Chapter 4 - Classification
Rafael S Toledo
November 4, 2016
Applied Exercises
Q10. This question should be answered using the Weekly data set, which is part of the ISLR package. This data is similar in nature to the Smarket data from this chapter’s lab, except that is contains 1,089 weekly returns for 21 years, from the beginning of the 1990 to the end of the 2010.
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
require(ISLR)## Loading required package: ISLRdata(Weekly)
summary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## Today Direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260pairs(Weekly)
Seemingly, the only evidence is at \(Volume \times Year\), where shows a logarithmic pattern.
(b) Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
glm.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data=Weekly, family="binomial")
summary(glm.fit)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = "binomial", data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4Lag2.
(c) Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
glm.probs <- predict(glm.fit, type="response")
glm.preds <- ifelse(glm.probs>.5, "Up", "Down")
cm <- table(Weekly$Direction, glm.preds)
cm## glm.preds
## Down Up
## Down 54 430
## Up 48 557There are a predominance of Up prediction. The model predicts well the Up direction, but it predict poorly the Down direction.
# Here i consider the Null/(-) as `Down` and the Non-null/(+) as `Up`
FP_rate <- cm["Down", "Up"]/(cm["Down", "Up"] + cm["Down", "Down"])
TP_rate <- cm["Up", "Up"]/(cm["Up", "Up"] + cm["Up", "Down"])
precision <- cm["Up", "Up"]/(cm["Up", "Up"] + cm["Down", "Up"])
data.frame("measurements"=c("specificity", "sensibility", "precision"), "rate"=c(FP_rate, TP_rate, precision))## measurements rate
## 1 specificity 0.8884298
## 2 sensibility 0.9206612
## 3 precision 0.5643364(d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
trainset = (Weekly$Year<=2008)
testset = Weekly[!trainset,]
glm.fit.d <- glm(Direction ~ Lag2, data=Weekly, subset=trainset, family="binomial")
glm.probs.d <- predict(glm.fit.d, type="response", newdata=testset)
glm.preds.d <- ifelse(glm.probs.d>.5, "Up", "Down")
cm.d <- table(testset$Direction, glm.preds.d)
cm.d## glm.preds.d
## Down Up
## Down 9 34
## Up 5 56# compute overall of correct predictions
acc.d <- (cm.d["Down", "Down"] + cm.d["Up", "Up"])/sum(cm.d)
acc.d## [1] 0.625(e) Repeat (d) using LDA.
library(MASS)
lda.fit.e <- lda(Direction ~ Lag2, data=Weekly, subset=trainset)
lda.preds.e <- predict(lda.fit.e, newdata=testset)
cm.e <- table(testset$Direction, lda.preds.e$class)
cm.e##
## Down Up
## Down 9 34
## Up 5 56# compute overall of correct predictions
acc.e<- (cm.e["Down", "Down"] + cm.e["Up", "Up"])/sum(cm.e)
acc.e## [1] 0.625(f) Repeat (d) using QDA.
qda.fit.f <- qda(Direction ~ Lag2, data=Weekly, subset=trainset)
qda.preds.f <- predict(qda.fit.f, newdata=testset)
cm.f <- table(testset$Direction, qda.preds.f$class)
cm.f##
## Down Up
## Down 0 43
## Up 0 61# compute overall of correct predictions
acc.f<- (cm.f["Down", "Down"] + cm.f["Up", "Up"])/sum(cm.f)
acc.f## [1] 0.5865385(g) Repeat (d) using KNN with \(K\) =1.
library(class)
set.seed(1)
train.g = Weekly[trainset, c("Lag2", "Direction")]
knn.pred = knn(train=data.frame(train.g$Lag2), test=data.frame(testset$Lag2), cl=train.g$Direction, k=1)
cm.g <- table(testset$Direction, knn.pred)
cm.g## knn.pred
## Down Up
## Down 21 22
## Up 30 31# compute overall of correct predictions
acc.g <- (cm.g["Down", "Down"] + cm.g["Up", "Up"])/sum(cm.g)
acc.g## [1] 0.5(h) Which of these methods appears to provide the best results on this data?
rbind(acc.d, acc.e, acc.f, acc.g)## [,1]
## acc.d 0.6250000
## acc.e 0.6250000
## acc.f 0.5865385
## acc.g 0.5000000The models from letter d and e, respectively Logistic Regression and LDA.
(i) Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for \(K\) in the KNN classifier.
I try many values of K.
set.seed(1)
results <- data.frame(k=1:50, acc=NA)
for(i in 1:50){
knn.pred = knn(train=data.frame(train.g$Lag2), test=data.frame(testset$Lag2), cl=train.g$Direction, k=i)
cm <- table(testset$Direction, knn.pred)
acc <- (cm["Down", "Down"] + cm["Up", "Up"])/sum(cm)
results$acc[i] <- acc
}
plot(x=results$k, y=results$acc, type="l", xlab="K", ylab="accuracy", ylim=c(.4,.65))
The K doesn’t seem to affect the accuracy values.
Now, using a QDA model with all Lags predictors plus Volume.
qda.fit <- qda(Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, subset=trainset)
qda.preds <- predict(qda.fit, testset)
# show accuracy
print( sum(qda.preds$class==testset$Direction)/length(qda.preds$class))## [1] 0.4326923It had a worse performance than using only the Lag2 predictor shown in g.
Again on QDA model, i try with interactive variables between all Lags predictors.
qda.fit <- qda(Direction ~ Lag1*Lag2*Lag3*Lag4*Lag5 + Volume, data=Weekly, subset=trainset)
qda.preds <- predict(qda.fit, testset)
# show accuracy
print( sum(qda.preds$class==testset$Direction)/length(qda.preds$class))## [1] 0.4134615The accuracy was even worse than before.
For last, i try the same predictors schema using LDA.
lda.fit <- lda(Direction ~ Lag1*Lag2*Lag3*Lag4*Lag5 + Volume, data=Weekly, subset=trainset)
lda.preds <- predict(lda.fit, testset)
# show accuracy
print( sum(lda.preds$class==testset$Direction)/length(lda.preds$class))## [1] 0.4423077The LDA performance kept similar of the QDA.
Q11. In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
remove(list=ls())
data(Auto)(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
Auto$mpg01 <- with(ifelse(mpg>median(mpg), "1", "0"), data=Auto)(b) Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
attach(Auto)
# Boxplots
par(mfrow=c(2,3))
for(i in names(Auto)){
# excluding the own mpgs variables and others categorical variables
if( grepl(i, pattern="^mpg|cylinders|origin|name")){ next}
boxplot(eval(parse(text=i)) ~ mpg01, ylab=i, col=c("red", "blue"))
}
As shown in the boxplot, all variables present some trend with mpg01.
# for the categorical variables i do barplots
colors = c("red", "yellow", "green", "violet", "orange", "blue", "pink", "cyan")
par(mfrow=c(1,2))
for(i in c("cylinders", "origin")){
aux <- table(eval(parse(text=i)), mpg01)
cols <- colors[1:nrow(aux)]
barplot(aux, xlab="mpg01", ylab=i, beside=T, legend=rownames(aux), col=cols)
}
At the barplots, cylinders and origin also show relation with mpg01. For instance, on dataset cars of lower mpg are majoraty from origin 1, which is American.
(c) Split the data into a training set and a test set.
# splitting the train and test set into 75% and 25%
set.seed(1)
rows <- sample(x=nrow(Auto), size=.75*nrow(Auto))
trainset <- Auto[rows, ]
testset <- Auto[-rows, ](d) Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
I use all variables, but name, since all appears correlation with mpg01.
library(MASS)
lda.fit <- lda(mpg01 ~ displacement+horsepower+weight+acceleration+year+cylinders+origin, data=trainset)
lda.pred <- predict(lda.fit, testset)
table(testset$mpg01, lda.pred$class)##
## 0 1
## 0 43 7
## 1 4 44# test-error
round(sum(lda.pred$class!=testset$mpg01)/nrow(testset)*100,2)## [1] 11.22(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit <- qda(mpg01 ~ displacement+horsepower+weight+acceleration+year+cylinders+origin, data=trainset)
qda.pred <- predict(qda.fit, testset)
table(testset$mpg01, qda.pred$class)##
## 0 1
## 0 44 6
## 1 4 44# test-error
round(sum(qda.pred$class!=testset$mpg01)/nrow(testset)*100,2)## [1] 10.2(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lr.fit <- glm(as.factor(mpg01) ~ displacement+horsepower+weight+acceleration+year+cylinders+origin, data=trainset, family="binomial")
lr.probs <- predict(lr.fit, testset, type="response")
lr.pred <- ifelse(lr.probs>0.5, "1", "0")
table(testset$mpg01, lr.pred)## lr.pred
## 0 1
## 0 43 7
## 1 6 42# test-error
round(sum(lr.pred!=testset$mpg01)/nrow(testset)*100,2)## [1] 13.27(g) Perform KNN on the training data, with serveral values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
library(class)
sel.variables <- which(names(trainset)%in%c("mpg01", "displacement", "horsepower", "weight", "acceleration", "year", "cylinders", "origin"))
set.seed(1)
accuracies <- data.frame("k"=1:10, acc=NA)
for(k in 1:10){
knn.pred <- knn(train=trainset[, sel.variables], test=testset[, sel.variables], cl=trainset$mpg01, k=k)
# test-error
accuracies$acc[k]= round(sum(knn.pred!=testset$mpg01)/nrow(testset)*100,2)
}
accuracies## k acc
## 1 1 16.33
## 2 2 20.41
## 3 3 14.29
## 4 4 14.29
## 5 5 13.27
## 6 6 13.27
## 7 7 12.24
## 8 8 14.29
## 9 9 14.29
## 10 10 13.27The k=7 was the best response, outperformed all others.
Q12. This problem involves writing functions.
remove(list=ls())(a) Write a function, Power(), that prints out the result of raising 2 to the 3rd power. In other words, your function should compute 2³ and print out the results.
Power <- function(){ print( 2^3)}
Power()## [1] 8(b) Create a new function, Power2(), that allows you to pass any two numbers, x and a, and prints out the value of x^a. You can do this by beginning your function with the line.
> Power2 = function(x,a){
You should be able to call your function by entering, for instance,
> Power2(3,8)
on the command line. This should output the value of \(3^8\), namely, \(6,561\).
Power2 <- function(x,a){
print( x^a)
}
Power2(3,8)## [1] 6561(c) Using the Power2() function that you just wrote, compute \(10^3\), \(8^{17}\), \(131^3\).
Power2(10,3)## [1] 1000Power2(8,17)## [1] 2.2518e+15Power2(131,3)## [1] 2248091(d) Now create a new function, Power3, that actually returns the result x^a as an R object, rather than simply printing it to the screen. That is, if you store the value x^a in an object called result within your function, then you can simply return() this result, using the following line:
return(result)
Power3 <- function(x,a){
return( x^a)
}(e) Now using the Power3() function, create a plot of \(f(x) = x^2\). The x-axis should display a range of integers from 1 to 10, and the y-axis should display \(x^2\). Label the axes appropriately, and use an appropriate title for the figure. Consider displaying either the x-axis, the y-axis, or both on the log-scale. You can do this by using log="x", log="y", or log="xy" as arguments to the plot() function.
par(mfrow=c(2,2))
plot(x = x<-1:10, y= y<-Power3(x,2), xlab="x", ylab="x²")
plot(x,y,log="x", xlab="log(x) scale", ylab="x²")
plot(x,y,log="y", xlab="x", ylab="log(x²) scale")
plot(x,y,log="xy", xlab="log(x) scale", ylab="log(x²) scale")
(f) Create a function, PlotPower(), that allows you to create a plot of x against x^a for a fixed a and for a range of values of x. For instance, if you call
> PlotPower(1:10,3)
then a plot should be create with an x-axis taking on values 1,2,…,10, and a y-axis taking on values \(1^3,2^3,...,10^3\).
par(mfrow=c(1,1))
PlotPower <- function(x,a){
plot(x = x, y= y<-Power3(x,a), xlab="x", ylab=paste0("x^",a))
}
PlotPower(1:10,3)
Q13. Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or bellow the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findinds.
rm(list=ls())
data("Boston")
Boston$crim01 <- ifelse(Boston$crim > median(Boston$crim), "1", "0")attach(Boston)
par(mfrow=c(2,6))
for(i in names(Boston)){
# excluding the own crime variables and the chas variable
if( grepl(i, pattern="^crim|^chas")){ next}
boxplot(eval(parse(text=i)) ~ crim01, ylab=i, col=c("red", "blue"), varwidth=T)
}
All variable shows trend to crim01, exceptrm which has some difference among the crimes situation but its most population lies in the same range values.
For Chas variable, i do a barplot, it is a dummy variable to if the tract bounds the river.
par(mfrow=c(1,1))
aux <- table(chas, crim01)
barplot(aux, beside = T, legend=rownames(aux), col=c("red", "blue"))
The chas doesn’t show much difference for crime situation.
Selecting the relevant variables, i use the: zn, indus, nox, age, dis, rad, tax, ptratio, black, lstat and medv.
set.seed(1)
vars = c("zn", "indus", "nox", "age", "dis", "rad", "tax", "ptratio", "black", "lstat", "medv", "crim01")
rows = sample(x=nrow(Boston), size=.75*nrow(Boston))
trainset = Boston[rows, vars]
testset = Boston[-rows, vars]Modelling
# LOGISTIC REGRESSION
lr.fit <- glm(as.factor(crim01) ~ ., data=trainset, family="binomial")
lr.probs <- predict(lr.fit, testset, type="response")
lr.pred <- ifelse(lr.probs>.5, "1","0")
test.err.lr <- mean(lr.pred!=testset$crim01)
# LINEAR DISCRIMINANT ANALYSIS
lda.fit <- lda(crim01 ~ ., data=trainset)
lda.pred <- predict(lda.fit, testset)
test.err.lda <- mean(lda.pred$class!=testset$crim01)
# QUADRATIC DISCRIMINANT ANALYSIS
qda.fit <- qda(crim01 ~ ., data=trainset)
qda.pred <- predict(qda.fit, testset)
test.err.qda <- mean(qda.pred$class!=testset$crim01)
# KNN-1
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=1)
test.err.knn_1 <- mean(knn.pred!=testset$crim01)
# KNN-CV
err.knn_cv <- rep(NA,9)
for(i in 2:10){
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=i)
err.knn_cv[i-1] <- mean(knn.pred!=testset$crim01)
}
test.err_knn_CV <- min(err.knn_cv)
round1 = data.frame("method"=c("LR", "LDA", "QDA", "KNN-1", "KNN-CV"), test.err=c(test.err.lr, test.err.lda, test.err.qda, test.err.knn_1, test.err_knn_CV))
round1## method test.err
## 1 LR 0.08661417
## 2 LDA 0.14173228
## 3 QDA 0.13385827
## 4 KNN-1 0.08661417
## 5 KNN-CV 0.07874016Both KNN methods outperforms the others, maybe it’s related to the form of the data, which can be more non-linear and either differs more from a gaussian shape. The logistic regression performs better than LDA and QDA, that enhances the assumption of a non Gaussian distribution from the data. And as QDA performs better than LDA, i can imagine that the non-linear decision boundary helps this decision. So the non-parametric method presents the best results.
Doing a second round of modelling, this time choosing only the predictors which seemed more relevants by the logistic regression coefficients. Cheking the p-values:
coefs <- summary(lr.fit)$coefficients
coefs[order(coefs[,"Pr(>|z|)"], decreasing=F),]## Estimate Std. Error z value Pr(>|z|)
## nox 46.268552442 8.367479623 5.5295686 3.210193e-08
## (Intercept) -34.234056447 7.272602370 -4.7072636 2.510642e-06
## rad 0.605101758 0.171668295 3.5248312 4.237528e-04
## age 0.040524724 0.012981423 3.1217475 1.797811e-03
## dis 0.720872986 0.253963152 2.8384944 4.532691e-03
## zn -0.083465982 0.037217704 -2.2426419 2.491992e-02
## tax -0.007250196 0.003264131 -2.2211721 2.633931e-02
## medv 0.096672482 0.048899905 1.9769462 4.804771e-02
## ptratio 0.216821863 0.128042186 1.6933627 9.038645e-02
## indus -0.064414786 0.051403490 -1.2531209 2.101617e-01
## black -0.006369013 0.005175739 -1.2305513 2.184907e-01
## lstat -0.001571527 0.056064484 -0.0280307 9.776377e-01I choose nox, rad, ptratio, black and medv.
vars <- c("nox", "rad", "ptratio", "black", "medv", "dis", "crim01")
trainset = Boston[rows, vars]
testset = Boston[-rows, vars]Modelling
# LOGISTIC REGRESSION
lr.fit <- glm(as.factor(crim01) ~ ., data=trainset, family="binomial")
lr.probs <- predict(lr.fit, testset, type="response")
lr.pred <- ifelse(lr.probs>.5, "1","0")
test.err.lr <- mean(lr.pred!=testset$crim01)
# LINEAR DISCRIMINANT ANALYSIS
lda.fit <- lda(crim01 ~ ., data=trainset)
lda.pred <- predict(lda.fit, testset)
test.err.lda <- mean(lda.pred$class!=testset$crim01)
# QUADRATIC DISCRIMINANT ANALYSIS
qda.fit <- qda(crim01 ~ ., data=trainset)
qda.pred <- predict(qda.fit, testset)
test.err.qda <- mean(qda.pred$class!=testset$crim01)
# KNN-1
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=1)
test.err.knn_1 <- mean(knn.pred!=testset$crim01)
# KNN-CV
err.knn_cv <- rep(NA,9)
for(i in 2:10){
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=i)
err.knn_cv[i-1] <- mean(knn.pred!=testset$crim01)
}
test.err_knn_CV <- min(err.knn_cv)
round2 = data.frame("method"=c("LR", "LDA", "QDA", "KNN-1", "KNN-CV"), test.err=c(test.err.lr, test.err.lda, test.err.qda, test.err.knn_1, test.err_knn_CV))
round2## method test.err
## 1 LR 0.07874016
## 2 LDA 0.13385827
## 3 QDA 0.15748031
## 4 KNN-1 0.08661417
## 5 KNN-CV 0.10236220On round 2, the general performance was worse for all approachs, so probably there are relevent information in the excluded variables.
Now, i try again, using the most 6 variable that seemed, in my observation from the graphs shown before, more associated with crime index. They are zn, indus, nox, dis, rad and tax.
vars <- c("zn","indus", "nox", "dis", "rad", "tax", "crim01")
trainset = Boston[rows, vars]
testset = Boston[-rows, vars]Modelling
# LOGISTIC REGRESSION
lr.fit <- glm(as.factor(crim01) ~ ., data=trainset, family="binomial")
lr.probs <- predict(lr.fit, testset, type="response")
lr.pred <- ifelse(lr.probs>.5, "1","0")
test.err.lr <- mean(lr.pred!=testset$crim01)
# LINEAR DISCRIMINANT ANALYSIS
lda.fit <- lda(crim01 ~ ., data=trainset)
lda.pred <- predict(lda.fit, testset)
test.err.lda <- mean(lda.pred$class!=testset$crim01)
# QUADRATIC DISCRIMINANT ANALYSIS
qda.fit <- qda(crim01 ~ ., data=trainset)
qda.pred <- predict(qda.fit, testset)
test.err.qda <- mean(qda.pred$class!=testset$crim01)
# KNN-1
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=1)
test.err.knn_1 <- mean(knn.pred!=testset$crim01)
# KNN-CV
err.knn_cv <- rep(NA,9)
for(i in 2:10){
knn.pred <- knn(train=trainset, test=testset, cl=trainset$crim01, k=i)
err.knn_cv[i-1] <- mean(knn.pred!=testset$crim01)
}
test.err_knn_CV <- min(err.knn_cv)
round3 = data.frame("method"=c("LR", "LDA", "QDA", "KNN-1", "KNN-CV"), test.err=c(test.err.lr, test.err.lda, test.err.qda, test.err.knn_1, test.err_knn_CV))
round3## method test.err
## 1 LR 0.09448819
## 2 LDA 0.14960630
## 3 QDA 0.08661417
## 4 KNN-1 0.00000000
## 5 KNN-CV 0.04724409Surprisingly, the third round of my chosen variable, based on the boxplot, had the greatest performance of the previous rounds. Mainly the QDA and KNNs approachs. KNN-1 had showed test error of 0.7%. The linear approachs were very bad.
When i eliminate some variables, it helped for the non-linear approachs did better models. Seeing the three rounds on the graph bellow.
performances <- rbind(cbind(round="round_1", round1), cbind(round="round_2", round2), cbind(round="round_3", round3))
library(reshape2)
dcast(data=performances, method ~ round, value.var="test.err")## method round_1 round_2 round_3
## 1 KNN-1 0.08661417 0.08661417 0.00000000
## 2 KNN-CV 0.07874016 0.10236220 0.04724409
## 3 LDA 0.14173228 0.13385827 0.14960630
## 4 LR 0.08661417 0.07874016 0.09448819
## 5 QDA 0.13385827 0.15748031 0.08661417library(ggplot2)
ggplot(data=performances, aes(x=method,y=test.err)) + geom_bar(stat="identity", aes(fill=method)) + coord_flip() + facet_grid(round ~ .)