Problem 13
This question should be answered using the Weekly data set, which is part of the ISLR2 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.
knitr::opts_chunk$set(echo = TRUE)
library("stats")
library("ISLR2")
attach(Weekly)(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
#head(Weekly)
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 1.000000000Response: As expected, there are not many patterns as trading it notoriously difficult to predict. The only obvious pattern is the relationship between volume of shares a year. Telling us that as the years have gone on, the amount of shares traded have increased.
(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?
weekly_fits<-glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(weekly_fits)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## 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: 4Response: Based on the logistic regression summary, Lag2 appears to be 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.
weekly_probs <- predict(weekly_fits, type = "response")
weekly_pred <- rep("Down", 1089)
weekly_pred[weekly_probs >.5]= "Up"
table(weekly_pred, Weekly$Direction)##
## weekly_pred Down Up
## Down 54 48
## Up 430 557(557+54)/1089## [1] 0.5610652557/(557+48)## [1] 0.9206612(54)/(54+430)## [1] 0.1115702Response: Based on the results of the confusion matrix, in general we predicted the weekly trend correctly 56.11% of the time. However, we correctly predicted when the trend would go up 92.07% of the time, while only correctly predicting when it would go down 11.16% 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_2009 <-Weekly[!train,]
Weekly_fits<-glm(Direction~Lag2, data=Weekly,family=binomial, subset=train)
Weekly_prob= predict(weekly_fits, Weekly_2009, type = "response")
Weekly_pred <- rep("Down", length(Weekly_prob))
Weekly_pred[Weekly_prob > 0.5] = "Up"
Direction_2009 = Direction[!train]
table(Weekly_pred, Direction_2009)## Direction_2009
## Weekly_pred Down Up
## Down 17 13
## Up 26 48mean(Weekly_pred == Direction_2009)## [1] 0.62556/(56+5)## [1] 0.91803289/(9+34)## [1] 0.2093023(e) Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## BostonWeeklylda_fit<-lda(Direction~Lag2, data=Weekly,family=binomial, subset=train)
Weeklylda_pred<-predict(Weeklylda_fit, Weekly_2009)
table(Weeklylda_pred$class, Direction_2009)## Direction_2009
## Down Up
## Down 9 5
## Up 34 56mean(Weeklylda_pred$class == Direction_2009)## [1] 0.625(f) Repeat (d) using QDA.
Weeklyqda_fit <- qda(Direction ~ Lag2, data = Weekly, subset = train)
Weeklyqda_pred <- predict(Weeklyqda_fit, Weekly_2009)$class
table(Weeklyqda_pred, Direction_2009)## Direction_2009
## Weeklyqda_pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda_pred == Direction_2009)## [1] 0.5865385(g) Repeat (d) using KNN with K = 1.
library(class)
Week_train <- as.matrix(Lag2[train])
Week_test <- as.matrix(Lag2[!train])
train_Direction <- Direction[train]
set.seed(1)
Weekknn_pred=knn(Week_train,Week_test,train_Direction,k=1)
table(Weekknn_pred,Direction_2009)## Direction_2009
## Weekknn_pred Down Up
## Down 21 30
## Up 22 31mean(Weekknn_pred == Direction_2009)## [1] 0.5(h) Repeat (d) using naive Bayes.
library(e1071)
weeklynb_fit <- naiveBayes(Direction~Lag2 ,data=Weekly ,subset=train)
weeklynb_fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485weeklynb_class <- predict(weeklynb_fit ,Weekly_2009)
table(weeklynb_class ,Direction_2009)## Direction_2009
## weeklynb_class Down Up
## Down 0 0
## Up 43 61mean (weeklynb_class == Direction_2009)## [1] 0.5865385(i) Which of these methods appears to provide the best results on this data?
Response: The logistic regression model appears to provide the best results with being able to correctly predict the outcome 62.5% of the time.
(j) 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.
set.seed(1)
Weekknn_pred2 <- knn(Week_train,Week_test,train_Direction,k=20)
table(Weekknn_pred2,Direction_2009)## Direction_2009
## Weekknn_pred2 Down Up
## Down 21 21
## Up 22 40mean(Weekknn_pred2 == Direction_2009)## [1] 0.5865385Weeklyqda_fit2 <- qda(Direction ~ Lag2^2, data = Weekly, subset = train)
Weeklyqda_pred2 <- predict(Weeklyqda_fit2, Weekly_2009)$class
table(Weeklyqda_pred2, Direction_2009)## Direction_2009
## Weeklyqda_pred2 Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda_pred2 == Direction_2009)## [1] 0.5865385Weeklylda_fit2<-lda(Direction~Lag2:Lag3, data=Weekly,family=binomial, subset=train)
Weeklylda_pred2<-predict(Weeklylda_fit2, Weekly_2009)
table(Weeklylda_pred2$class, Direction_2009)## Direction_2009
## Down Up
## Down 0 0
## Up 43 61mean(Weeklylda_pred2$class == Direction_2009)## [1] 0.5865385detach(Weekly)Response: After creating a model using KNN with K=20, we able to increase the accuracy to 58.65%, up from 50% from the K=1 model created in part (g). Using Lag2^2 in a QDA model, gave us an accuracy of 58.65%. This is the same accuracy as the QDA model created in part (f). This LDA model with Lag2:Lag3 as the predictor, gave us an accuracy of 58.65% which is lower than the previous LDA model created in part (e).
Problem 14
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(ISLR2)
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.
mpg01 <- ifelse(Auto$mpg > median(Auto$mpg),1,0)
Auto_mpg01 <- data.frame(Auto, mpg01)(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.
Response: The graph as well as the correlation table indicate that ‘cylinder’, ‘displacement’, and ‘weight’ seem to be the most useful in predicting mpg01.
pairs(Auto_mpg01[,-9])
pairs(Auto_mpg01[,-9])
par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto_mpg01, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto_mpg01, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto_mpg01, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto_mpg01, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto_mpg01, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto_mpg01, main = "Year vs mpg01")
cor(Auto_mpg01[, -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 1.0000000cor(Auto_mpg01[, -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 1.0000000(c) Split the data into a training set and a test set.
set.seed(2)
train_x <- sample(1:nrow(Auto_mpg01), 0.8*nrow(Auto_mpg01))
test_x <- Auto_mpg01[,-train_x](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?
Response: After creating a LDA model using the variables that seemed most associated with mpg01, it gave a test accuracy of 89.54%.
library(MASS)
lda_autofit <- lda(mpg01 ~ cylinders + displacement + weight, data=Auto_mpg01)
lda_autopred <-predict(lda_autofit, test_x)$class
table(lda_autopred, test_x$mpg01)##
## lda_autopred 0 1
## 0 168 13
## 1 28 183mean(lda_autopred == test_x$mpg01)## [1] 0.8954082(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Response: After creating a QDA model using the variables that seemed most associated with mpg01, it gave a test accuracy of 90.31%.
qda_autofit <- qda(mpg01 ~ cylinders + displacement + weight, data=Auto_mpg01)
qda_autopred <- predict(qda_autofit, test_x)$class
table(qda_autopred, test_x$mpg01)##
## qda_autopred 0 1
## 0 175 17
## 1 21 179(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Response: After performing a logistic regression using the variables that seemed most associated with mpg01, it gave a test accuracy of 89.54%.
autoglm_fit <- glm(mpg01 ~ cylinders + displacement + weight, family=binomial, data=Auto_mpg01)
glm_autoprobs <- predict(autoglm_fit,test_x,type="response")
glm_autopred <- rep(0,nrow(test_x))
glm_autopred[glm_autoprobs > 0.50]=1
table(glm_autopred, test_x$mpg01)##
## glm_autopred 0 1
## 0 170 15
## 1 26 181mean(glm_autopred == test_x$mpg01)## [1] 0.8954082(g) Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Response: After performing naive Bayes on the training data using cylinders, displacement, and weight to predict mpg01, we got a accuracy of 88.01%.
library(e1071)
nb_autofit <- naiveBayes(mpg01~ cylinders + displacement + weight, data=Auto_mpg01)
nb_autoclass <- predict(nb_autofit, test_x)## Warning in predict.naiveBayes(nb_autofit, test_x): Type mismatch between
## training and new data for variable 'cylinders'. Did you use factors with
## numeric labels for training, and numeric values for new data?## Warning in predict.naiveBayes(nb_autofit, test_x): Type mismatch between
## training and new data for variable 'displacement'. Did you use factors with
## numeric labels for training, and numeric values for new data?table(nb_autoclass, test_x$mpg01)##
## nb_autoclass 0 1
## 0 165 16
## 1 31 180mean(nb_autoclass == test_x$mpg01)## [1] 0.880102(h) Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
Response: After performing KNN on the training data with K=29, gives the an accuracy of 86.08%, whereas K=1 gives 84.81%
library(class)
set.seed(3)
idx <- sample(1:nrow(Auto), 0.8*nrow(Auto))
train_auto <- Auto[idx,]
test_auto <- Auto[-idx,]
y_train <- train_auto$mpg
x_train <- train_auto[,-1]
x_test <- test_auto[,-1]
x_train <- x_train[,-8]
x_test <- x_test[,-8]
y_test<- ifelse(test_auto$mpg>=23,1,0)
high_low <- ifelse(y_train>=23,1,0)
knn_pred <- knn (train=x_train, test=x_test, cl=high_low, k=1)
table(knn_pred, y_test)## y_test
## knn_pred 0 1
## 0 33 3
## 1 9 34mean(knn_pred == y_test)## [1] 0.8481013set.seed(3)
knn_pred <- knn (train=x_train, test=x_test, cl=high_low, k=29)
table(knn_pred, y_test)## y_test
## knn_pred 0 1
## 0 34 3
## 1 8 34mean(knn_pred == y_test)## [1] 0.8607595Problem 16
16. Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
library(ISLR2)
attach(Boston)Crime <- rep(0, length(crim))
Crime[crim > median(crim)] <- 1
Boston01 <- data.frame(Boston, Crime)train <- 1:(dim(Boston01)[1]/2)
test <- (dim(Boston01)[1]/2 + 1):dim(Boston01)[1]
Boston_train <- Boston01[train, ]
Boston_test <- Boston01[test, ]
Crime_test <- Crime[test]
cor(Boston01)## 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
## Crime 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
## Crime -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv Crime
## 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
## Crime -0.35121093 0.4532627 -0.2630167 1.00000000Using the logistic regression model with the predictor variables ‘nox’, ‘rad’, ‘tax’, ‘indus’, and ‘chas’, we got an accuracy of 91.70%.
set.seed(1)
glm_bostonfit <-glm(Crime~ nox+rad+tax+indus+chas, data = Boston_train, family = 'binomial')
Boston_probs <- predict(glm_bostonfit, Boston_test, type = 'response')
Boston_pred <- rep(0, length(Boston_probs))
Boston_pred[Boston_probs > 0.5] = 1
table(Boston_pred, Crime_test)## Crime_test
## Boston_pred 0 1
## 0 75 6
## 1 15 157mean(Boston_pred == Crime_test)## [1] 0.916996Using the Linear Discriminant Analysis model, we got an accuracy of 88.93%, which is lower than the logistic regression model.
library(MASS)
lda_bostonfit <-lda(Crime~ nox+rad+tax+indus+chas, data= Boston_train)
Bostonlda_pred <- predict(lda_bostonfit, Boston_test)
table(Bostonlda_pred$class, Crime_test)## Crime_test
## 0 1
## 0 80 18
## 1 10 145mean(Bostonlda_pred$class == Crime_test)## [1] 0.8893281After testing the QDA model using a few different predictor variables, I got an accuracy of 45.85%, which is extremely lower than the accuracy of the LDA and Logistic regression model.
qda_bostonfit <-qda(Crime~ nox*rad+tax+indus*chas, data= Boston_train)
Bostonqda_pred <- predict(qda_bostonfit, Boston_test)
table(Bostonqda_pred$class, Crime_test)## Crime_test
## 0 1
## 0 83 130
## 1 7 33Using the KNN model, with the variables ‘indus’, ‘nox’, ‘age’, ‘dis’, ‘rad’, and ‘tax’ with K=10, we got an accuracy of 78.66%, which is lower than the LDA and Logistic regression model but better than the QDA model.
library(class)
set.seed(1)
train_knn <- cbind(indus,nox,age,dis,rad,tax)[train,]
test_knn <- cbind(indus,nox,age,dis,rad,tax)[test,]
Bostonknn_pred <- knn(train_knn, test_knn, Crime_test, k=10)
table(Bostonknn_pred, Crime_test)## Crime_test
## Bostonknn_pred 0 1
## 0 44 8
## 1 46 155mean(Bostonknn_pred == Crime_test)## [1] 0.7865613