Assignment #3
Yutae Lee
2021 3 1
Problem 10
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 it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
library(ISLR)
attach(Weekly)(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
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 Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## pairs(Weekly)
cor(Weekly[,-9])## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1 -0.03228927 1.000000000 -0.07485305 0.05863568 -0.071273876
## Lag2 -0.03339001 -0.074853051 1.00000000 -0.07572091 0.058381535
## Lag3 -0.03000649 0.058635682 -0.07572091 1.00000000 -0.075395865
## Lag4 -0.03112792 -0.071273876 0.05838153 -0.07539587 1.000000000
## Lag5 -0.03051910 -0.008183096 -0.07249948 0.06065717 -0.075675027
## Volume 0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today -0.03245989 -0.075031842 0.05916672 -0.07124364 -0.007825873
## Lag5 Volume Today
## Year -0.030519101 0.84194162 -0.032459894
## Lag1 -0.008183096 -0.06495131 -0.075031842
## Lag2 -0.072499482 -0.08551314 0.059166717
## Lag3 0.060657175 -0.06928771 -0.071243639
## Lag4 -0.075675027 -0.06107462 -0.007825873
## Lag5 1.000000000 -0.05851741 0.011012698
## Volume -0.058517414 1.00000000 -0.033077783
## Today 0.011012698 -0.03307778 1.000000000plot(Volume)
(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.fits<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(glm.fits)##
## 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: 4We can notice that predictor Lag2(0.0296<0.05) is the only significant predictor.
(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.fits, type = "response")
contrasts(Direction)## Up
## Down 0
## Up 1glm.pred=rep("Down",1089)
glm.pred[glm.probs > .5] = "Up"
table(glm.pred ,Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557(54+557)/1089## [1] 0.5610652mean(glm.pred==Direction)## [1] 0.561065254/(54+430)## [1] 0.1115702557/(557+48)## [1] 0.9206612The diagonal elements of the confusion matrix indicate correct predictions, while the off-diagonals represent incorrect predictions. Our overall fraction of correct prediction is 611/1089= .5610652. For Down the fraction of correct prediction is .1115702 and for Up the fraction of correct prediction is .9206612
(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).
train=(Year<2009)
Weekly.2009= Weekly[!train,]
Direction.2009= Direction[!train]
dim(Weekly.2009)## [1] 104 9Direction.2009= Direction[!train]glm.fits=glm(Direction~Lag2, data=Weekly, family=binomial ,subset=train)
glm.probs=predict (glm.fits,Weekly.2009, type="response")glm.pred=rep('Down',104)
glm.pred[glm.probs >.5]= 'Up'
table(glm.pred ,Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 9 5
## Up 34 56mean(glm.pred==Direction.2009)## [1] 0.625Using Lag 2 as our only predictor, we get fraction of correct prediction of 0.625. We can notice it is greater than the last test data.
(e) Repeat (d) using LDA.
library(MASS)
lda.fit=lda(Direction~Lag2 ,data= `Weekly` ,subset=train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162plot(lda.fit)
lda.pred=predict (lda.fit , Weekly.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class=lda.pred$class
table(lda.class ,Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class==Direction.2009)## [1] 0.625By looking at the result we can see that we get same fraction of correct prediction as when we did with logistic regression.
(f) Repeat (d) using QDA.
qda.fit=qda(Direction~Lag2 ,data=Weekly ,subset=train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class=predict(qda.fit ,Weekly.2009)$class
table(qda.class ,Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==Direction.2009)## [1] 0.5865385The fraction of correct prediction this was .5865385. I noticed that in qda only had Up variable.
(g) Repeat (d) using KNN with K = 1.
library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction=Direction[train]
dim(train.X)## [1] 985 1dim(test.X)## [1] 104 1dim(train.Direction)## NULLlength(train.Direction)## [1] 985set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction ,k=1)
table(knn.pred ,Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred==Direction.2009)## [1] 0.5In our prediction using KNN with K = 1. We get fraction of correct prediction of 0.5.
(h) Which of these methods appears to provide the best results on this data?
We can notice that logistic regression model and LDA model has highest fraction of correct prediction with 0.625.
(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.
glm.fit2=glm(Direction~Lag1*Lag2, data=Weekly, family=binomial)
glm.probs=predict(glm.fit2, Weekly.2009, type='response')
glm.pred=rep('Down', 104)
glm.pred[glm.probs>0.5]='Up'
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 5 6
## Up 38 55mean(glm.pred==Direction.2009)## [1] 0.5769231This is the correct prediction of Lag1 and Lag2 using logistic regression
glm.fit3=glm(Direction~(sqrt(Lag2)), data=Weekly, family=binomial)
glm.probs=predict(glm.fit3, Weekly.2009, type='response')
glm.pred=rep('Down', 104)
glm.pred[glm.probs>0.5]='Up'
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 21 28
## Up 22 33mean(glm.pred==Direction.2009)## [1] 0.5192308This is the correct prediction of square root of Lag2 using logistic regression
glm.fit4=glm(Direction~Lag1+I(Lag2^2), data=Weekly, family=binomial)
glm.probs=predict(glm.fit4, Weekly.2009, type='response')
glm.pred=rep('Down', 104)
glm.pred[glm.probs>0.5]='Up'
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 3 3
## Up 40 58mean(glm.pred==Direction.2009)## [1] 0.5865385This is the correct prediction of Lag1 and Lag2^2 using logistic regression
lda.fit2=lda(Direction~Lag1 + I(Lag2^2), data = Weekly, subset = train)
lda.fit2## Call:
## lda(Direction ~ Lag1 + I(Lag2^2), data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 I(Lag2^2)
## Down 0.289444444 4.828121
## Up -0.009213235 5.428657
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.42334484
## I(Lag2^2) 0.02078431lda.pred2=predict(lda.fit2, Weekly.2009)
lda.class2=lda.pred2$class
table(lda.class2, Direction.2009)## Direction.2009
## lda.class2 Down Up
## Down 6 6
## Up 37 55mean(lda.class2==Direction.2009)## [1] 0.5865385This is the fraction of correct prediction of Lag1 and Lag2^2 using LDA
qda.fit2=qda(Direction~Lag1 + I(Lag2^2), data = Weekly, subset = train)
qda.fit2## Call:
## qda(Direction ~ Lag1 + I(Lag2^2), data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 I(Lag2^2)
## Down 0.289444444 4.828121
## Up -0.009213235 5.428657qda.pred=predict(qda.fit2, Weekly.2009)
qda.class=qda.pred$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 4 6
## Up 39 55mean(qda.class==Direction.2009)## [1] 0.5673077set.seed(1)
knn.pred=knn(train.X, test.X, train.Direction, k=3)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 16 20
## Up 27 41mean(knn.pred==Direction.2009)## [1] 0.5480769This is fraction of correct prediction using KNN with k = 3
set.seed(1)
knn.pred=knn(train.X, test.X, train.Direction, k=5)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 16 21
## Up 27 40mean(knn.pred==Direction.2009)## [1] 0.5384615This is fraction of correct prediction using KNN with K = 5
Using different values and different methods, lda and logistic regression of Lag1 and Lag2^2 has the greatest correct prediction.
Problem 11
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.
library(ISLR)
attach(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.
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365mpg01=rep(0, length(mpg))
mpg01[mpg > median(mpg)] = 1
Auto2 = data.frame(Auto, mpg01)
summary(Auto2)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
## mpg01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## (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.
pairs(Auto2[,-9])
cor(Auto2[,-9])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000par(mfrow=c(2,2))
boxplot(cylinders~mpg01)
boxplot(displacement~mpg01)
boxplot(horsepower~mpg01)
boxplot(weight~mpg01)
pairs() graph would not be helpful because the mpg01 variable only consists of 0 and 1. Looking at correlation those with negative correlation cylinders, displacement, horsepower, weight would be the useful predictor.
(c) Split the data into a training set and a test set.
set.seed(1)
Train=sample(1:nrow(Auto2), nrow(Auto2)*0.7)
training = Auto2[Train, -1]
summary(training)## cylinders displacement horsepower weight acceleration
## Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1649 Min. : 8.00
## 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 78.0 1st Qu.:2224 1st Qu.:14.00
## Median :4.000 Median :151.0 Median : 93.5 Median :2790 Median :15.50
## Mean :5.464 Mean :193.2 Mean :103.8 Mean :2967 Mean :15.66
## 3rd Qu.:8.000 3rd Qu.:261.5 3rd Qu.:123.8 3rd Qu.:3612 3rd Qu.:17.27
## Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140 Max. :24.80
##
## year origin name
## Min. :70.00 Min. :1.00 amc gremlin : 4
## 1st Qu.:73.00 1st Qu.:1.00 amc hornet : 4
## Median :76.00 Median :1.00 amc matador : 3
## Mean :76.06 Mean :1.54 chevrolet caprice classic: 3
## 3rd Qu.:79.00 3rd Qu.:2.00 chevrolet chevette : 3
## Max. :82.00 Max. :3.00 chevrolet impala : 3
## (Other) :254
## mpg01
## Min. :0.0000
## 1st Qu.:0.0000
## Median :1.0000
## Mean :0.5073
## 3rd Qu.:1.0000
## Max. :1.0000
## dim(training)## [1] 274 9testing = Auto2[-Train, -1]
dim(testing)## [1] 118 9dim(Auto2)## [1] 392 10test.mpg01=mpg01[-Train]
train.mpg01=mpg01[Train]
length(test.mpg01)## [1] 118length(train.mpg01)## [1] 274(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?
library(MASS)
lda.fit=lda(mpg01~cylinders+displacement+horsepower+weight, data = Auto2, subset=Train)
lda.fit## Call:
## lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto2,
## subset = Train)
##
## Prior probabilities of groups:
## 0 1
## 0.4927007 0.5072993
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.777778 271.9333 129.13333 3611.052
## 1 4.187050 116.8129 79.27338 2342.165
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.3962357999
## displacement -0.0047630097
## horsepower 0.0061919395
## weight -0.0008321338plot(lda.fit)
lda.pred=predict(lda.fit, testing)
names(lda.pred)## [1] "class" "posterior" "x"lda.class=lda.pred$class
table(lda.class, test.mpg01)## test.mpg01
## lda.class 0 1
## 0 50 3
## 1 11 54mean(lda.class==test.mpg01)## [1] 0.88135591-0.8813559## [1] 0.1186441Using LDA method we get test error of the model of 0.1186441.
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit=qda(mpg01~cylinders+displacement+horsepower+weight, data = Auto2, subset=Train)
qda.fit## Call:
## qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto2,
## subset = Train)
##
## Prior probabilities of groups:
## 0 1
## 0.4927007 0.5072993
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.777778 271.9333 129.13333 3611.052
## 1 4.187050 116.8129 79.27338 2342.165qda.pred=predict(qda.fit, testing)
qda.class=qda.pred$class
table(qda.class, test.mpg01)## test.mpg01
## qda.class 0 1
## 0 52 5
## 1 9 52mean(qda.class==test.mpg01)## [1] 0.88135591-0.8813559## [1] 0.1186441Using the QDA method we can see that test error of the model is 0.1186441
(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?
glm.fits=glm(mpg01~cylinders + displacement + horsepower + weight, data = Auto2, subset=Train, family=binomial)
glm.probs=predict(glm.fits, testing, type='response')
glm.pred=rep(0,length(glm.probs))
glm.pred[glm.probs>0.5]=1
table(glm.pred, test.mpg01)## test.mpg01
## glm.pred 0 1
## 0 53 3
## 1 8 54mean(glm.pred==test.mpg01) ## [1] 0.90677971-0.9067796## [1] 0.0932204Using logistic regression on the training data, the test error of the model is 0.0932204
(g) Perform KNN on the training data, with several 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)
train.X=cbind(training$cylinders, training$weight, training$displacement, training$horsepower)
test.X=cbind(testing$cylinders, testing$weight, testing$displacement, testing$horsepower)
knn.pred=knn(train.X, test.X, train.mpg01, k=1)
mean(knn.pred==test.mpg01)## [1] 0.8644068knn.pred=knn(train.X, test.X, train.mpg01, k=3)
mean(knn.pred==test.mpg01)## [1] 0.8898305knn.pred=knn(train.X, test.X, train.mpg01, k=5)
mean(knn.pred==test.mpg01)## [1] 0.8728814knn.pred=knn(train.X, test.X, train.mpg01, k=10)
mean(knn.pred==test.mpg01)## [1] 0.85593221-0.8898305## [1] 0.1101695Using the KNN prediction, K=3 seems to perform best, which gives us lowest test error value of 0.1101695
Problem 13
Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
attach(Boston)summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00cor(Boston)## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## black lstat medv
## crim -0.38506394 0.4556215 -0.3883046
## zn 0.17552032 -0.4129946 0.3604453
## indus -0.35697654 0.6037997 -0.4837252
## chas 0.04878848 -0.0539293 0.1752602
## nox -0.38005064 0.5908789 -0.4273208
## rm 0.12806864 -0.6138083 0.6953599
## age -0.27353398 0.6023385 -0.3769546
## dis 0.29151167 -0.4969958 0.2499287
## rad -0.44441282 0.4886763 -0.3816262
## tax -0.44180801 0.5439934 -0.4685359
## ptratio -0.17738330 0.3740443 -0.5077867
## black 1.00000000 -0.3660869 0.3334608
## lstat -0.36608690 1.0000000 -0.7376627
## medv 0.33346082 -0.7376627 1.0000000attach(Boston)
crim_1=rep(0, length(crim))
crim_1[crim>median(crim)]=1
Boston1 = data.frame(Boston, crim_1)
summary(Boston1)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv crim_1
## Min. : 1.73 Min. : 5.00 Min. :0.0
## 1st Qu.: 6.95 1st Qu.:17.02 1st Qu.:0.0
## Median :11.36 Median :21.20 Median :0.5
## Mean :12.65 Mean :22.53 Mean :0.5
## 3rd Qu.:16.95 3rd Qu.:25.00 3rd Qu.:1.0
## Max. :37.97 Max. :50.00 Max. :1.0set.seed(1)
train = sample(1:nrow(Boston1), 0.7*nrow(Boston1))
Boston1.train = Boston1[train,]
Boston1.test = Boston1[-train,]
dim(Boston1.train)## [1] 354 15crim_1.test=crim_1[-train]
crim_1.test## [1] 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [38] 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 0 0
## [75] 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1
## [112] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
## [149] 1 0 0 0glm.fit=glm(crim_1~zn+indus+chas+nox+rm+age+dis+rad+tax+ptratio+black+lstat+medv, data=Boston1, subset=train, family=binomial)
summary(glm.fit)##
## Call:
## glm(formula = crim_1 ~ zn + indus + chas + nox + rm + age + dis +
## rad + tax + ptratio + black + lstat + medv, family = binomial,
## data = Boston1, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1363 -0.1504 -0.0009 0.0018 3.5797
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -42.731502 8.009551 -5.335 9.55e-08 ***
## zn -0.105323 0.045585 -2.310 0.020862 *
## indus -0.066308 0.057649 -1.150 0.250057
## chas 0.187896 0.810675 0.232 0.816711
## nox 56.344292 10.073420 5.593 2.23e-08 ***
## rm -0.234402 0.900292 -0.260 0.794585
## age 0.022052 0.014304 1.542 0.123141
## dis 1.143918 0.303909 3.764 0.000167 ***
## rad 0.657976 0.184373 3.569 0.000359 ***
## tax -0.006299 0.003239 -1.944 0.051849 .
## ptratio 0.292158 0.155755 1.876 0.060689 .
## black -0.008703 0.005212 -1.670 0.094933 .
## lstat 0.112414 0.060168 1.868 0.061715 .
## medv 0.191745 0.087816 2.184 0.028999 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 490.65 on 353 degrees of freedom
## Residual deviance: 140.41 on 340 degrees of freedom
## AIC: 168.41
##
## Number of Fisher Scoring iterations: 9glm.fit2=glm(crim_1~ zn+nox+dis+rad+medv, data=Boston1, subset=train, family=binomial)
summary(glm.fit2)##
## Call:
## glm(formula = crim_1 ~ zn + nox + dis + rad + medv, family = binomial,
## data = Boston1, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0766 -0.3226 -0.0034 0.0069 3.2715
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -30.10487 4.91615 -6.124 9.14e-10 ***
## zn -0.09780 0.03537 -2.765 0.005698 **
## nox 42.65837 7.07436 6.030 1.64e-09 ***
## dis 0.84586 0.23756 3.561 0.000370 ***
## rad 0.49353 0.13194 3.741 0.000184 ***
## medv 0.08028 0.02906 2.763 0.005730 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 490.65 on 353 degrees of freedom
## Residual deviance: 164.99 on 348 degrees of freedom
## AIC: 176.99
##
## Number of Fisher Scoring iterations: 8glm.probs=predict(glm.fit2, Boston1.test, type='response')
glm.pred=rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
table(glm.pred, crim_1.test)## crim_1.test
## glm.pred 0 1
## 0 60 11
## 1 13 68mean(glm.pred == crim_1.test)## [1] 0.8421053lda.fit=lda(crim_1~zn+nox+dis+rad+medv, data=Boston1, subset=train)
lda.pred = predict(lda.fit, Boston1.test)
names(lda.pred)## [1] "class" "posterior" "x"lda.class=lda.pred$class
table(lda.class, crim_1.test)## crim_1.test
## lda.class 0 1
## 0 71 19
## 1 2 60mean(lda.class == crim_1.test)## [1] 0.8618421qda.fit=qda(crim_1~zn+nox+dis+rad+medv, data=Boston1, subset=train)
qda.pred = predict(qda.fit, Boston1.test)
names(qda.pred)## [1] "class" "posterior"qda.class=qda.pred$class
table(qda.class, crim_1.test)## crim_1.test
## qda.class 0 1
## 0 66 15
## 1 7 64mean(qda.class== crim_1.test)## [1] 0.8552632library(class)
train.X=cbind(nox,rad,ptratio,medv)[train,]
test.X=cbind(nox,rad,ptratio,medv)[-train,]
train.crim_1 = crim_1[train]
dim(train.X)## [1] 354 4dim(test.X)## [1] 152 4dim(train.crim_1)## NULLknn.pred=knn(train.X, test.X, train.crim_1, k=1)
table(knn.pred, crim_1.test)## crim_1.test
## knn.pred 0 1
## 0 57 6
## 1 16 73mean(knn.pred== crim_1.test)## [1] 0.8552632knn.pred=knn(train.X, test.X, train.crim_1, k=3)
table(knn.pred, crim_1.test)## crim_1.test
## knn.pred 0 1
## 0 59 11
## 1 14 68mean(knn.pred== crim_1.test)## [1] 0.8355263knn.pred=knn(train.X, test.X, train.crim_1, k=5)
table(knn.pred, crim_1.test)## crim_1.test
## knn.pred 0 1
## 0 61 12
## 1 12 67mean(knn.pred== crim_1.test)## [1] 0.8421053knn.pred=knn(train.X, test.X, train.crim_1, k=10)
table(knn.pred, crim_1.test)## crim_1.test
## knn.pred 0 1
## 0 66 21
## 1 7 58mean(knn.pred== crim_1.test)## [1] 0.8157895knn.pred=knn(train.X, test.X, train.crim_1, k=100)
table(knn.pred, crim_1.test)## crim_1.test
## knn.pred 0 1
## 0 73 39
## 1 0 40mean(knn.pred== crim_1.test)## [1] 0.7434211I see that LDA has the highest fraction of correct prediction value.