# [R] building random matrices from vectors of random parameters

Duncan Murdoch murdoch.duncan at gmail.com
Thu Sep 28 17:55:17 CEST 2017

```On 28/09/2017 9:10 AM, Evan Cooch wrote:
> Thanks for both the mapply and array approaches! However, although
> intended to generate the same result, they don't:
>
> # mapply approach
>
> n = 3
> sa <- rnorm(n,0.8,0.1)
> so <- rnorm(n,0.5,0.1)
> m <- rnorm(n,1.2,0.1)
> mats = mapply(function(sa1, so1, m1)
> matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE)
>
> print(mats)
>
> []
>            [,1]      [,2]
> [1,] 0.0000000 0.8643679
> [2,] 0.4731249 0.7750431
>
> []
>            [,1]      [,2]
> [1,] 0.0000000 0.8838286
> [2,] 0.5895258 0.7880983
>
> []
>            [,1]      [,2]
> [1,] 0.0000000 1.1491560
> [2,] 0.4947322 0.9744166
>
>
> Now, the array approach:
>
> # array approach
>
> ms <- array(c(rep(0, 3),sa*m,so,sa), c(3, 2, 2))
>
> for (i in 1:n) { print(ms[i,,])
>
>            [,1]      [,2]
> [1,] 0.0000000 0.4731249
> [2,] 0.8643679 0.7750431
>
>            [,1]      [,2]
> [1,] 0.0000000 0.5895258
> [2,] 0.8838286 0.7880983
>
>           [,1]      [,2]
> [1,] 0.000000 0.4947322
> [2,] 1.149156 0.9744166
>
>
> These matrices are the transpose of those returned by the mapply
> approach. To see if one approach or the other is 'confused', I simply
> rerun setting sd=0 for the parameters -- thus, every matrix will be the
> same. The correct matrix would be:
>
>       [,1] [,2]
> [1,]  0.0 0.96
> [2,]  0.5 0.80
>
>
> In fact, this is what is returned by the mapply approach, while the
> array approach returns the transpose. I gather the 'missing step' is to
> use aperm, but haven't figured out how to get that to work...yet.
>
>
> On 9/28/2017 5:11 AM, Duncan Murdoch wrote:
>> ms <- array(c(rep(0, 5),sa*m,so,sa), c(5, 2, 2))
>

Sorry about that -- I didn't notice the "byrow = T" in your original.

Duncan Murdoch

```