ISLR第五章重采样方法应用练习题
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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
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)a)
> set.seed(1)> glm.fit=glm(default~income+balance,data=Default,family=binomial)b)
> 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, ]$default))+ }> FiveB()[1] 0.02362.36%的错误率
c)
> FiveB()[1] 0.028> FiveB()[1] 0.0268> FiveB()[1] 0.0252错误率在2.6%上下波动。
d)
> train=sample(dim(Default)[1],dim(Default)[1]/2)> glm.fit = glm(default ~ income + balance + student, data=Default, family = binomial, subset = train)> glm.pred = rep("No",dim(Default)[1]/2)> glm.probs = predict(glm.fit, Default[-train,],type="response")> glm.pred[glm.probs > 0.5] = "Yes"> mean(glm.pred != Default[-train,]$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)a)
> 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 freedomResidual deviance: 1579.0 on 9997 degrees of freedomAIC: 1585Number of Fisher Scoring iterations: 8b)
> boot.fn = function(data, index) return(coef(glm(default ~ income + balance, data = data, family = binomial, subset = index)))c)
> library(boot)> boot(Default, boot.fn, 50)ORDINARY NONPARAMETRIC BOOTSTRAPCall:boot(data = Default, statistic = boot.fn, R = 50)Bootstrap Statistics : original bias std. errort1* -1.154047e+01 1.181200e-01 4.202402e-01t2* 2.080898e-05 -5.466926e-08 4.542214e-06t3* 5.647103e-03 -6.974834e-05 2.282819e-04d)
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)a)
> 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 freedom Residual deviance: 1488.2 on 1086 degrees of freedom AIC: 1494.2 Number of Fisher Scoring iterations: 4b)
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 freedom Residual deviance: 1486.5 on 1085 degrees of freedom AIC: 1492.5 Number of Fisher Scoring iterations: 4c)
> predict.glm(glm.fit,Weekly[1, ],type = "response") > 0.5 1 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, ]$Direction == "Up" + if (is_up != is_true_up) + count[i] = 1 + } > sum(count) [1] 490有490个错误。
e)
> mean(count)[1] 0.4499541LOOCV估计错误率为45%
8.在一个假数据集上做交叉估计
a)
> set.seed(1)> y=rnorm(100)> x=rnorm(100)> y=x-2 * x^2 + rnorm(100)n=100,p=2
Y=X-2*X^2+ε
b)
> plot(x,y)
x与y成二次关系
c)
> library(boot)> Data = data.frame(x, y)> set.seed(1)> #1> glm.fit = glm(y ~ x)> cv.glm(Data, glm.fit)$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)$delta[1] 1.114772 1.114334d)
> set.seed(10)> #1> glm.fit = glm(y ~ x)> cv.glm(Data, glm.fit)$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 freedomResidual deviance: 103.04 on 95 degrees of freedomAIC: 298.78Number of Fisher Scoring iterations: 2p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)a)
> medv.mean = mean(medv)> medv.mean[1] 22.53281b)
> medv.err = sd(medv)/sqrt(length(medv))> medv.err[1] 0.4088611c)
> 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. errort1* 22.53281 0.008517589 0.4119374d)
> t.test(medv) One Sample t-testdata: medvt = 55.1111, df = 505, p-value < 2.2e-16alternative hypothesis: true mean is not equal to 095 percent confidence interval: 21.72953 23.33608sample 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] 21.2f)
> 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. errort1* 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.tenth 10% 12.75 h)
> 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. errort1* 12.75 0.00515 0.5113487与实际有很小的标准差
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