R: Efficient way to convert factor into binary matrix

Question

I'd like to convert a size n factor into a n×n binary matrix whose (i, j) element is 1 if i-th and j-th element of factor are same and 0 otherwise.

The following is a naive way to implement what I want to do but this code is quite slow. Is there any more efficient way to do the same thing?

size <- 100
id <- factor(sample(3, size, replace=TRUE))
mat <- matrix(0, nrow=size, ncol=size)

for(i in 1:size){
  for(j in 1:size){
    if(id[i] == id[j]){
      mat[i, j] <- 1
    }
  }
}

And instead of if(...) just do mat[i, j] <- (id[i] == id[j]) + 0L. — Rui Barradas
– Rui Barradas, Commented Jul 17, 2017 at 10:26

user20650 · Accepted Answer · 2017-07-17 11:24:12Z

4

Another alternative, which should be relatively fast

tcrossprod(model.matrix( ~ id + 0))

Similarly to Hong Ooi's answer you can use also sparse matrices

library(Matrix)
tcrossprod(sparse.model.matrix( ~ id + 0))

answered Jul 17, 2017 at 11:24

user20650

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Hong Ooi · Accepted Answer · 2017-07-17 10:22:32Z

3

outer can be used for this.

mat <- outer(id, id, "==")

Since the output is a binary matrix, and O(N^2) objects are kind of large, this is a good use case for sparse matrices:

library(Matrix)
mat <- Matrix(nrow=100, ncol=100)
mat[] <- outer(id, id, "==")  # [] means to assign into the existing 'mat' matrix

edited Jul 17, 2017 at 10:22

answered Jul 17, 2017 at 10:14

Hong Ooi

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2 Comments

myuuuuun Over a year ago

It's interesting way to use tensordot. Thanks!

nicola Over a year ago

Good answer, but you are not in any way making use of sparse matrices.

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