ISLR第四章分类应用练习题答案

2024-06-14 00:08

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ISLR;R语言; 机器学习 ;线性回归

一些专业词汇只知道英语的,中文可能不标准,请轻喷


10.Weekly数据集分析
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            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  > 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 -0.007825873Lag5      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

发现YearVolume具有相关性。
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: 1500.4Number of Fisher Scoring iterations: 4

Lag2在统计上更显著,Pr(>|z|) =0.0296
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  UpDown   54  48Up    430 557

预测准确率为(54+557)/(54+48+430+557)=0.56,当预测weak增加时,正确率为557/(557+48)=92.1%,当预测weak减少时正确率为54/(430+54)=11.2%
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)Direction.0910
glm.pred Down UpDown    9  5Up     34 56
> mean(glm.pred==Direction.0910)
[1] 0.625

e)

> library(MASS)
> lda.fit=lda(Direction~Lag2,data=Weekly,subset=train)
> lda.pred=predict(lda.fit,Weekly.0910)
> table(lda.pred$class,Direction.0910)Direction.0910Down UpDown    9  5Up     34 56
> mean(lda.pred$class == Direction.0910)
[1] 0.625

f)

  > qda.fit=qda(Direction~Lag2,data=Weekly,subset=train)> qda.class=predict(qda.fit,Weekly.0910)$class> table(qda.class,Direction.0910)Direction.0910qda.class Down UpDown    0  0Up     43 61> mean(qda.class==Direction.0910)[1] 0.5865385

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)Direction.0910
knn.pred Down UpDown   21 30Up     22 31
> mean(knn.pred==Direction.0910)
[1] 0.5

h)
逻辑回归和LDA具有最高的正确率
i)
逻辑回归 Lag2与Lag1相关

> 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 UpDown    1  1Up     42 60
> mean(glm.pred==Direction.0910)
[1] 0.5865385

LDA Lag2与Lag1相关

> lda.fit=lda(Direction~Lag2:Lag1,data=Weekly,subset=train)
> lda.pred=predict(lda.fit,Weekly.0910)
> table(lda.pred$class,Direction.0910)Direction.0910Down UpDown    0  1Up     43 60
> mean(lda.pred$class==Direction.0910)
[1] 0.5769231

QDA Lag2与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 UpDown   12 13Up     31 48
> mean(qda.class==Direction.0910)
[1] 0.5769231

K=10

> knn.pred=knn(train.X,test.X,train.Direction,k=10)
> table(knn.pred,Direction.0910)Direction.0910
knn.pred Down UpDown   17 18Up     26 43
> mean(knn.pred==Direction.0910)
[1] 0.5769231

k=100

> knn.pred=knn(train.X,test.X,train.Direction,k=100)
> table(knn.pred,Direction.0910)Direction.0910
knn.pred Down UpDown    9 12Up     34 49
> mean(knn.pred==Direction.0910)
[1] 0.5576923

依旧是原来的LDA和逻辑回归正确率最高


11.Auto数据集 汽车与每公里油耗高低预测
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)

b)

 > cor(Auto[,-9])mpg  cylinders displacement horsepower     weightmpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054mpg01         0.8369392 -0.7591939   -0.7534766 -0.6670526 -0.7577566acceleration       year     origin      mpg01mpg             0.4233285  0.5805410  0.5652088  0.8369392cylinders      -0.5046834 -0.3456474 -0.5689316 -0.7591939displacement   -0.5438005 -0.3698552 -0.6145351 -0.7534766horsepower     -0.6891955 -0.4163615 -0.4551715 -0.6670526weight         -0.4168392 -0.3091199 -0.5850054 -0.7577566acceleration    1.0000000  0.2903161  0.2127458  0.3468215year            0.2903161  1.0000000  0.1815277  0.4299042origin          0.2127458  0.1815277  1.0000000  0.5136984mpg01           0.3468215  0.4299042  0.5136984  1.0000000> pairs(Auto)


cylinders, weight, displacement, horsepower 有很大的可能相关。
c)

> train=(year%%2==0)
> 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)
[1] 0.1263736

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)
[1] 0.1318681

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)
[1] 0.1208791

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)

k=1

knn.pred=knn(train.X,test.X,train.mpg01,k=1)
mean(knn.pred != mpg01.test)
[1] 0.1538462
k=5
knn.pred=knn(train.X,test.X,train.mpg01,k=5)
mean(knn.pred != mpg01.test)
[1] 0.1483516
k=10
knn.pred=knn(train.X,test.X,train.mpg01,k=10)
mean(knn.pred != mpg01.test)
[1] 0.1648352
k=5时可能最好


12.写函数
a)

> Power=function(){ 2^3}
> print(Power())
[1] 8

b)

> Power2=function(x,a){x^a}
> Power2(3,8)
[1] 6561

c)

> Power2(10,3)
[1] 1000
> Power2(8,17)
[1] 2.2518e+15
> Power2(131,3)
[1] 2248091

d)

> Power3=function(x,a){
+ result=x^a
+ return(result)
+ }

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")


f)

> PlotPower=function(x,a){
+ plot(x,Power3(x,a))
+ }
> PlotPower(1:10,3)


13.Boston数据集

> 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]

逻辑回归

 > glm.fit=glm(crime01~.-crime01-crim,data=Boston,family=binomial,subset=train)Warning message:glm.fit:拟合機率算出来是数值零或一 > 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> glm.fit=glm(crime01~.-crime01-crim-chas-tax,data=Boston,family=binomial,subset=train)Warning message:glm.fit:拟合機率算出来是数值零或一 > 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

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

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)

k=1

 > knn.preb=knn(train.X,test.X,train.crime01,k=1)> mean(knn.preb!=crime01.test)[1] 0.458498

k=5

 > knn.preb=knn(train.X,test.X,train.crime01,k=5)> mean(knn.preb!=crime01.test)[1] 0.1699605

k=10

 > knn.preb=knn(train.X,test.X,train.crime01,k=10)> mean(knn.preb!=crime01.test)[1] 0.1146245

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