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
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
