STA6543 - Assignment 3
Felicia Villela
Due Date: 6/26/2020
ISLR Chapter 4: Classification
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.
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)
attach(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 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
##
##
##
## str(Weekly)## 'data.frame': 1089 obs. of 9 variables:
## $ Year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
## $ Lag1 : num 0.816 -0.27 -2.576 3.514 0.712 ...
## $ Lag2 : num 1.572 0.816 -0.27 -2.576 3.514 ...
## $ Lag3 : num -3.936 1.572 0.816 -0.27 -2.576 ...
## $ Lag4 : num -0.229 -3.936 1.572 0.816 -0.27 ...
## $ Lag5 : num -3.484 -0.229 -3.936 1.572 0.816 ...
## $ Volume : num 0.155 0.149 0.16 0.162 0.154 ...
## $ Today : num -0.27 -2.576 3.514 0.712 1.178 ...
## $ Direction: Factor w/ 2 levels "Down","Up": 1 1 2 2 2 1 2 2 2 1 ...pairs(Weekly)
cor(Weekly[1:8])## 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 1.000000000Volume is the only predictor that shows a correlation to Year of .84194. The other variables do not show a relationship.
plot(Volume)
The plot of Volume indicates an increasing relationship.
(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, family = binomial)
summary(glm_fit)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial)
##
## 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: 4Based on the model, Lag2 appears to be statistically significant with a p-value of 0.0296. It is the only variable that is statistically significant.
(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_pred<-rep("Down", length(glm_probs))
glm_pred[glm_probs > 0.5] = "Up"
table(glm_pred,Direction)## Direction
## glm_pred Down Up
## Down 54 48
## Up 430 557Based on the glm_pred output, the percentage of correct predictions is \((\frac{54+557}{1089}) = 56.1065\)%. The confusion matrix shows the logistic regression model erroneously predicted a Up week, when it was Down, and only accurately predicted a Down week 11.1157% of the time. Therefore when the market is up, the model is correct 92.066% of the time.
(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_20092010<-Weekly[!train,]
Direction_20092010<- Direction[!train]glm_fit2<-glm(Direction ~ Lag2, family = binomial, subset = train)
summary(glm_fit2)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4glm_probs2<-predict(glm_fit2, Weekly_20092010, type = "response")
glm_pred2<-rep("Down", length(glm_probs2))
glm_pred2[glm_probs2>0.5] = "Up"
table(glm_pred2,Direction_20092010)## Direction_20092010
## glm_pred2 Down Up
## Down 9 5
## Up 34 56mean(glm_pred2 == Direction_20092010)## [1] 0.625The percentage of correct prediction for the held out data is \(\frac{9+56}{104}=62.5\)%.
(e) Repeat (d) using LDA.
library(MASS)lda_fit<- lda(Direction~ Lag2, subset = train)
lda_fit## Call:
## lda(Direction ~ Lag2, 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 0.4414162lda_pred<- predict(lda_fit, Weekly_20092010)
table(lda_pred$class, Direction_20092010)## Direction_20092010
## Down Up
## Down 9 5
## Up 34 56mean(lda_pred$class == Direction_20092010)## [1] 0.625The percentage of correct prediction for the held out data remains \(\frac{9+56}{104}=62.5\)%.
(f) Repeat (d) using QDA.
qda_fit<- qda(Direction ~ Lag2, subset = train)
qda_fit## Call:
## qda(Direction ~ Lag2, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda_pred<- predict(qda_fit, Weekly_20092010)
table(qda_pred$class, Direction_20092010)## Direction_20092010
## Down Up
## Down 0 0
## Up 43 61mean(qda_pred$class == Direction_20092010)## [1] 0.5865385The percentage of correct prediction for the held out data is \(\frac{0+61}{104}=58.654\)%.
(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]
set.seed(1)
knn_pred <- knn(train.x, test.x, train.Direction, k=1)
table(knn_pred, Direction_20092010)## Direction_20092010
## knn_pred Down Up
## Down 21 30
## Up 22 31mean(knn_pred == Direction_20092010)## [1] 0.5The percentage of correct prediction for the held out data is \(\frac{21+31}{104}=50\)%.
(h) Which of these methods appears to provide the best results on this data?
The logistic regression and the LDA models have the highest rates of corrext prediction, both at 62.5%, which would be considered the better models.
(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.
#Logistic regression with Lag2:Lag1
glm_fit3 <- glm(Direction ~ Lag2:Lag1, family = binomial, subset = train)
glm_probs3 <- predict(glm_fit3, Weekly_20092010, type = "response")
glm_pred3 <- rep("Down", length(glm_probs3))
glm_pred3[glm_probs3 > 0.5] = "Up"
Direction2_20092010 <- Direction[!train]
table(glm_pred3, Direction2_20092010)## Direction2_20092010
## glm_pred3 Down Up
## Down 1 1
## Up 42 60mean(glm_pred3 == Direction2_20092010)## [1] 0.5865385#LDA with Lag2 interaction with Lag1
lda_fit2 <- lda(Direction ~ Lag2:Lag1, subset = train)
lda_pred2 <- predict(lda_fit2, Weekly_20092010)
lda_fit2## Call:
## lda(Direction ~ Lag2:Lag1, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2:Lag1
## Down -0.8014495
## Up -0.1393632
##
## Coefficients of linear discriminants:
## LD1
## Lag2:Lag1 0.1013404mean(lda_pred2$class == Direction2_20092010)## [1] 0.5769231#KNN k =10
knn_pred2 <- knn(train.x, test.x, train.Direction, k = 10)
table(knn_pred2, Direction2_20092010)## Direction2_20092010
## knn_pred2 Down Up
## Down 17 18
## Up 26 43mean(knn_pred2 == Direction2_20092010)## [1] 0.5769231The original logistic regression model glm_pred2 and original lda_pred remain the highest in percentage of correct prediction at 62.5%.
detach(Weekly)Problem 11.
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
library(ISLR)
data(Auto)
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg01 <- rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] = 1
Auto2 <- data.frame(Auto,mpg01)
attach(Auto2)(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)
par(mfrow=c(2,2))
boxplot(mpg~cylinders)
plot(mpg~horsepower)
plot(mpg~acceleration)
(c) Split the data into a training set and a test set.
train <- (year%%2 ==0)
test <- !train
Auto2.train <- Auto2[train,]
Auto2.test <- Auto2[test,]
mpg01.test <- mpg01[test](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_fit3 <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, subset = train)
lda_pred3 = predict(lda_fit3, Auto2.test)
mean(lda_pred3$class != mpg01.test)## [1] 0.1263736The test error of this LDA model is 12.64%.
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda_fit3<- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
qda_pred3 <- predict(qda_fit3, Auto2.test)
mean(qda_pred3$class != mpg01.test)## [1] 0.1318681The test error of this QDA model is 13.19%.
(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_fit4 <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, family = binomial, subset = train)
glm_probs4 <- predict(glm_fit4, Auto2.test, type = "response")
glm_pred4 <- rep(0, length(glm_probs4))
glm_pred4[glm_probs4 > 0.5] = 1
mean(glm_pred4 != mpg01.test)## [1] 0.1208791The test error of this logistic regression model is 12.09%.
(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.x2 <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.x2 <- cbind(cylinders, weight, displacement, horsepower)[test, ]
train.mpg01 <- mpg01[train]
set.seed(1)
#KNN(k=1)
knn_pred3 = knn(train.x2, test.x2, train.mpg01, k = 1)
table(knn_pred3, mpg01.test)## mpg01.test
## knn_pred3 0 1
## 0 83 11
## 1 17 71mean(knn_pred3 != mpg01.test)## [1] 0.1538462#KNN(k=10)
knn_pred4 = knn(train.x2, test.x2, train.mpg01, k = 10)
table(knn_pred4, mpg01.test)## mpg01.test
## knn_pred4 0 1
## 0 77 7
## 1 23 75mean(knn_pred4 != mpg01.test)## [1] 0.1648352#KNN(k=100)
knn_pred5 = knn(train.x2, test.x2, train.mpg01, k = 100)
table(knn_pred5, mpg01.test)## mpg01.test
## knn_pred5 0 1
## 0 81 7
## 1 19 75mean(knn_pred5 != mpg01.test)## [1] 0.1428571Based on the results of the KNN method, for k=1, there is a 15.38% test error rate. For k=10, there is a 16.48% test error rate. For k=100, there is a 14.29% test error rate, which is the lowest error rate and therefore would be the best.
detach(Auto2)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)
data(Boston)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 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. :50.00crime01 <- rep(0, length(Boston$crim))
crime01[Boston$crim > median(Boston$crim)] = 1
Boston <-data.frame(Boston, crime01)
attach(Boston)pairs(Boston)
cor(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
## crime01 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## 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
## crime01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crime01
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crime01 -0.35121093 0.4532627 -0.2630167 1.00000000train2 <- 1:(dim(Boston)[1]/2)
test2 <- (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston_train <- Boston[train2, ]
Boston_test <- Boston[test2, ]
crime01_test <- crime01[test2]#logistic regression
glm_fit5 <- glm(crime01 ~ . - crime01 - crim, family = binomial, data = Boston, subset = train2)#logistic regression
glm_probs5 <- predict(glm_fit5, Boston_test, type = "response")
glm_pred5 <- rep(0, length(glm_probs5))
glm_pred5[glm_probs5 > 0.5] = 1
mean(glm_pred5 != crime01_test)## [1] 0.1818182#logistic regression
glm_fit6 <- glm(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, family = binomial, subset = train2)#logistic regression
glm_probs6 <- predict(glm_fit6, Boston_test, type = "response")
glm_pred6 <- rep(0, length(glm_probs6))
glm_pred6[glm_probs6 > 0.5] = 1
mean(glm_pred6 != crime01_test)## [1] 0.1857708#LDA
lda_fit4 <- lda(crime01 ~ . - crime01 - crim, data = Boston, subset = train2)
lda_pred4 <- predict(lda_fit4, Boston_test)
mean(lda_pred4$class != crime01_test)## [1] 0.1343874#LDA
lda_fit5 <- lda(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, subset = train2)
lda_pred5 <- predict(lda_fit5, Boston_test)
mean(lda_pred5$class != crime01_test)## [1] 0.1225296#LDA
lda_fit6 <- lda(crime01 ~ . - crime01 - crim - chas - tax - lstat - indus - age, data = Boston, subset = train2)
lda_pred6 <- predict(lda_fit6, Boston_test)
mean(lda_pred6$class != crime01_test)## [1] 0.1185771#KNN
library(class)
train.X3 <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train2, ]
test.X3 <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test2, ]
train_crime01 <- crime01[train2]
set.seed(1)
#KNN(k=1)
knn_pred6 <- knn(train.X3, test.X3, train_crime01, k = 1)
mean(knn_pred6 != crime01_test)## [1] 0.458498#KNN(k=10)
knn_pred7 <- knn(train.X3, test.X3, train_crime01, k = 10)
mean(knn_pred7 != crime01_test)## [1] 0.1185771#KNN(k=100)
knn_pred8 <- knn(train.X3, test.X3, train_crime01, k = 100)
mean(knn_pred8 != crime01_test)## [1] 0.4901186Based on the model output, lda_fit6 and KNN of k=10 are the lowest error rate at 11.86%.
detach(Boston)