[R] building random matrices from vectors of random parameters

[Duncan Murdoch]

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]]
>            [,1]      [,2]
> [1,] 0.0000000 0.8643679
> [2,] 0.4731249 0.7750431
> 
> [[2]]
>            [,1]      [,2]
> [1,] 0.0000000 0.8838286
> [2,] 0.5895258 0.7880983
> 
> [[3]]
>            [,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