ISLR第五章重采样方法应用练习题

本文主要是介绍ISLR第五章重采样方法应用练习题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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

5.Default数据分析

> library(ISLR)
> summary(Default)
default    student       balance           income     No :9667   No :7056   Min.   :   0.0   Min.   :  772  Yes: 333   Yes:2944   1st Qu.: 481.7   1st Qu.:21340  Median : 823.6   Median :34553  Mean   : 835.4   Mean   :33517  3rd Qu.:1166.3   3rd Qu.:43808  Max.   :2654.3   Max.   :73554  
> attach(Default)

> set.seed(1)
> glm.fit=glm(default~income+balance,data=Default,family=binomial)

> FiveB=function(){
+ #i.
+ train=sample(dim(Default)[1],dim(Default)[1]/2)
+ #ii.
+ glm.fit = glm(default ~ income + balance, data=Default, family = binomial,subset=train)
+ #iii.
+ glm.pred = rep("No",dim(Default)[1]/2)
+ glm.probs=predict(glm.fit,Default[-train, ],type="response")
+ glm.pred[glm.probs > 0.5]="Yes"
+ #iv.
+ return(mean(glm.pred != Default[-train, ]$default))
+ }
> FiveB()
[1] 0.0236

2.36%的错误率
c)

> FiveB()
[1] 0.028
> FiveB()
[1] 0.0268
> FiveB()
[1] 0.0252

错误率在2.6%上下波动。
d)

> train=sample(dim(Default)[1],dim(Default)[1]/2)
> glm.fit = glm(default ~ income + balance + student, data=Default, family = binomial, subset = train)
> glm.pred = rep("No",dim(Default)[1]/2)
> glm.probs = predict(glm.fit, Default[-train,],type="response")
> glm.pred[glm.probs > 0.5] = "Yes"
> mean(glm.pred != Default[-train,]$default)
[1] 0.0246

错误率为2.46%，增加student变量并没有减少错误率

6.Default数据集

> library(ISLR)
> summary(Default)default    student       balance           income     No :9667   No :7056   Min.   :   0.0   Min.   :  772  Yes: 333   Yes:2944   1st Qu.: 481.7   1st Qu.:21340  Median : 823.6   Median :34553  Mean   : 835.4   Mean   :33517  3rd Qu.:1166.3   3rd Qu.:43808  Max.   :2654.3   Max.   :73554  
> attach(Default)

> set.seed(1)
> glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
> summary(glm.fit)Call:
glm(formula = default ~ income + balance, family = binomial, 
data = Default)Deviance Residuals: Min       1Q   Median       3Q      Max  
-2.4725  -0.1444  -0.0574  -0.0211   3.7245  Coefficients:Estimate Std. Error z value Pr(>|z|)    
(Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
balance      5.647e-03  2.274e-04  24.836  < 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: 2920.6  on 9999  degrees of freedom
Residual deviance: 1579.0  on 9997  degrees of freedom
AIC: 1585Number of Fisher Scoring iterations: 8

> boot.fn = function(data, index) return(coef(glm(default ~ income + balance, data = data, family = binomial, subset = index)))

> library(boot)
> boot(Default, boot.fn, 50)ORDINARY NONPARAMETRIC BOOTSTRAPCall:
boot(data = Default, statistic = boot.fn, R = 50)Bootstrap Statistics :original        bias     std. error
t1* -1.154047e+01  1.181200e-01 4.202402e-01
t2*  2.080898e-05 -5.466926e-08 4.542214e-06
t3*  5.647103e-03 -6.974834e-05 2.282819e-04

d)
比较接近

7.Weekly数据集分析

> library(ISLR)
> summary(Weekly)Year           Lag1               Lag2         Min.   :1990   Min.   :-18.1950   Min.   :-18.1950  1st Qu.:1995   1st Qu.: -1.1540   1st Qu.: -1.1540  Median :2000   Median :  0.2410   Median :  0.2410  Mean   :2000   Mean   :  0.1506   Mean   :  0.1511  3rd Qu.:2005   3rd Qu.:  1.4050   3rd Qu.:  1.4090  Max.   :2010   Max.   : 12.0260   Max.   : 12.0260  Lag3               Lag4               Lag5         Min.   :-18.1950   Min.   :-18.1950   Min.   :-18.1950  1st Qu.: -1.1580   1st Qu.: -1.1580   1st Qu.: -1.1660  Median :  0.2410   Median :  0.2380   Median :  0.2340  Mean   :  0.1472   Mean   :  0.1458   Mean   :  0.1399  3rd Qu.:  1.4090   3rd Qu.:  1.4090   3rd Qu.:  1.4050  Max.   : 12.0260   Max.   : 12.0260   Max.   : 12.0260  Volume            Today          Direction Min.   :0.08747   Min.   :-18.1950   Down:484  1st Qu.:0.33202   1st Qu.: -1.1540   Up  :605  Median :1.00268   Median :  0.2410             Mean   :1.57462   Mean   :  0.1499             3rd Qu.:2.05373   3rd Qu.:  1.4050             Max.   :9.32821   Max.   : 12.0260  > set.seed(1)> attach(Weekly)

 > glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly, family = binomial)> summary(glm.fit)Call:glm(formula = Direction ~ Lag1 + Lag2, family = binomial, data = Weekly)Deviance Residuals: Min      1Q  Median      3Q     Max  -1.623  -1.261   1.001   1.083   1.506  Coefficients:Estimate Std. Error z value Pr(>|z|)    (Intercept)  0.22122    0.06147   3.599 0.000319 ***Lag1        -0.03872    0.02622  -1.477 0.139672    Lag2         0.06025    0.02655   2.270 0.023232 *  ---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 freedomResidual deviance: 1488.2  on 1086  degrees of freedomAIC: 1494.2Number of Fisher Scoring iterations: 4

glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly[-1, ], family = binomial)
summary(glm.fit)

 Call:glm(formula = Direction ~ Lag1 + Lag2, family = binomial, data = Weekly[-1, 
])Deviance Residuals: Min       1Q   Median       3Q      Max  -1.6258  -1.2617   0.9999   1.0819   1.5071  Coefficients:Estimate Std. Error z value Pr(>|z|)    (Intercept)  0.22324    0.06150   3.630 0.000283 ***Lag1        -0.03843    0.02622  -1.466 0.142683    Lag2         0.06085    0.02656   2.291 0.021971 *  ---Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for binomial family taken to be 1)Null deviance: 1494.6  on 1087  degrees of freedomResidual deviance: 1486.5  on 1085  degrees of freedomAIC: 1492.5Number of Fisher Scoring iterations: 4

 > predict.glm(glm.fit,Weekly[1, ],type = "response") > 0.51 TRUE

预测方向为UP，实际方向为DOWN
d)

 > count = rep(0, dim(Weekly)[1])> for (i in 1:(dim(Weekly)[1])){+ glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ], family = binomial)+ is_up = predict.glm(glm.fit, Weekly[i, ], type="response") > 0.5+ is_true_up = Weekly[i, ]$Direction == "Up"+ if (is_up != is_true_up)+ count[i] = 1+ }> sum(count)[1] 490

有490个错误。
e)

> mean(count)
[1] 0.4499541

LOOCV估计错误率为45%

8.在一个假数据集上做交叉估计
a)

> set.seed(1)
> y=rnorm(100)
> x=rnorm(100)
> y=x-2 * x^2 + rnorm(100)

n=100,p=2
Y=X-2*X^2+ε
b)

> plot(x,y)

x与y成二次关系
c)

> library(boot)
> Data = data.frame(x, y)
> set.seed(1)
> #1
> glm.fit = glm(y ~ x)
> cv.glm(Data, glm.fit)$delta
[1] 5.890979 5.888812
> #2
> glm.fit = glm(y ~ poly(x,2))
> cv.glm(Data, glm.fit)$delta
[1] 1.086596 1.086326
> #3
> glm.fit = glm(y ~ poly(x,3))
> cv.glm(Data, glm.fit)$delta
[1] 1.102585 1.102227
> glm.fit = glm(y ~ poly(x,4))
> cv.glm(Data, glm.fit)$delta
[1] 1.114772 1.114334

> set.seed(10)
> #1
> glm.fit = glm(y ~ x)
> cv.glm(Data, glm.fit)$delta
[1] 5.890979 5.888812
> #2
> glm.fit = glm(y ~ poly(x,2))
> cv.glm(Data, glm.fit)$delta
[1] 1.086596 1.086326
> #3
> glm.fit = glm(y ~ poly(x,3))
> cv.glm(Data, glm.fit)$delta
[1] 1.102585 1.102227
> #4
> glm.fit = glm(y ~ poly(x,4))
> cv.glm(Data, glm.fit)$delta
[1] 1.114772 1.114334

基本相同，因为LOOCV每次就除去了一个数据
e)
2次的，因为最近真实的关系
f)

> summary(glm.fit)Call:
glm(formula = y ~ poly(x, 4))Deviance Residuals: Min       1Q   Median       3Q      Max  
-2.8914  -0.5244   0.0749   0.5932   2.7796  Coefficients:Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -1.8277     0.1041 -17.549   <2e-16 ***
poly(x, 4)1   2.3164     1.0415   2.224   0.0285 *  
poly(x, 4)2 -21.0586     1.0415 -20.220   <2e-16 ***
poly(x, 4)3  -0.3048     1.0415  -0.293   0.7704    
poly(x, 4)4  -0.4926     1.0415  -0.473   0.6373    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for gaussian family taken to be 1.084654)Null deviance: 552.21  on 99  degrees of freedom
Residual deviance: 103.04  on 95  degrees of freedom
AIC: 298.78Number of Fisher Scoring iterations: 2

p值显示的统计上显著关系与cv结果相同

9.Boston数据集分析

> library(MASS)
> attach(Boston)
> 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  
> set.seed(1)

> medv.mean = mean(medv)
> medv.mean
[1] 22.53281

> medv.err = sd(medv)/sqrt(length(medv))
> medv.err
[1] 0.4088611

> boot.fn = function(data, index) return(mean(data[index]))
> library(boot)
> bstrap = boot(medv, boot.fn, 1000)
> bstrapORDINARY NONPARAMETRIC BOOTSTRAPCall:
boot(data = medv, statistic = boot.fn, R = 1000)Bootstrap Statistics :original      bias    std. error
t1* 22.53281 0.008517589   0.4119374

> t.test(medv)One Sample t-testdata:  medv
t = 55.1111, df = 505, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:21.72953 23.33608
sample estimates:
mean of x 22.53281 > c(bstrap$t0 - 2 * 0.4119, bstrap$t0 + 2 * 0.4119)
[1] 21.70901 23.35661

引导估计与t估计仅相差0.02
e)

> medv.med = median(medv)
> medv.med
[1] 21.2

> boot.fn = function(data, index) return(median(data[index]))
> boot(medv, boot.fn, 1000)ORDINARY NONPARAMETRIC BOOTSTRAPCall:
boot(data = medv, statistic = boot.fn, R = 1000)Bootstrap Statistics :original  bias    std. error
t1*     21.2 -0.0098   0.3874004

引导学习结果与实际相同，标准差也很小
g)

> medv.tenth = quantile(medv, c(0.1))
> medv.tenth10% 
12.75

> boot.fn = function(data, index) return(quantile(data[index],c(0.1)))
> boot(medv, boot.fn, 1000)ORDINARY NONPARAMETRIC BOOTSTRAPCall:
boot(data = medv, statistic = boot.fn, R = 1000)Bootstrap Statistics :original  bias    std. error
t1*    12.75 0.00515   0.5113487

与实际有很小的标准差

这篇关于ISLR第五章重采样方法应用练习题的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！