> On Dec 22, 2015, at 4:42 PM, Dimitri Liakhovitski
> <dimitri.liakhovit...@gmail.com> wrote:
>
> # I have a matrix x:
>
> k <- 20
> N <- 5
> set.seed(123)
> x <- matrix(c(sample(1:k, N, replace = F),
> sample(1:k, N, replace = F),
> sample(1:k, N, replace = F),
> sample(1:k, N, replace = F),
> sample(1:k, N, replace = F),
> sample(1:k, N, replace = F)), byrow = T, ncol = 5)
> colnames(x) <- paste0("column", 1:5)
> (x)
>
> # I want to reshape it into a matrix 'result' with k columns (not N).
> # 'result' should contain the same number of rows as x, and it should have
> # in each row 1s in those columns that correspond to the entries in x
> in the same row.
> # For example, the first row of 'result' should contain 1s in columns
> 6, 8, 15, 16, and 17, etc.
> # The remaining entries should be zeros.
>
> result <- matrix(rep(0, nrow(x) * k), nrow = nrow(x))
> colnames(result) <- paste0("item", 1:k)
>
> # I can see how to do it by looping through rows of result.
> # But I need to do it fast (not using a loop).
> # I feel like I should subset 'result' with 'x', but I am not sure how.
>
> Thank you very much!
Something like this might work for you:
result <- matrix(0, nrow(x), k)
> result
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0 0 0 0 0 0
[,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
[1,] 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0 0
result[cbind(as.vector(row(x)), as.vector(x))] <- 1
> result
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 0 0 0 0 0 1 0 1 0 0 0 0
[2,] 1 0 0 0 0 0 0 1 0 1 1 0
[3,] 0 1 0 0 0 0 0 0 1 1 0 0
[4,] 1 0 0 0 1 1 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0 0 0 0 1 1
[6,] 0 0 1 0 1 0 0 0 0 0 1 0
[,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
[1,] 0 0 1 1 1 0 0 0
[2,] 0 0 0 0 1 0 0 0
[3,] 1 0 0 0 0 0 0 1
[4,] 0 0 0 1 0 1 0 0
[5,] 0 1 0 0 1 1 0 0
[6,] 0 0 1 0 0 0 1 0
See ?row
Regards,
Marc Schwartz
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