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 no
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)
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 dimension
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) {
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 i
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