给定数据集cars.txt，我们希望使用变量马力、长度、行李箱、掉头、轴距和宽度为中档价格制定一个好的回归模型。两个都：
- 使用所有可能的子集选择，以及
- 使用自动选择技术。
对于第一部分，我们在 R 中进行：
cars <- read.table(file=file.choose(), header=TRUE)
names(cars)
#regression
attach(cars)
leap <- leaps(x=cbind(cars$Horsepower, cars$Length, cars$Luggage, cars$Uturn, cars$Wheelbase, cars$Width),
y=cars$MidrangePrice, method=c("r2"), nbest=3)
combine <- cbind(leap$which,leap$size, leap$r2)
n <- length(leap$size)
dimnames(combine) <- list(1:n,c("horsep","length","Luggage","Uturn","Wheelbase","Width","size","r2"))
round(combine, digits=3)
leap.cp <- leaps(x=cbind(cars$Horsepower, cars$Length, cars$Luggage, cars$Uturn, cars$Wheelbase, cars$Width),
y=cars$MidrangePrice, nbest=3)
combine.cp <- cbind(leap.cp$which,leap.cp$size, leap.cp$Cp)
dimnames(combine.cp) <- list(1:n,c("horsep","length","Luggage","Uturn","Wheelbase","Width","size","cp"))
round(combine.cp, digits=3)
plot(leap.cp$size, leap.cp$Cp, ylim=c(1,7))
abline(a=0, b=1)
我的解释是否正确，即最合适的模型是具有 4 个参数（三个变量马力、轴距和宽度）的模型，因为它具有最低的 Mallows 的 Cp 值？
对于第二部分，我们可以在前向、后向或逐步选择模型之间进行选择：
#stepwise selection methods
#forward
slm.foward <- step(lm(cars$MidrangePrice ~1, data=cars), scope=~cars$Horsepower + cars$Length + cars$Luggage + cars$Uturn + cars$Wheelbase + cars$Horsepower+ cars$Width, direction="forward")
#backward
reg.lm1 <- lm(cars$MidrangePrice ~ cars$Horsepower + cars$Length + cars$Luggage + cars$Uturn + cars$Wheelbase + cars$Horsepower + cars$Width)
slm.backward <- step(reg.lm1, direction="backward")
#stepwise
reg.lm1 <- lm(cars$MidrangePrice ~ cars$Horsepower + cars$Length + cars$Luggage + cars$Uturn + cars$Wheelbase + cars$Horsepower + cars$Width)
slm.stepwise <- step(reg.lm1,direction="both")
我如何解释我从这个 R 代码中得到的结果？