统计学习导论：基于R应用——第四章习题

时间：2023-12-15 12:06:14

第四章习题，部分题目未给出答案

1.

这个题比较简单，有高中生推导水平的应该不难。

2~3证明题，略

4.

(a)

这个问题问我略困惑，答案怎么直接写出来了，难道不是10%么

(b)

这个答案是(0.1*0.1)/(1*1)，所以答案是1%

(c)

其实就是个空间所占比例，所以这题是(0.1**100)*100 = 0.1**98%

(d)

这题答案显而易见啊，而且是指数级别下降

(e)

答案是0.1**(1)、0.1**(1/2)、0.1**(1/3)...0.1**(1/100)

5.

这题在中文版的104页的偏差-方差权衡说的听清楚。

(a)

当贝叶斯决策边界是线性的时候，训练集上当然是QDA效果好，因为拟合的更好。而测试集上是LDA更好，因为更接近实际。

(b)

当贝叶斯决策边界是非线性的时候，QDA在训练集和测试集都比LDA好

(c)

相比于LDA，QDA的预测率变得更好。因为当样本量n提升时，一个*度更高的模型会产生更好的效果，因为方差会被大的样本抵消一点

(d)

不对。因为当样本很少时，QDA会过拟合。

6.

(a)

由公式统计学习导论：基于R应用——第四章习题直接带入p(X)=37.75%

(b)

还是带入上述公式，反求X1为50hours

7.

其实就是贝叶斯公式+中文版书97页公式4-12。。。有点繁琐，最后答案是75.2%

8.

文字题。。当你用K=1的KNN时，在训练集上的错误率是0%，所以测试集上错误率实际是36%。我们当然选逻辑回归啦

9.

参见92页公式4-3。。。就是带入公式而已，第一题是27%，第二题是0.19

10.

(a)

感觉题目里面让我们进行数值和图像描述统计时，大概就三条命令：summary()、pairs()、cor()。不过pairs()在特征很多的时候，跑的真心慢，cor()在使用前也要把定性的变量去掉。

library(ISLR)

summary(Weekly)

pairs(Weekly)

cor(Weekly[, -9])

(b)

attach(Weekly)

glm.fit = glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly,  family = binomial)

summary(glm.fit)

(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)

(d)

train = (Year < 2009)

Weekly.0910 = Weekly[!train, ]

glm.fit = glm(Direction ~ Lag2, 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)
mean(glm.pred == Direction.0910)

(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)

mean(lda.pred$class == Direction.0910)

(f)

qda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)

qda.class = predict(qda.fit, Weekly.0910)$class

table(qda.class, Direction.0910)

mean(qda.class == Direction.0910)

(g)

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.0910)

mean(knn.pred == Direction.0910)

(h)

两种方法的准确率一样。。。

(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)

mean(glm.pred == Direction.0910)

## [1] 0.5865

# 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] 0.5769

# 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)

mean(qda.class == Direction.0910)

## [1] 0.5769

# KNN k =10

knn.pred = knn(train.X, test.X, train.Direction, k = 10)

table(knn.pred, Direction.0910)

mean(knn.pred == Direction.0910)

## [1] 0.5769

# KNN k = 100

knn.pred = knn(train.X, test.X, train.Direction, k = 100)

table(knn.pred, Direction.0910)

mean(knn.pred == Direction.0910)

## [1] 0.5577

结果在代码注释中，逻辑回归效果最好

11.

(a)

library(ISLR)

summary(Auto)

attach(Auto)

mpg01 = rep(0, length(mpg))

mpg01[mpg > median(mpg)] = 1

Auto = data.frame(Auto, mpg01)

(b)

cor(Auto[, -9])

pairs(Auto)

(c)

train = (year%%2 == 0)  # if the year is even

test = !train

Auto.train = Auto[train, ]

Auto.test = Auto[test, ]

mpg01.test = mpg01[test]

(d)

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)

(e)

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)

(f)

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)

(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)

# KNN(k=10)

knn.pred = knn(train.X, test.X, train.mpg01, k = 10)

mean(knn.pred != mpg01.test)

# KNN(k=100)

knn.pred = knn(train.X, test.X, train.mpg01, k = 100)

mean(knn.pred != mpg01.test)

13题和11题类似，就是用这几个函数。所以13题略。

12.

(a)~(b)

Power = function() {

  2^3

}

print(Power())

Power2 = function(x, a) {

  x^a

}

Power2(3, 8)

(c)

Power2(10, 3)

Power2(8, 17)

Power2(131, 3)

(d)~(f)

Power3 = function(x, a) {

  result = x^a

  return(result)

}

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")

PlotPower = function(x, a) {

  plot(x, Power3(x, a))

}

PlotPower(1:10, 3)

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