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,
apparently you forgot the commented, minimal, self-contained,
reproducible code part...
L = 10
M = 20
N = 30
P = 40
set.seed(1)
A = array(rnorm(L*M*N*P), dim=c(L, M, N, P))
B = array(rnorm(L*M*N), dim=c(L, M, N))
B[sample(100, 10)] = 0
C = array(0, dim=c(L, M, N, P))
for (i in 1:L) {
for (j in
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
Hum you are rigth, I forgot of 'else'.
On 06/03/2008, Benilton Carvalho [EMAIL PROTECTED] wrote:
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
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)
Thank both of you for the suggestions!
Gang
On Mar 6, 2008, at 2:12 PM, Henrique Dallazuanna wrote:
Hum you are rigth, I forgot of 'else'.
On 06/03/2008, Benilton Carvalho [EMAIL PROTECTED] wrote:
no, it won't.
you're doing the right math on the valid subset... but you're not
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