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可以先自定义函数,也可以用的时候再定义。
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> 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