Dearc
b <- NULL
a <- c('A' ,'B' ,'C', 'a' ,'b' ,'c' ,'d' ,'e' ,'f')
for(i in 1:3)
{
b[[i]] <- rep(a[i],length(a))
x <- do.call(cbind,b)
x1 <- cbind(x,a)[-c(seq(1:i)),]
}
> x1
a
[1,] "A" "B" "C" "a"
[2,] "A" "B" "C" "b"
[3,] "A" "B" "C" "c"
[4,] "A" "B" "C" "d"
[5,] "A" "B" "C" "e"
[6,] "A" "B" "C" "f"
mengchen
[未知用户]
a <- c('A' ,'B' ,'C', 'a' ,'b' ,'c' ,'d' ,'e' ,'f')
d <- data.frame(c1=a[1], c2=a[2], c3=a[3], c4=a[-(1:3)])
Dearc
[未知用户]
测试了下可以办到，
`
b <- NULL
a <- c('A' ,'B' ,'C', 'a' ,'b' ,'c' ,'d' ,'e' ,'f')
a2 <- c('D', 'E', 'F', 'd', 'e')
a3 <- c('G', 'H', 'I', 'g', 'h',' i')
df <- list(a,a2,a3)
fun_1 <- function(z){for(i in 1:3) {
b［［i］］ <- rep(z,length(z))
x <- do.call(cbind,b)
x1 <- cbind(x,z)[-c(seq(1:i)),]
}
return (x1)}
do.call(rbind,sapply(df,FUN = fun_1))｀
最上面的回答不知道为什么代码放进code里面［［会自动变成［。注意下b［［i］］是两个方括号
z
[1,] "A" "B" "C" "a"
[2,] "A" "B" "C" "b"
[3,] "A" "B" "C" "c"
[4,] "A" "B" "C" "d"
[5,] "A" "B" "C" "e"
[6,] "A" "B" "C" "f"
[7,] "D" "E" "F" "d"
[8,] "D" "E" "F" "e"
[9,] "G" "H" "I" "g"
[10,] "G" "H" "I" "h"
[11,] "G" "H" "I" " i"
