10.(a)
library(ISLR)
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
## 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. : 12.0260
pairs(Weekly)
cor(Weekly[,-9])#除去第9层
## 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 1.000000000
Year and Volume appear to have a relationship.No other patterns are discernible.
10.(b)
attach(Weekly)
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: 4
Lag2 appears to have some statistical significance with a Pr(>|z|)=3%.
10.(c)
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 430 557
Percentage of currect predictions: (54+557)/(54+557+48+430) = 56.1%. Weeks the market goes up the logistic regression is right most of the time, 557/(557+48) = 92.1%. Weeks the market goes down the logistic regression is wrong most of the time 54/(430+54) = 11.2%.
10.(d)
train=(Year<2009)
Weekly.0910=Weekly[!train,]
glm.fit=glm(Direction~Lag2,data=Weekly,family=binomial,subset=train)#parameter 'subset'
glm.probs=predict(glm.fit,Weekly.0910,type="response")
glm.pred=rep("Down",length(glm.probs))
glm.pred[glm.probs>0.5]="Up"
Direction.0910=Direction[!train]
table(glm.pred,Direction.0910)
## Direction.0910
## glm.pred Down Up
## Down 9 5
## Up 34 56
mean(glm.pred==Direction.0910)
## [1] 0.625
10.(e)
library(MASS)
lda.fit = lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly.0910)
table(lda.pred$class, Direction.0910)
## Direction.0910
## Down Up
## Down 9 5
## Up 34 56
mean(lda.pred$class == Direction.0910)
## [1] 0.625
10.(f)
qda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.class = predict(qda.fit, Weekly.0910)$class
table(qda.class, Direction.0910)
## Direction.0910
## qda.class Down Up
## Down 0 0
## Up 43 61
mean(qda.class == Direction.0910)
## [1] 0.5865385
A correctness of 58.7% even though it picked Up the whole time!
10.(g)
library(class)#KNN
train.X = as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction=Direction[train]
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=1)
table(knn.pred, Direction.0910)
## Direction.0910
## knn.pred Down Up
## Down 21 30
## Up 22 31
mean(knn.pred==Direction.0910)
## [1] 0.5
10.(h)
Logistic Regression and LDA methods provide similar test error rates.
10.(i)
# Logistic regression with Lag2:Lag1
glm.fit = glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
glm.probs = predict(glm.fit, Weekly.0910, type = "response")
glm.pred = rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(glm.pred, Direction.0910)
## Direction.0910
## glm.pred Down Up
## Down 1 1
## Up 42 60
mean(glm.pred == Direction.0910)
## [1] 0.5865385
# LDA with Lag2 interaction with Lag1
lda.fit = lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly.0910)
mean(lda.pred$class == Direction.0910)
## [1] 0.5769231
# QDA with sqrt(abs(Lag2))
qda.fit = qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
qda.class = predict(qda.fit, Weekly.0910)$class
table(qda.class, Direction.0910)
## Direction.0910
## qda.class Down Up
## Down 12 13
## Up 31 48
mean(qda.class == Direction.0910)
## [1] 0.5769231
# KNN k =10
knn.pred = knn(train.X, test.X, train.Direction, k = 10)
table(knn.pred, Direction.0910)
## Direction.0910
## knn.pred Down Up
## Down 17 18
## Up 26 43
mean(knn.pred == Direction.0910)
## [1] 0.5769231
# KNN k = 100
knn.pred = knn(train.X, test.X, train.Direction, k = 100)
table(knn.pred, Direction.0910)
## Direction.0910
## knn.pred Down Up
## Down 9 12
## Up 34 49
mean(knn.pred == Direction.0910)
## [1] 0.5576923
Out of these permutations, the original LDA and logistic regression have better performance in terms of test error rate.
11.(a)
library(ISLR)
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
## amc matador : 5
## ford pinto : 5
## toyota corolla : 5
## amc gremlin : 4
## amc hornet : 4
## chevrolet chevette: 4
## (Other) :365
attach(Auto)
mpg01=rep(0,length(mpg))
mpg01[mpg>median(mpg)]=1
Auto=data.frame(Auto,mpg01)
11.(b)
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 1.0000000
pairs(Auto) # doesn't work well since mpg01 is 0 or 1
Anti-correlated with cylinders, weight, displacement, horsepower. (mpg, of course)
11.(c)
train = (year%%2 == 0) # if the year is even
test = !train
Auto.train = Auto[train, ]
Auto.test = Auto[test, ]
mpg01.test = mpg01[test]
11.(d)
# LDA
library(MASS)
lda.fit = lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
subset = train)
lda.pred = predict(lda.fit, Auto.test)
mean(lda.pred$class != mpg01.test)
## [1] 0.1263736
12.6% test error rate.
11.(e)
# QDA
qda.fit = qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
subset = train)
qda.pred = predict(qda.fit, Auto.test)
mean(qda.pred$class != mpg01.test)
## [1] 0.1318681
13.2% test error rate.
11.(f)
# Logistic regression
glm.fit = glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
family = binomial, subset = train)
glm.probs = predict(glm.fit, Auto.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != mpg01.test)
## [1] 0.1208791
11.(g)
library(class)
train.X = cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X = cbind(cylinders, weight, displacement, horsepower)[test, ]
train.mpg01 = mpg01[train]
set.seed(1)
# KNN(k=1)
knn.pred = knn(train.X, test.X, train.mpg01, k = 1)
mean(knn.pred != mpg01.test)
## [1] 0.1538462
# KNN(k=10)
knn.pred = knn(train.X, test.X, train.mpg01, k = 10)
mean(knn.pred != mpg01.test)
## [1] 0.1648352
# KNN(k=100)
knn.pred = knn(train.X, test.X, train.mpg01, k = 100)
mean(knn.pred != mpg01.test)
## [1] 0.1428571
k=1, 15.4% test error rate. k=10, 16.5% test error rate. k=100, 14.3% test error rate. K of 100 seems to perform the best. 100 nearest neighbors.
12.(a)
Power = function() {
2^3
}
print(Power())
## [1] 8
12.(b)
Power2 = function(x, a) {
x^a
}
Power2(3, 8)
## [1] 6561
12.(c)
Power2(10, 3)
## [1] 1000
Power2(8, 17)
## [1] 2.2518e+15
Power2(131, 3)
## [1] 2248091
12.(d)
Power3 = function(x, a) {
result = x^a
return(result)
}
12.(e)
x = 1:10
plot(x, Power3(x, 2), log = "xy", ylab = "Log of y = x^2", xlab = "Log of x",
main = "Log of x^2 versus Log of x")
12.(f)
PlotPower = function(x, a) {
plot(x, Power3(x, a))
}
PlotPower(1:10, 3)
library(MASS)
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. :50.00
attach(Boston)
crime01 = rep(0, length(crim))
crime01[crim > median(crim)] = 1
Boston = data.frame(Boston, crime01)
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]
# logistic regression
glm.fit = glm(crime01 ~ . - crime01 - crim, data = Boston, family = binomial,
subset = train)
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
glm.probs = predict(glm.fit, Boston.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != crime01.test)
## [1] 0.1818182
#18.2% test error rate
glm.fit = glm(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, family = binomial,
subset = train)
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
glm.probs = predict(glm.fit, Boston.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != crime01.test)
## [1] 0.1857708
18.6 test error rate
# LDA
lda.fit = lda(crime01 ~ . - crime01 - crim, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)
## [1] 0.1343874
13.4% test error rate
lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)
## [1] 0.1225296
12.3% test error rate
lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax - lstat - indus - age,
data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)
## [1] 0.1185771
11.9% test error rate
#KNN
library(class)
train.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black,
lstat, medv)[train, ]
test.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black,
lstat, medv)[test, ]
train.crime01 = crime01[train]
set.seed(1)
# KNN(k=1)
knn.pred = knn(train.X, test.X, train.crime01, k = 1)
mean(knn.pred != crime01.test)
## [1] 0.458498
45.8% test error rate.
# KNN(k=10)
knn.pred = knn(train.X, test.X, train.crime01, k = 10)
mean(knn.pred != crime01.test)
## [1] 0.1185771
11.1% test error rate.
# KNN(k=100)
knn.pred = knn(train.X, test.X, train.crime01, k = 100)
mean(knn.pred != crime01.test)
## [1] 0.4901186
# KNN(k=10) with subset of variables
train.X = cbind(zn, nox, rm, dis, rad, ptratio, black, medv)[train, ]
test.X = cbind(zn, nox, rm, dis, rad, ptratio, black, medv)[test, ]
knn.pred = knn(train.X, test.X, train.crime01, k = 10)
mean(knn.pred != crime01.test)
## [1] 0.2727273
28.5% test error rate.
```