可以先自定义函数，也可以用的时候再定义。
> mat <- matrix(c(1:3,7:9,4:6), byrow = T, nc = 3)
> mat
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 7 8 9
[3,] 4 5 6
> apply(mat, 2, function(x){order(x, decreasing=T)[1]}) # 查找每一列
[1] 2 2 2
> apply(mat, 1, function(x){order(x, decreasing=T)[1]}) # 查找每一行
[1] 3 3 3
> apply(mat, 1, function(x){which.max(x)}) # 查找每一行
[1] 3 3 3
> n <- letters[1:5]
> n
[1] "a" "b" "c" "d" "e"
> t <- apply(mat, 1, function(x){which.max(x)})
> n[t]
[1] "c" "c" "c"
另一个例子：
MaxVar <- function(x, na.rm = FALSE) {
## compute `max`
maxx <- max(x, na.rm = na.rm)
## which equal the max
wmax <- which(x == max(x))
## how many equal the max
nmax <- length(wmax)
## return
out <- if(nmax > 1L) {
c(999, NA)
} else {
c(maxx, wmax)
}
out
}
And use it like this:
> new <- apply(Mydata[, -1], 1, MaxVar)
> new
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 3 4 999 999 999 4 4 2 4 999
[2,] 1 4 NA NA NA 4 2 3 4 NA
> Mydata <- cbind(Mydata, Max = new[1, ], Var = new[2, ])
> Mydata
ID X1 X2 X3 X4 Max Var
1 1 3 1 1 1 3 1
2 2 1 2 1 4 4 4
3 3 1 1 1 1 999 NA
4 4 1 3 3 1 999 NA
5 5 2 2 2 1 999 NA
6 6 1 2 3 4 4 4
7 7 2 4 3 3 4 2
8 8 1 1 2 1 2 3
9 9 3 2 1 4 4 4
10 10 4 4 4 4 999 NA
参考： http://stackoverflow.com/questions/29683339/number-of-maximums-in-each-row-and-more/29686201#29686201
时间： 2024-10-13 21:59:33