1

如果问题有 10 个变量

通过恰好使用两个变量，找到为您提供最大调整 R 平方值的模型。

fit(i,j)=lm(y~xi+xj,data=data)

其中 xi 和 xj 可以是 x1,x2,...,x10 之间的任何给定变量

例如，我想比较调整后的 R 平方

fit(1,2)=lm(y~x1+x2,data=data)

fit(1,3)=lm(y~x1+x3,data=data)

. . .

fit(9,10)=lm(y~x9+x10,data=data)

有没有办法使用“for loop”命令比较所有结果？

4

2 回答 2

1

假设您的结果变量被调用outcome并且您的数据框被调用df，我们首先可以自定义一个函数来返回调整后的正方形。之后，我们应用该combn功能。请注意，要使其正常工作，您需要将结果（如果因子）转换为数字。-df$outcome <- as.numeric(as.character(df$outcome))

R.squared <- function(y, x, z){
  summary(lm(y ~ x+z, df))$adj.r.squared
}
combn(ncol(df[,-1]), 2, function(i) R.squared(df$outcome, df[,i[1]], df[,i[2]]))
 #[1]  1.00000000  1.00000000  1.00000000  1.00000000  1.00000000  1.00000000  1.00000000  1.00000000  1.00000000 -0.97583296 -0.61915873 -1.31151020 -1.51437504
#[14] -1.51135538  0.79397030 -1.21025638 -1.46657250  0.98277557 -0.53936636 -0.63855221 -0.02568424  0.78512289  0.71934837 -0.31817844 -0.14891020  0.68253538
#[27] -1.05545863  0.85541926  0.67673403 -1.09460547 -1.70138478  0.75931881  0.98464144 -1.55739495 -0.05148017 -1.26050288  0.70467265  0.68822770 -1.24740025
#[40]  0.99877169 -1.78165575 -1.21522704  0.77518005  0.98376700 -1.53121019

如您所见，我们得到了 45 个正确的结果（10C2 = 45）。

数据

dput(df)
structure(list(outcome = structure(c(2L, 1L, 1L, 2L), .Label = c("0", 
"1"), class = "factor"), X1 = c(-0.086580111257948, 1.3225244296403, 
0.63970203781302, 1.17478656505647), X2 = c(0.116290308776141, 
-2.93084636363391, 0.67750806223535, 1.11777194347258), X3 = c(1.38404752146435, 
1.2839408555363, -0.976479813387477, 0.990836347961829), X4 = c(-1.53428156591653, 
-1.81700160188474, 0.35563308328848, 0.863904683601422), X5 = c(-0.0805126064587461, 
-0.962480324796481, 0.112310964386636, -0.257651852496691), X6 = c(1.48342629539586, 
0.677600299153581, -0.718621221409107, -0.547872283010696), X7 = c(1.52752065946695, 
-0.039941426401065, 0.384087275444754, 2.23916461213194), X8 = c(1.753974300534, 
1.22050988486485, 2.61512874217525, 1.76150083091101), X9 = c(-0.786009592713507, 
-0.176356977987529, 0.0947058204731415, 0.127134850846526), X10 = c(0.510517865869084, 
-1.24821415198133, 0.963011806720543, 0.307956641660821)), .Names = c("outcome", 
"X1", "X2", "X3", "X4", "X5", "X6", "X7", "X8", "X9", "X10"), row.names = c(NA, 
-4L), class = "data.frame")
于 2016-04-22T12:22:57.813 回答
0

你可以做

set.seed(42)
data <- as.data.frame(matrix(rnorm(110), 10, 11))
names(data) <- c("y", paste0("x", 1:10))

fit.R2 <- function(i,j, dat) summary.lm(lm(as.formula(paste0("y ~ x", i, " + x", j)), data=dat))$adj.r.squared

n <- 10
i <- 1:(n-1)
result <- data.frame(I=rep(i, n-i), J=unlist(sapply(2:n, ':', to=n)))
result$R2 <- apply(result, 1, function(ij) fit.R2(ij["I"], ij["J"], data))
于 2016-04-22T12:34:02.173 回答