Exercise 2
Load data
# install.packages('mlbench')
library(mlbench)
data(Glass)
str(Glass)## 'data.frame': 214 obs. of 10 variables:
## $ RI : num 1.52 1.52 1.52 1.52 1.52 ...
## $ Na : num 13.6 13.9 13.5 13.2 13.3 ...
## $ Mg : num 4.49 3.6 3.55 3.69 3.62 3.61 3.6 3.61 3.58 3.6 ...
## $ Al : num 1.1 1.36 1.54 1.29 1.24 1.62 1.14 1.05 1.37 1.36 ...
## $ Si : num 71.8 72.7 73 72.6 73.1 ...
## $ K : num 0.06 0.48 0.39 0.57 0.55 0.64 0.58 0.57 0.56 0.57 ...
## $ Ca : num 8.75 7.83 7.78 8.22 8.07 8.07 8.17 8.24 8.3 8.4 ...
## $ Ba : num 0 0 0 0 0 0 0 0 0 0 ...
## $ Fe : num 0 0 0 0 0 0.26 0 0 0 0.11 ...
## $ Type: Factor w/ 6 levels "1","2","3","5",..: 1 1 1 1 1 1 1 1 1 1 ...a) Utilize suitable visualizations (employ any types of data visualization you deem appropriate) to explore the predictor variables, aiming to understand their distributions and relationships among them.
# Histograms
par(mfrow=c(3,3))
hist(Glass$RI)
hist(Glass$Na)
hist(Glass$Mg)
hist(Glass$Al)
hist(Glass$Si)
hist(Glass$K)
hist(Glass$Ca)
hist(Glass$Ba)
hist(Glass$Fe)
b) Do there appear to be any outliers in the data? Are any predictors skewed? Show all the work!
# Boxplots
boxplot(Glass[1], xlab = "RI")
boxplot(Glass[2:9], xlab = "Elements")
Outliers are present in all predictors except Mg. Strong right skew is observed in K, Ba, and Fe; Al and Ca exhibit moderate right skew, and RI shows a slight right skew. Mg and Si appear approximately symmetric, and Mg displays a bimodal distribution.
c) Are there any relevant transformations of one or more predictors that might improve the classification model? Show all the work!
# Resolve skewness via box-cox transformation (K, Ba, Fe)
library(caret)## Loading required package: ggplot2## Loading required package: latticeKBoxCox = BoxCoxTrans(Glass$K)
BaBoxCox = BoxCoxTrans(Glass$Ba)
FeBoxCox = BoxCoxTrans(Glass$Fe)
KBoxCox$lambda## [1] NABaBoxCox$lambda## [1] NAFeBoxCox$lambda## [1] NA# Center & scale
GlassCS = scale(Glass[1:9], center = TRUE, scale = TRUE)
colMeans(GlassCS)## RI Na Mg Al Si
## 2.637535e-13 -4.240637e-15 -4.233374e-16 -7.896072e-16 9.426520e-16
## K Ca Ba Fe
## 3.413676e-16 -3.134824e-16 -2.490220e-16 3.206158e-16covariance = cov(GlassCS)
diag(covariance)## RI Na Mg Al Si K Ca Ba Fe
## 1 1 1 1 1 1 1 1 1# Resolve outliers via spatial sign
GlasSS = spatialSign(GlassCS, center = TRUE, shape = TRUE)A Box-Cox transformation may improve skewness in the three predictors (K, Ba, Fe) with a strong right skew. However, these predictors contain many zero values, which resulted in estimated λ values of NA. Therefore, there was little change from the original distribution. All predictors were centered and scaled to have mean 0 and variance 1. In order to resolve outliers, a spatial sign transformation was applied. Overall, standardization and the spatial sign transformation are expected to improve the performance of the classification model.
d) Fit SVM model (You may refer to Chapter 4 material for details) using the following R codes: (This code will be discussed in detail in the following chapters)
# install.packages('kernlab')
library(kernlab)##
## Attaching package: 'kernlab'## The following object is masked from 'package:ggplot2':
##
## alphaset.seed(231)
sigDist <- sigest(Type~ ., data = Glass, frac = 1)
sigDist## 90% 50% 10%
## 0.03407935 0.11297847 0.62767315svmTuneGrid <- data.frame(sigma = as.vector(sigDist)[1], C = 2^(-2:10))
svmTuneGrid## sigma C
## 1 0.03407935 0.25
## 2 0.03407935 0.50
## 3 0.03407935 1.00
## 4 0.03407935 2.00
## 5 0.03407935 4.00
## 6 0.03407935 8.00
## 7 0.03407935 16.00
## 8 0.03407935 32.00
## 9 0.03407935 64.00
## 10 0.03407935 128.00
## 11 0.03407935 256.00
## 12 0.03407935 512.00
## 13 0.03407935 1024.00# install.packages('AppliedPredictiveModeling')
library(AppliedPredictiveModeling)
library(caret)
set.seed(1056)
svmFit <- train(Type~ ., data = Glass, method = "svmRadial",
preProc = c("center", "scale"),tuneGrid = svmTuneGrid,
trControl = trainControl(method = "repeatedcv", repeats = 5))
plot(svmFit, scales = list(x = list(log = 2)))