Statistical Data Analysis 과제3
chaemj
2024-04-16
Weekly
- 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.
library(ISLR2)## Warning: 패키지 'ISLR2'는 R 버전 4.4.0에서 작성되었습니다library(MASS)##
## 다음의 패키지를 부착합니다: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonhead(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Down- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
#기초 통계량
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
##
##
##
## #scatter plot
pairs(Weekly)
#상관계수
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 1.000000000그래프와 상관계수수 보면 year와 volume을 제외한 다른 변수들은 서로 상관관계가 강하지 않음을 알 수 있음. volume&year의 그래프를 보면 시간이 지날수록 증가하는 추세를 보임. 상관계수도 0.842로 1과 가까움. 다른 변수들 간의 상관계수의 절대값들은 0.1을 넘지 않음.
- 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?
wlogis<-glm(Direction~.-Year-Today,data=Weekly,family=binomial)
summary(wlogis)##
## Call:
## glm(formula = Direction ~ . - Year - Today, 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: 4- 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.
wlogis.probs<-predict(wlogis, type="response")
wlogis.pred<-rep("Down",1089)
wlogis.pred[wlogis.probs>.5]<-"Up"
table(wlogis.pred,Weekly$Direction)##
## wlogis.pred Down Up
## Down 54 48
## Up 430 557#정확도
mean(wlogis.pred==Weekly$Direction)## [1] 0.5610652# Up일 때의 정확도
557/(430+557)## [1] 0.5643364#Down일 때의 정확도
54/(54+48)## [1] 0.5294118전체 정확도는 56.1%이고 Up을 예측할 때의 정확도는 56.4%로 Down을 예측할 때의 정확도인 53%보다 정확도가 더 높다.
- 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).
weekly_train=(Weekly$Year<2009)
wlogis_split=glm(Direction~ Lag2, data=Weekly, subset=weekly_train,family="binomial")
summary(wlogis_split)##
## Call:
## glm(formula = Direction ~ Lag2, family = "binomial", data = Weekly,
## subset = weekly_train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4wlogis_split.probs<-predict(wlogis_split,Weekly[!weekly_train,], type="response")
wlogis_split.pred<-rep("Down",dim(Weekly[!weekly_train,])[1])
wlogis_split.pred[wlogis_split.probs>.5]<-"Up"
table(wlogis_split.pred,Weekly[!weekly_train, ]$Direction)##
## wlogis_split.pred Down Up
## Down 9 5
## Up 34 56#정확도
mean(wlogis_split.pred==Weekly[!weekly_train, ]$Direction)## [1] 0.625# Up일 때의 정확도
56/(56+34)## [1] 0.6222222#Down일 때의 정확도
9/(9+5)## [1] 0.6428571전체 정확도는 62.5%이고 Up을 예측할 때의 정확도는 62.2%로 Down을 예측할 때의 정확도인 64.3%보다 정확도가 더 낮다.
- Repeat (d) using LDA
wlda<-lda(Direction~ Lag2, data=Weekly, subset=weekly_train)
wlda## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = weekly_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162wlda.pred<-predict(wlda,Weekly[!weekly_train,])
table(wlda.pred$class,Weekly[!weekly_train,]$Direction)##
## Down Up
## Down 9 5
## Up 34 56#정확도
mean(wlda.pred$class==Weekly[!weekly_train, ]$Direction)## [1] 0.625- Repeat (d) using QDA
wqda<-qda(Direction~ Lag2, data=Weekly, subset=weekly_train)
wqda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = weekly_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581wqda.pred<-predict(wqda,Weekly[!weekly_train,])
table(wqda.pred$class,Weekly[!weekly_train,]$Direction)##
## Down Up
## Down 0 0
## Up 43 61#정확도
mean(wqda.pred$class==Weekly[!weekly_train, ]$Direction)## [1] 0.5865385- Which of these methods appears to provide the best results on this data?
cat('Logistic Regression',mean(wlogis_split.pred==Weekly[!weekly_train, ]$Direction))## Logistic Regression 0.625cat('LDA',mean(wlda.pred$class==Weekly[!weekly_train, ]$Direction))## LDA 0.625cat('QDA',mean(wqda.pred$class==Weekly[!weekly_train, ]$Direction))## QDA 0.5865385정확도가 가장 높은 로지스틱 회귀분석과 LDA가 가장 좋은 결과를 내는 방법이다.
- 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.
wlogis_inter<-glm(Direction ~ Lag1 * Lag2, data = Weekly, family = "binomial", subset = weekly_train)
summary(wlogis_inter)##
## Call:
## glm(formula = Direction ~ Lag1 * Lag2, family = "binomial", data = Weekly,
## subset = weekly_train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.211419 0.064589 3.273 0.00106 **
## Lag1 -0.051505 0.030727 -1.676 0.09370 .
## Lag2 0.053471 0.029193 1.832 0.06700 .
## Lag1:Lag2 0.001921 0.007460 0.257 0.79680
## ---
## 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: 1346.9 on 981 degrees of freedom
## AIC: 1354.9
##
## Number of Fisher Scoring iterations: 4wlogis_inter.probs<-predict(wlogis_inter,Weekly[!weekly_train,], type="response")
wlogis_inter.pred<-rep("Down",dim(Weekly[!weekly_train,])[1])
wlogis_inter.pred[wlogis_inter.probs>.5]<-"Up"
table(wlogis_inter.pred,Weekly[!weekly_train, ]$Direction)##
## wlogis_inter.pred Down Up
## Down 7 8
## Up 36 53mean(wlogis_inter.pred==Weekly[!weekly_train, ]$Direction)## [1] 0.5769231wlda_inter<-lda(Direction~ Lag1*Lag2, data=Weekly, subset=weekly_train)
wlda_inter## Call:
## lda(Direction ~ Lag1 * Lag2, data = Weekly, subset = weekly_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2 Lag1:Lag2
## Down 0.289444444 -0.03568254 -0.8014495
## Up -0.009213235 0.26036581 -0.1393632
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.285484602
## Lag2 0.295080109
## Lag1:Lag2 0.009629381wlda_inter.pred<-predict(wlda_inter,Weekly[!weekly_train,])
table(wlda_inter.pred$class,Weekly[!weekly_train,]$Direction)##
## Down Up
## Down 7 8
## Up 36 53mean(wlda_inter.pred$class==Weekly[!weekly_train, ]$Direction)## [1] 0.5769231wqda_inter<-qda(Direction~ Lag1*Lag2, data=Weekly, subset=weekly_train)
wqda_inter## Call:
## qda(Direction ~ Lag1 * Lag2, data = Weekly, subset = weekly_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2 Lag1:Lag2
## Down 0.289444444 -0.03568254 -0.8014495
## Up -0.009213235 0.26036581 -0.1393632wqda_inter.pred<-predict(wqda_inter,Weekly[!weekly_train,])
table(wqda_inter.pred$class,Weekly[!weekly_train,]$Direction)##
## Down Up
## Down 23 36
## Up 20 25mean(wqda_inter.pred$class==Weekly[!weekly_train, ]$Direction)## [1] 0.4615385로지스틱 회귀분석에서는 interaction term의 유의성이 높지않고, 3개 모델 모두 interaction term을 포함시킨 것의 정확도가 더 낮다.
Auto
- 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)
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500변수 설명
mpg: 연비
cylinders: 엔진 실린더 개수
displacement: 배기량
horsepower: 마력(hp)
weight: 자동차 무게
acceleration: 초당 가속력
year: 출시 년도
origin: 제조국(1: 미국, 2: 유럽, 3: 일본)
car name: 자동차 이름
- 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<-Auto[,-9] # 자동차 이름 열 제거
Autoplus<- data.frame(Auto, mpg01)
head(Autoplus)## mpg cylinders displacement horsepower weight acceleration year origin mpg01
## 1 18 8 307 130 3504 12.0 70 1 0
## 2 15 8 350 165 3693 11.5 70 1 0
## 3 18 8 318 150 3436 11.0 70 1 0
## 4 16 8 304 150 3433 12.0 70 1 0
## 5 17 8 302 140 3449 10.5 70 1 0
## 6 15 8 429 198 4341 10.0 70 1 0- 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.
summary(Autoplus)## 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 mpg01
## Min. : 8.00 Min. :70.00 Min. :1.000 Min. :0.0
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 1st Qu.:0.0
## Median :15.50 Median :76.00 Median :1.000 Median :0.5
## Mean :15.54 Mean :75.98 Mean :1.577 Mean :0.5
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 3rd Qu.:1.0
## Max. :24.80 Max. :82.00 Max. :3.000 Max. :1.0cor(Autoplus)## 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 1.0000000pairs(Autoplus)
par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Autoplus, main = "Cylinders & mpg01")
boxplot(displacement ~ mpg01, data = Autoplus, main = "Displacement & mpg01")
boxplot(horsepower ~ mpg01, data = Autoplus, main = "Horsepower & mpg01")
boxplot(weight ~ mpg01, data = Autoplus, main = "Weight & mpg01")
boxplot(acceleration ~ mpg01, data = Autoplus, main = "Acceleration & mpg01")
boxplot(year ~ mpg01, data = Autoplus, main = "Year & mpg01")
위의 수치들과 그래프들을 종합해보면 mpg01을 예측할 때 가장 유용한 변수는
mpg, cylinders, displacement, horsepower, weight이다.
- Split the data into a training set and test set.
set.seed(2)
train<-sample(nrow(Autoplus)*0.7)
auto_train<-Autoplus[train, ]
auto_test<-Autoplus[-train, ]- 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?
alda <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Autoplus, subset = train)
alda## Call:
## lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Autoplus,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.6423358 0.3576642
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.784091 275.6250 131.68750 3637.381
## 1 4.040816 106.4541 78.04082 2227.612
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.446091128
## displacement -0.002075782
## horsepower 0.011483399
## weight -0.001146243alda.pred<-predict(alda,auto_test)
table(alda.pred$class,auto_test$mpg01)##
## 0 1
## 0 18 16
## 1 2 82mean(alda.pred$class==auto_test$mpg01)## [1] 0.8474576- 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?
aqda <- qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Autoplus, subset = train)
aqda## Call:
## qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Autoplus,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.6423358 0.3576642
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.784091 275.6250 131.68750 3637.381
## 1 4.040816 106.4541 78.04082 2227.612aqda.pred<-predict(aqda,auto_test)
table(aqda.pred$class,auto_test$mpg01)##
## 0 1
## 0 18 19
## 1 2 79mean(aqda.pred$class==auto_test$mpg01)## [1] 0.8220339- 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?
alogis <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, data = Autoplus, subset = train, family="binomial")
summary(alogis)##
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower +
## weight, family = "binomial", data = Autoplus, subset = train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 15.550194 2.878745 5.402 6.6e-08 ***
## cylinders -0.631193 0.639513 -0.987 0.3236
## displacement -0.007109 0.014815 -0.480 0.6313
## horsepower -0.044299 0.023190 -1.910 0.0561 .
## weight -0.002898 0.001142 -2.538 0.0112 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 357.33 on 273 degrees of freedom
## Residual deviance: 105.85 on 269 degrees of freedom
## AIC: 115.85
##
## Number of Fisher Scoring iterations: 8alogis.probs<-predict(alogis,auto_test, type="response")
alogis.pred<-rep(0,nrow(auto_test))
alogis.pred[alogis.probs>.5]=1
table(alogis.pred,auto_test$mpg01)##
## alogis.pred 0 1
## 0 20 21
## 1 0 77mean(alogis.pred==auto_test$mpg01)## [1] 0.8220339LDA 방법의 정확도가 84.7%로 가장 높음.
Boston
- 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. Hint: You will have to create the response variable yourself, using the variables that are contained in Boston data set.
library(ISLR2)
head(Boston)## crim zn indus chas nox rm age dis rad tax ptratio black lstat
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 396.90 4.98
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 396.90 9.14
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 392.83 4.03
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 394.63 2.94
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 396.90 5.33
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 394.12 5.21
## medv
## 1 24.0
## 2 21.6
## 3 34.7
## 4 33.4
## 5 36.2
## 6 28.7변수 설명
crim: 마을별 1인당 범죄율
zn: 25,000 평방피트이상인 주거용 토지의 비율
indus: 마을 당 비소매상업부지의의 비율
chas: 찰스강에 대한 더미변수(강 경계에 위치한 경우 1, 아니면 0)
nox: 10ppm 당 일산화질소 농도
rm: 거주지당 평균 방의 개수
age:1940년 이전에 건축된 주택의 자가거주 비율
dis: 5개의 보스턴 직업센터까지의 접근성 지수
rad: 방사형 도로까지의 접근성 지수
tax: 10,000 달러 당 재산세율
ptratio: 마을별 학생/교사 비율
black: 마을별 흑인의 비율율 lstat: 저소득계층의 비율
medv: 자가주택의 가격의 중앙값
- Create a binary variable, crim01, that contains a 1 if crim contains a value above its median, and a 0 if crim contains a value below its median.
crim01 <- ifelse(Boston$crim > median(Boston$crim),1,0)
bostonplus<- data.frame(Boston, crim01)
head(bostonplus)## crim zn indus chas nox rm age dis rad tax ptratio black lstat
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 396.90 4.98
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 396.90 9.14
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 392.83 4.03
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 394.63 2.94
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 396.90 5.33
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 394.12 5.21
## medv crim01
## 1 24.0 0
## 2 21.6 0
## 3 34.7 0
## 4 33.4 0
## 5 36.2 0
## 6 28.7 0- Explore the data graphically in order to investigate the association between crim01 and the other features.
summary(bostonplus)## 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 crim01
## 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. :1.0cor(bostonplus)## 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
## crim01 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
## crim01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crim01
## 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
## crim01 -0.35121093 0.4532627 -0.2630167 1.00000000pairs(bostonplus)
par(mfrow=c(3,5))
boxplot(zn ~ crim01, data = bostonplus, main = "zn & crim01")
boxplot(indus ~ crim01, data = bostonplus, main = "indus & crim01")
boxplot(chas ~ crim01, data = bostonplus, main = "chas & crim01")
boxplot(nox ~ crim01, data = bostonplus, main = "nox & crim01")
boxplot(rm ~ crim01, data = bostonplus, main = "rm & crim01")
boxplot(age ~ crim01, data = bostonplus, main = "age & crim01")
boxplot(dis ~ crim01, data = bostonplus, main = "dis & crim01")
boxplot(rad ~ crim01, data = bostonplus, main = "rad & crim01")
boxplot(tax ~ crim01, data = bostonplus, main = "tax & crim01")
boxplot(ptratio ~ crim01, data = bostonplus, main = "ptratio & crim01")
boxplot(black ~ crim01, data = bostonplus, main = "black & crim01")
boxplot(lstat ~ crim01, data = bostonplus, main = "lstat & crim01")
boxplot(medv ~ crim01, data = bostonplus, main = "medv & crim01")
위의 수치들과 그래프들을 종합해보면 crim01을 예측할 때 가장 유용한 변수는 indus, nox, age, dis, rad, tax이다.
- Split the data into a training set and test set.
set.seed(2)
btrain<-sample(nrow(bostonplus)*0.7)
boston_train<-bostonplus[btrain, ]
boston_test<-bostonplus[-btrain, ]- Perform LDA on the training data in order to predict crim01 using all variables.
#Logistic Regression
blogis_a <- glm(crim01 ~ ., data = bostonplus, subset = btrain, family="binomial")## Warning: glm.fit: 알고리즘이 수렴하지 않았습니다## Warning: glm.fit: 적합된 확률값들이 0 또는 1 입니다summary(blogis_a)##
## Call:
## glm(formula = crim01 ~ ., family = "binomial", data = bostonplus,
## subset = btrain)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.604e+16 1.044e+08 -153663445 <2e-16 ***
## crim 4.369e+15 9.794e+06 446063648 <2e-16 ***
## zn -1.952e+13 2.074e+05 -94116146 <2e-16 ***
## indus -4.772e+12 9.287e+05 -5139070 <2e-16 ***
## chas -1.914e+14 1.398e+07 -13689582 <2e-16 ***
## nox 4.185e+15 7.905e+07 52945376 <2e-16 ***
## rm -5.043e+13 1.382e+07 -3650155 <2e-16 ***
## age -9.634e+11 2.176e+05 -4427362 <2e-16 ***
## dis -6.701e+13 3.301e+06 -20302018 <2e-16 ***
## rad 3.366e+14 2.428e+06 138666512 <2e-16 ***
## tax -5.552e+12 6.564e+04 -84583168 <2e-16 ***
## ptratio 2.224e+14 2.210e+06 100596897 <2e-16 ***
## black 2.189e+13 1.064e+05 205791141 <2e-16 ***
## lstat 2.303e+13 1.061e+06 21703614 <2e-16 ***
## medv 1.708e+13 1.189e+06 14365726 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 450.65 on 353 degrees of freedom
## Residual deviance: 648.79 on 339 degrees of freedom
## AIC: 678.79
##
## Number of Fisher Scoring iterations: 25blogis_a.probs<-predict(blogis_a,boston_test, type="response")
blogis_a.pred<-rep(0,nrow(boston_test))
blogis_a.pred[blogis_a.probs>.5]=1
table(blogis_a.pred,boston_test$crim01)##
## blogis_a.pred 0 1
## 0 12 0
## 1 5 135mean(blogis_a.pred==boston_test$crim01)## [1] 0.9671053모든 변수의 추정 계수가 통계적으로 유의하다고 나옴. 정확도도 96.7%로 높게 나옴.
#LDA
blda_a <- lda(crim01 ~ ., data = bostonplus, subset = btrain)
blda_a## Call:
## lda(crim01 ~ ., data = bostonplus, subset = btrain)
##
## Prior probabilities of groups:
## 0 1
## 0.6666667 0.3333333
##
## Group means:
## crim zn indus chas nox rm age
## 0 0.09301674 22.398305 6.444746 0.05508475 0.4639606 6.426911 49.83305
## 1 1.01016424 2.576271 12.248814 0.11864407 0.6017458 6.357703 82.00847
## dis rad tax ptratio black lstat medv
## 0 5.225892 4.190678 295.8051 17.73305 388.9124 9.134195 25.45847
## 1 2.982705 5.033898 340.0847 17.66102 363.6945 12.840763 24.25085
##
## Coefficients of linear discriminants:
## LD1
## crim 1.7379398567
## zn 0.0002354896
## indus -0.0102358034
## chas 0.0336838021
## nox 3.9093069414
## rm 0.4472964829
## age 0.0098605632
## dis -0.0532252212
## rad 0.1345181399
## tax 0.0006061110
## ptratio 0.1990744467
## black 0.0030884882
## lstat 0.0030285219
## medv 0.0157608224blda_a.pred<-predict(blda_a,boston_test)
table(blda_a.pred$class,boston_test$crim01)##
## 0 1
## 0 17 3
## 1 0 132mean(blda_a.pred$class==boston_test$crim01)## [1] 0.9802632정확도가 98%로 높게 나옴.
#QDA
bqda_a <- qda(crim01 ~ ., data = bostonplus, subset = btrain)
bqda_a## Call:
## qda(crim01 ~ ., data = bostonplus, subset = btrain)
##
## Prior probabilities of groups:
## 0 1
## 0.6666667 0.3333333
##
## Group means:
## crim zn indus chas nox rm age
## 0 0.09301674 22.398305 6.444746 0.05508475 0.4639606 6.426911 49.83305
## 1 1.01016424 2.576271 12.248814 0.11864407 0.6017458 6.357703 82.00847
## dis rad tax ptratio black lstat medv
## 0 5.225892 4.190678 295.8051 17.73305 388.9124 9.134195 25.45847
## 1 2.982705 5.033898 340.0847 17.66102 363.6945 12.840763 24.25085bqda_a.pred<-predict(bqda_a,boston_test)
table(bqda_a.pred$class,boston_test$crim01)##
## 0 1
## 0 16 0
## 1 1 135mean(bqda_a.pred$class==boston_test$crim01)## [1] 0.9934211정확도가 99.3%로 매우 높게 나옴. 모든 변수들을 포함한 모델을 사용할 때 QDA 방법의 정확도가 가장 높음.
- Perform LDA on the training data in order to predict crim01 using the variables that seemed most associated with crim01 in (b).
#Logistic Regression
blogis <- glm(crim01 ~ indus + nox + age + dis + rad + tax, data = bostonplus, subset = btrain, family="binomial")
summary(blogis)##
## Call:
## glm(formula = crim01 ~ indus + nox + age + dis + rad + tax, family = "binomial",
## data = bostonplus, subset = btrain)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -28.270990 4.318242 -6.547 5.88e-11 ***
## indus -0.088496 0.049471 -1.789 0.0736 .
## nox 50.296106 8.470018 5.938 2.88e-09 ***
## age 0.003096 0.009206 0.336 0.7366
## dis 0.168815 0.151746 1.112 0.2659
## rad 0.556877 0.129065 4.315 1.60e-05 ***
## tax -0.003906 0.003162 -1.235 0.2167
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 450.65 on 353 degrees of freedom
## Residual deviance: 214.13 on 347 degrees of freedom
## AIC: 228.13
##
## Number of Fisher Scoring iterations: 7blogis.probs<-predict(blogis,boston_test, type="response")
blogis.pred<-rep(0,nrow(boston_test))
blogis.pred[blogis.probs>.5]=1
table(blogis.pred,boston_test$crim01)##
## blogis.pred 0 1
## 0 12 0
## 1 5 135mean(blogis.pred==boston_test$crim01)## [1] 0.9671053모든 변수를 포함한 모델과 정확도는 동일하지만, 각 계수들의 통계적 유의성이 더 감소함.(p-value가 더 큼.)
#LDA
blda <- lda(crim01 ~ indus + nox + age + dis + rad + tax, data = bostonplus, subset = btrain)
blda## Call:
## lda(crim01 ~ indus + nox + age + dis + rad + tax, data = bostonplus,
## subset = btrain)
##
## Prior probabilities of groups:
## 0 1
## 0.6666667 0.3333333
##
## Group means:
## indus nox age dis rad tax
## 0 6.444746 0.4639606 49.83305 5.225892 4.190678 295.8051
## 1 12.248814 0.6017458 82.00847 2.982705 5.033898 340.0847
##
## Coefficients of linear discriminants:
## LD1
## indus 0.002900596
## nox 8.572070230
## age 0.011460414
## dis -0.063449167
## rad 0.185990960
## tax 0.001364278blda.pred<-predict(blda,boston_test)
table(blda.pred$class,boston_test$crim01)##
## 0 1
## 0 7 0
## 1 10 135mean(blda.pred$class==boston_test$crim01)## [1] 0.9342105정확도가 93.2%로 모든 변수를 포함한 모델보다 정확도가 감소함.
#QDA
bqda <- qda(crim01 ~ indus + nox + age + dis + rad + tax, data = bostonplus, subset = btrain)
bqda## Call:
## qda(crim01 ~ indus + nox + age + dis + rad + tax, data = bostonplus,
## subset = btrain)
##
## Prior probabilities of groups:
## 0 1
## 0.6666667 0.3333333
##
## Group means:
## indus nox age dis rad tax
## 0 6.444746 0.4639606 49.83305 5.225892 4.190678 295.8051
## 1 12.248814 0.6017458 82.00847 2.982705 5.033898 340.0847bqda.pred<-predict(bqda,boston_test)
table(bqda.pred$class,boston_test$crim01)##
## 0 1
## 0 12 132
## 1 5 3mean(bqda.pred$class==boston_test$crim01)## [1] 0.09868421정확도가 9.87%로 모든 변수를 포함한 모델보다 정확도가 확연히 감소함. 세 가지 방법 모두 변수들을 선별한 모델보다 모든 변수를 포함한 모델의 정확도와 통계적 유의성이 더 높기 때문에 모든 변수를 포함한 모델을 사용하는 것이 더 좋다.