Vectorizing order in R

Question

I am trying to order each row in a matrix with few columns and many rows. Is there a vectorized version of this in R? More concretely, let's set our seed to 10 and make an example matrix:

set.seed(10)
example.matrix = replicate(12,runif(500000))

To order example.matrix, I would,

ordered.example = apply(example.matrix,1,order)

But that is very slow and I would love something faster. As an analogy,

rowSums(example.matrix)

Is preferable to,

apply(example.matrix,1,sum)

Much appreciated.

Answer 1

Here's a way of speeding it up 10x. It's specifically tailored to your example and depending on what your real data is like this method may or may not work.

The idea is to add 0 to first row, 1 to second and so on, then collapse it to a vector, sort that and then recombine into a matrix:

N = 12; M = 500000; d = replicate(N,runif(M))

system.time(d1<-t(apply(d, 1, order)))
#   user  system elapsed 
#  11.26    0.06   11.34 

system.time(d2<-matrix(order(as.vector(t(matrix(as.vector(d) + 0:(M-1), nrow = M)))) -
                       rep(0:(M-1), each = N)*N, nrow = M, byrow = T))
#   user  system elapsed 
#   1.39    0.14    1.53 

# Note: for some reason identical() fails, but the two are in fact the same
sum(abs(d1-d2))
# 0