homework5

1. In homework 3 problem 4 ridge regression was performed on the wine quality data set : winequality-red.csv. Now use k-nearest neighbors to classify this data. Use cross validation to choose the best value for k. Round the results from the ridge regression to the nearest integer to form the classification with ridge regression. Using the best values for k (nearest neighbor) and lambda (ridge regression), compare and contrast the results of these two classification techniques.

rm(list = ls())
library(class)
library(e1071)
winequality <- read.csv("/home/archana/ML_works_ucsc/winequality-red.csv", header = TRUE,sep = ";")
train.labels <- winequality[1:1400, 12]
train <- winequality[1:1400, -12]

test.labels <- winequality[1401:1599, 12]
test <- winequality[1401:1599, -12]

Err <- rep(0, 20)
for (i in seq(from = 1, to = 20)) {
    out <- knn.cv(train, train.labels, k = i)
    Error <- 1 - sum(abs(train.labels == out))/length(out)
    Err[i] <- Error
}
Err

##  [1] 0.3807 0.4550 0.4621 0.4664 0.4821 0.4979 0.4836 0.4950 0.5007 0.4943
## [11] 0.4964 0.5079 0.5043 0.4957 0.4821 0.4771 0.4757 0.4843 0.4764 0.4686

plot(Err)

bestk = which.min(Err)
bestk

## [1] 1

The best k-nearsest-neighbor is 1, which cooresponds to error of 0.38. 2) Use k-nearest neighbors to classify the Iris data set. Compare the k- nearest neighbor results with the results obtained in class using the Naive Bayes Classifier.

rm(list = ls())
library(class)
library(e1071)

library(klaR)

## Loading required package: MASS

library(MASS)

train.labels <- iris[1:130, 5]
train <- iris[1:130, -5]

test.labels <- iris[131:150, 12]
test <- iris[131:150, -12]

Err <- rep(0, 20)
for (kk in seq(from = 1, to = 20)) {
    out <- knn.cv(train, train.labels, k = kk)
    Error <- 1 - sum(abs(train.labels == out))/length(out)
    Err[kk] <- Error
}
Err

##  [1] 0.03846 0.03077 0.03077 0.03846 0.04615 0.03846 0.04615 0.04615
##  [9] 0.04615 0.03846 0.04615 0.04615 0.05385 0.05385 0.05385 0.05385
## [17] 0.06154 0.05385 0.06154 0.05385

plot(Err)

bestk = which.min(Err)
bestk 

## [1] 2

data(iris)
m <- NaiveBayes(Species ~ ., data = iris)
out <- predict(m)

Err <- 1 - sum(out$class == iris$Species)/length(iris$Species)
Err

## [1] 0.04

#3) Classify the wine quality data using Naive Bayes. Compare the results with the two methods described in problem 1 of this homework set. Think about why one of the methods used works better than the other
winequality <- read.csv("/home/archana/ML_works_ucsc/winequality-red.csv", header = TRUE,sep=";")
winequality$quality <- as.factor(winequality$quality)
m <- NaiveBayes(quality ~ ., data = winequality)
out<-predict(m)
Err <- 1 - sum(out$class == winequality$quality)/length(winequality$quality)
Err 

## [1] 0.4396

for the 2) K-Nearest when k = 3, gives a better error 0.030 compared to 0.040 with Naive Bayes. K = 3 > NB for Classifying species in the Iris

for the 3) NaiveBayes, does worse off 0.43197 error compared to k-nearestNeighbor with k = 1 having error of 0.38000

4. Classify the sonar data using Naive Bayes. Compare the results with the methods used in class and with the last homework set. Give reasons for any discrepancies between the results for these methods. (Either in class or in homework, the following methods have been used on this data set: Trees, Linear Regression, Ridge Regression, an Ensemble Method, and now Naive Bayes.)

rm(list = ls())
library(class)
library(e1071)
library(klaR)
library(MASS)
library(rpart)

sonar.train <- read.csv("/home/archana/ML_works_ucsc/sonar_train.csv", header = FALSE)
sonar.train$V61 <- as.factor(sonar.train$V61)
m <- NaiveBayes(V61 ~ ., data = sonar.train)
out <- predict(m)
Err <- 1 - sum(out$class == sonar.train$V61)/length(sonar.train$V61)
Err

## [1] 0.2385

train.labels <- sonar.train$V61
train <- sonar.train[, -61]

Err <- rep(0, 20)
for (kk in seq(from = 1, to = 20)) {
    out <- knn.cv(train, train.labels, k = kk)
    Error <- 1 - sum(abs(train.labels == out))/length(out)
    Err[kk] <- Error
}
Err

##  [1] 0.2000 0.2538 0.2615 0.2923 0.3000 0.3000 0.3000 0.2769 0.2923 0.2846
## [11] 0.3000 0.3154 0.3154 0.3077 0.3077 0.3077 0.3077 0.3231 0.3231 0.3308

plot(Err)

bestk = which.min(Err)
bestk

## [1] 1

K-neirest with k = 1, does the best with error of 0.2 verus NaiveBayes,which also perform well with an error of 0.2384. It depends on costs of error and ease of computation on ghd dataset that would make the determination for ‘best’ method to be used to classify the sonar data-set.