HW3
Problem 10
Question A: Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)## Warning: package 'ISLR' was built under R version 3.4.4summary(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
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## Today Direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260pairs(Weekly)
Question A answer
There seems to be a pattern at Volume and year
Question 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, data=Weekly, family="binomial")
summary(glm.fit)##
## 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: 4Question b answer
Lag 2 seems to be significant
Question 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.prob <- predict(glm.fit, type="response")
glm.pred <- ifelse(glm.prob>.5, "Up", "Down")
cmatrix <- table(Weekly$Direction, glm.pred)
cmatrix## glm.pred
## Down Up
## Down 54 430
## Up 48 557Question c answer
There is prevalence of Up prediction. The model predicts Up direction very well, but for Down direction it performs badly.
Question d: Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train = (Weekly$Year<=2008)
test = Weekly[!train,]
glm.fit2 <- glm(Direction ~ Lag2, data=Weekly, subset=train, family="binomial")
glm.prob2 <- predict(glm.fit2, type="response", newdata=test)
glm.pred2 <- ifelse(glm.prob2>.5, "Up", "Down")
cmatrix2 <- table(test$Direction, glm.pred2)
cmatrix2## glm.pred2
## Down Up
## Down 9 34
## Up 5 56oa <- (cmatrix2["Down", "Down"] + cmatrix2["Up", "Up"])/sum(cmatrix2)
oa## [1] 0.625Question d answer above
Question e: Repeat (d) using LDA
library(dplyr)## Warning: package 'dplyr' was built under R version 3.4.4##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(MASS)## Warning: package 'MASS' was built under R version 3.4.4##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectlda.fit <- lda(Direction ~ Lag2, data=Weekly, subset=train)
lda.pred <- predict(lda.fit, newdata=test)
cmatrix3 <- table(test$Direction, lda.pred$class)
cmatrix3##
## Down Up
## Down 9 34
## Up 5 56oa2<- (cmatrix3["Down", "Down"] + cmatrix3["Up", "Up"])/sum(cmatrix3)
oa2## [1] 0.625Question e answer above
Question f: Repeat (d) using QDA
qda.fit <- qda(Direction ~ Lag2, data=Weekly, subset=train)
qda.pred <- predict(qda.fit, newdata=test)
cmatrix4 <- table(test$Direction, qda.pred$class)
cmatrix4##
## Down Up
## Down 0 43
## Up 0 61oa3<- (cmatrix4["Down", "Down"] + cmatrix4["Up", "Up"])/sum(cmatrix4)
oa3## [1] 0.5865385Question f answer above
Question g: Repeat (d) using KNN with K = 1.
library(class)
train2 = Weekly[train, c("Lag2", "Direction")]
knn.pred = knn(train=data.frame(train2$Lag2), test=data.frame(test$Lag2), cl=train2$Direction, k=1)
cmatrix5 <- table(test$Direction, knn.pred)
cmatrix5## knn.pred
## Down Up
## Down 21 22
## Up 29 32oa4 <- (cmatrix5["Down", "Down"] + cmatrix5["Up", "Up"])/sum(cmatrix5)
oa4## [1] 0.5096154Question g answer above
Question h: Which of these methods appears to provide the best results on this data?
rbind(oa, oa2, oa3, oa4)## [,1]
## oa 0.6250000
## oa2 0.6250000
## oa3 0.5865385
## oa4 0.5096154Question h answer
models glm.fit2 oa from d and lda.fit oa2 from e
Question i:Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
qda.fit <- qda(Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, subset=train)
qda.pred <- predict(qda.fit, test)
print( sum(qda.pred$class==test$Direction)/length(qda.pred$class))## [1] 0.4326923qda.fit <- qda(Direction ~ Lag1*Lag2*Lag3*Lag4*Lag5 + Volume, data=Weekly, subset=train)
qda.pred <- predict(qda.fit, test)
print( sum(qda.pred$class==test$Direction)/length(qda.pred$class))## [1] 0.4134615one of Questions i answers
its worse than using only the Lag2 predictor shown in g and the second time even worse.
lda.fit <- lda(Direction ~ Lag1*Lag2*Lag3*Lag4*Lag5 + Volume, data=Weekly, subset=train)
lda.pred <- predict(lda.fit, test)
print( sum(lda.pred$class==test$Direction)/length(lda.pred$class))## [1] 0.4423077Question i’s other answer
the lda is just as bad as when I ran qda the first time
Problem 11
Question 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
Auto <- data.frame(Auto, mpg01)
summary(Auto)## mpg cylinders displacement horsepower
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0
##
## weight acceleration year origin
## Min. :1613 Min. : 8.00 Min. :70.00 Min. :1.000
## 1st Qu.:2225 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000
## Median :2804 Median :15.50 Median :76.00 Median :1.000
## Mean :2978 Mean :15.54 Mean :75.98 Mean :1.577
## 3rd Qu.:3615 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000
## Max. :5140 Max. :24.80 Max. :82.00 Max. :3.000
##
## name mpg01
## amc matador : 5 Min. :0.0
## ford pinto : 5 1st Qu.:0.0
## toyota corolla : 5 Median :0.5
## amc gremlin : 4 Mean :0.5
## amc hornet : 4 3rd Qu.:1.0
## chevrolet chevette: 4 Max. :1.0
## (Other) :365Question a answer above
Question 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 predictiong “mpg01” ? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings. ### Correlation
cor(Auto[, -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 1.0000000pairs(Auto)
par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")
## Question b answer some association between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower” exsist.
Question c: Split the data into a training set and a test set.
set.seed(123)
train <- sample(1:dim(Auto)[1], dim(Auto)[1]*.7, rep=FALSE)
test <- -train
trainingset <- Auto[train, ]
testingset = Auto[test, ]
mpg01.test <- mpg01[test]Question c answer above
Question 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 ?
lda.fit <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = trainingset)
lda.fit## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = trainingset)
##
## Prior probabilities of groups:
## 0 1
## 0.4817518 0.5182482
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.795455 3659.008 274.3333 130.88636
## 1 4.218310 2343.599 117.0528 79.71127
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.4483062756
## weight -0.0012201696
## displacement 0.0001078142
## horsepower 0.0046253024lda.pred = predict(lda.fit, testingset)
names(lda.pred)## [1] "class" "posterior" "x"lda.pred <- predict(lda.fit, testingset)
table(lda.pred$class, mpg01.test)## mpg01.test
## 0 1
## 0 53 3
## 1 11 51mean(lda.pred$class != mpg01.test)## [1] 0.1186441Question d answer
test error rate of 11.86%
Question e: Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit = qda(mpg01 ~ cylinders + horsepower + weight + acceleration, data=trainingset)
qda.fit## Call:
## qda(mpg01 ~ cylinders + horsepower + weight + acceleration, data = trainingset)
##
## Prior probabilities of groups:
## 0 1
## 0.4817518 0.5182482
##
## Group means:
## cylinders horsepower weight acceleration
## 0 6.795455 130.88636 3659.008 14.65303
## 1 4.218310 79.71127 2343.599 16.46620qda.class=predict(qda.fit, testingset)$class
table(qda.class, testingset$mpg01)##
## qda.class 0 1
## 0 56 3
## 1 8 51mean(qda.class != testingset$mpg01)## [1] 0.09322034Question e answer
test error rate of 0.093%
Question 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.fit <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = trainingset, family = binomial)
summary(glm.fit)##
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower,
## family = binomial, data = trainingset)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4870 -0.1763 0.1231 0.3633 3.2024
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 11.6661375 1.9885822 5.867 4.45e-09 ***
## cylinders 0.0365940 0.3817545 0.096 0.92363
## weight -0.0023706 0.0008319 -2.850 0.00438 **
## displacement -0.0106969 0.0096956 -1.103 0.26991
## horsepower -0.0327332 0.0154688 -2.116 0.03434 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 379.48 on 273 degrees of freedom
## Residual deviance: 147.50 on 269 degrees of freedom
## AIC: 157.5
##
## Number of Fisher Scoring iterations: 7probs <- predict(glm.fit, testingset, type = "response")
glm.pred <- rep(0, length(probs))
glm.pred[probs > 0.5] <- 1
table(glm.pred, mpg01.test)## mpg01.test
## glm.pred 0 1
## 0 54 3
## 1 10 51mean(glm.pred != mpg01.test)## [1] 0.1101695Question f answer
Test error rate is 11.01695%.
Question g: Perform KNN on the training data, with several values of K, in order to predict “mpg01” using 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 ?
str(Auto)## 'data.frame': 392 obs. of 10 variables:
## $ mpg : num 18 15 18 16 17 15 14 14 14 15 ...
## $ cylinders : num 8 8 8 8 8 8 8 8 8 8 ...
## $ displacement: num 307 350 318 304 302 429 454 440 455 390 ...
## $ horsepower : num 130 165 150 150 140 198 220 215 225 190 ...
## $ weight : num 3504 3693 3436 3433 3449 ...
## $ acceleration: num 12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
## $ year : num 70 70 70 70 70 70 70 70 70 70 ...
## $ origin : num 1 1 1 1 1 1 1 1 1 1 ...
## $ name : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
## $ mpg01 : num 0 0 0 0 0 0 0 0 0 0 ...data = scale(Auto[,-c(9,10)])
set.seed(1234)
train <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
test <- -train
trainingset = data[train,c("cylinders","horsepower","weight","acceleration")]
testingset = data[test, c("cylinders", "horsepower","weight","acceleration")]
train.mpg01 = Auto$mpg01[train]
test.mpg01= Auto$mpg01[test]
library(class)
set.seed(1234)
knn_pred_y = knn(trainingset, testingset, train.mpg01, k = 1)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 50 4
## 1 6 58mean(knn_pred_y != test.mpg01)## [1] 0.08474576Question g answer
Test error rate is 8.47%
13
Question a: 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")
crim01 <- rep(0, length(Boston$crim))
crim01[Boston$crim > median(Boston$crim)] <- 1
Boston <- data.frame(Boston, crim01)
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 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. :1.0set.seed(1234)
train <- sample(1:dim(Boston)[1], dim(Boston)[1]*.7, rep=FALSE)
test <- -train
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]glm.fit <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial)
glm.fit##
## Call: glm(formula = crim01 ~ . - crim01 - crim, family = binomial,
## data = Boston)
##
## Coefficients:
## (Intercept) zn indus chas nox
## -34.103704 -0.079918 -0.059389 0.785327 48.523782
## rm age dis rad tax
## -0.425596 0.022172 0.691400 0.656465 -0.006412
## ptratio black lstat medv
## 0.368716 -0.013524 0.043862 0.167130
##
## Degrees of Freedom: 505 Total (i.e. Null); 492 Residual
## Null Deviance: 701.5
## Residual Deviance: 211.9 AIC: 239.9library(corrplot)## Warning: package 'corrplot' was built under R version 3.4.4## corrplot 0.84 loadedcorrplot::corrplot.mixed(cor(Boston[, -1]), upper="shade")
glm.fit <- glm(crim01 ~ nox + indus + age + rad, data = Boston, family = binomial)
probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(probs))
glm.pred[probs > 0.5] <- 1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 65 15
## 1 4 68mean(glm.pred != crim01.test)## [1] 0.125For the logistic regression, we have a test error rate of 12.5%.
lda.fit <- lda(crim01 ~ nox + indus + age + rad , data = Boston)
lda.pred <- predict(lda.fit, Boston.test)
table(lda.pred$class, crim01.test)## crim01.test
## 0 1
## 0 66 20
## 1 3 63mean(lda.pred$class != crim01.test)## [1] 0.1513158For the LDA regression, we have a test error rate of 15.1%.
data = scale(Boston[,-c(1,15)])
set.seed(1234)
train <- sample(1:dim(Boston)[1], dim(Boston)[1]*.7, rep=FALSE)
test <- -train
trainingset = data[train, c("nox" , "indus" , "age" , "rad")]
testingset = data[test, c("nox" , "indus" , "age" , "rad")]
train.crime01 = Boston$crim01[train]
test.crime01= Boston$crim01[test]
library(class)
set.seed(1234)
knn.pred.y = knn(trainingset, testingset, train.crime01, k = 1)
table(knn.pred.y, test.crime01)## test.crime01
## knn.pred.y 0 1
## 0 62 7
## 1 7 76mean(knn.pred.y != test.crime01)## [1] 0.09210526For KNN (k=1), test error rate of 9.21%
knn_pred_y = NULL
error_rate = NULL
for(i in 1:dim(testingset)[1]){
set.seed(1234)
knn.pred.y = knn(trainingset,testingset,train.crime01,k=i)
error_rate[i] = mean(test.crime01 != knn.pred.y)}
min_error_rate = min(error_rate)
print(min_error_rate)## [1] 0.06578947K = which(error_rate == min_error_rate)
print(K)## [1] 3Minimum error rate is 6.57% at K=3
library(ggplot2)## Warning: package 'ggplot2' was built under R version 3.4.4qplot(1:dim(testingset)[1], error_rate, xlab = "K",
ylab = "Error Rate",
geom=c("point", "line"))
library(rsconnect)## Warning: package 'rsconnect' was built under R version 3.4.4