[R] filling the matrix row by row in the order from lower to larger elements
Rui Barradas
rui1174 at sapo.pt
Sat Apr 7 01:02:59 CEST 2012
Hello,
Oops!
What happened to the function 'f'?
Forgot to copy and pasted only the rest, now complete.
f <- function(x){
nr <- nrow(x)
result <- matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) <- colnames(x)
inp.ord <- order(x)[1:nr] - 1 # Keep only one per row, must be zero based
inx <- cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] <- 100
result
}
# Original example
input <- as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
(input)
desired.result <- as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100)))
(desired.result)
all.equal(f(input), desired.result)
# Two other examples
set.seed(123)
(x <- matrix(sample(10, 10), ncol=2))
f(x)
(y <- matrix(sample(40, 40), ncol=5))
f(y)
Note that there's no loops (or apply, which is also a loop.)
Rui Barradas
--
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