# [SciPy-user] filling array without loop...

fred fredmfp@gmail....
Mon Apr 23 08:06:40 CDT 2007

Anne Archibald a écrit :
> On 22/04/07, fred <fredmfp@gmail.com> wrote:
>
>
>> Forget that is a convolution (because in fact, it is really not),
>> simply a scalar product (inner product says scipy) in each cells of an
>> array, as I wrote in my example.
>>
>> Does it changes something ?
>>
>
> Uh, maybe I'm confused -
Ok, so let me explain a little more...

I have two 2D matrices, say A et B, with same dims = 571x876.
B is to be computed, A is known.
Each cell of the matrix B is computed from a "scalar product" with a few
cells of the matrix A and
a "weights" vector W:

B[i,j] = \sum_0^8 w_n a_n

where

- W = [w_n] is a 1D vector (computed from solve(a,b)) with dim = 9

- a_n = A[i,j]

a_n are selected from a submatrix of A (much smaller than A), say A':

a_2 -- a_5 -- a_8
|         |        |
A' = a_1 -- a_4 -- a_7
|         |        |
a_0 -- a_3 -- a_6

My idea is to write B[i,j] as a scalar product between W and A'
flattened to a 1D vector:

B[i,j] = dot (W,A')

The problem, from my own opinion, is once again, the two (three in
general case) loops to compute each cell
of the matrix B.

So my question is : is it possible to fill the B matrix without loops ?

I hope I more clear.