no, it won't.
you're doing the right math on the "valid" subset... but you're not
returning the zeros where needed.... therefore, the whole thing will
get recycled to match the dimensions.
b
On Mar 6, 2008, at 2:03 PM, Henrique Dallazuanna wrote:
I think this should work:
array(A[abs(B) > 10e-5]/B[abs(B) > 10e-5], dim=c(L, M, N, P))
On 06/03/2008, Gang Chen <[EMAIL PROTECTED]> wrote:
I have two arrays A and B with dimensions of (L, M, N, P) and (L, M,
N), and I want to do
for (i in 1:L) {
for (j in 1:M) {
for (k in 1:N) {
if (abs(B[i, j, k]) > 10e-5) C[i, j, k,] <- A[i, j, k,]/B[i, j, k]
else C[i, j, k,] <- 0
}
}
}
How can I get C more efficiently than looping?
Thanks,
Gang
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Henrique Dallazuanna
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25° 25' 40" S 49° 16' 22" O
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and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
