weights = w, ...): You are trying to do
## regression and your outcome only has two possible values Are you trying to
## do classification? If so, use a 2 level factor as your outcome column.

missClassTrain <- missClass(trainSA$chd,predict(regModel,newdata=trainSA))
missClassTest <- missClass(testSA$chd,predict(regModel,newdata=testSA))
missClassTrain

## [1] 0.2727273

missClassTest

## [1] 0.3116883

Question 5

Load the vowel.train and vowel.test data sets:

library(ElemStatLearn)
data(vowel.train)
data(vowel.test)

Set the variable y to be a factor variable in both the training and test set. Then set the seed to 33833. Fit a random forest predictor relating the factor variable y to the remaining variables. Read about variable importance in random forests here: http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm#ooberr The caret package uses by default the Gini importance.

Calculate the variable importance using the varImp function in the caret package. What is the order of variable importance?

[NOTE: Use randomForest() specifically, not caret, as there’s been some issues reported with that approach. 11/6/2016]

set.seed(33833)
str(vowel.train)

## 'data.frame':    528 obs. of  11 variables:
##  $ y   : int  1 2 3 4 5 6 7 8 9 10 ...
##  $ x.1 : num  -3.64 -3.33 -2.12 -2.29 -2.6 ...
##  $ x.2 : num  0.418 0.496 0.894 1.809 1.938 ...
##  $ x.3 : num  -0.67 -0.694 -1.576 -1.498 -0.846 ...
##  $ x.4 : num  1.779 1.365 0.147 1.012 1.062 ...
##  $ x.5 : num  -0.168 -0.265 -0.707 -1.053 -1.633 ...
##  $ x.6 : num  1.627 1.933 1.559 1.06 0.764 ...
##  $ x.7 : num  -0.388 -0.363 -0.579 -0.567 0.394 0.217 0.322 -0.435 -0.512 -0.466 ...
##  $ x.8 : num  0.529 0.51 0.676 0.235 -0.15 -0.246 0.45 0.992 0.928 0.702 ...
##  $ x.9 : num  -0.874 -0.621 -0.809 -0.091 0.277 0.238 0.377 0.575 -0.167 0.06 ...
##  $ x.10: num  -0.814 -0.488 -0.049 -0.795 -0.396 -0.365 -0.366 -0.301 -0.434 -0.836 ...

library(randomForest)

## randomForest 4.6-14

## Type rfNews() to see new features/changes/bug fixes.

## 
## Attaching package: 'randomForest'

## The following object is masked from 'package:rattle':
## 
##     importance

## The following object is masked from 'package:ggplot2':
## 
##     margin

vowel.train$y <- as.factor(vowel.train$y)
vowel.test$y <- as.factor(vowel.test$y)
modelRF <- randomForest(y~.,data=vowel.train)
order(varImp(modelRF),decreasing=TRUE)

##  [1]  1  2  5  6  8  4  3  9  7 10