Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as
opposed to really thinking hard about how it works ;-)
On 9/28/2017 11:55 AM, Duncan Murdoch wrote:
> 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
>
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