Separable filter on numpy array

Question

Say I have a numpy array a, and I want to create a new array, b such that b[i, j] is a function of, say:

a[i-1, j-1], a[i-1, j  ], a[i-1, j+1],
a[i  , j-1], a[i  , j  ], a[i  , j+1],
a[i+1, j-1], a[i+1, j  ], a[i+1, j+1]

What would be the fastest way to do this?

As this is a separable filter, is there any way to run this in multiple threads? (not processes, because I would have to copy the data back)

Or is writing C code to bypass the GIL mandatory?

Partial solutions (like assuming the function is linear) are welcome too.

Do you mean like a rolling/moving window or filter? the example in this link is for a summation for a 3x3 window over a 2D array johnvinyard.com/blog/?p=268 — user1121588
– user1121588, Commented Apr 25, 2015 at 0:14
Sounds like a number of other SO questions, most using the term sliding windows (or moving). Though most focus on iterating over the window, not on splitting up the task among threads or processes. — hpaulj
– hpaulj, Commented Apr 25, 2015 at 4:17

hpaulj · Accepted Answer · 2015-04-25 04:52:04Z

An idealized numpy way of working with a sliding window like this is to construct a 4D array

C.shape = (N,M,3,3)

where

C[i,j,:,:] = np.array([a[i-1, j-1], a[i-1, j  ], a[i-1, j+1],
                       a[i  , j-1], a[i  , j  ], a[i  , j+1],
                       a[i+1, j-1], a[i+1, j  ], a[i+1, j+1]])

and write your function do some sort of reduction on the last 2 dimensions. sum or mean would be typical, e.g.

B = C.sum(axis=(2,3))

Other SO questions show how to use np.lib.stride_tricks.as_strided to construct such an array. But with only a 3x3 subarray, it might be just as fast to do something like

C = np.zeros((N,M,3,3))
C[:,:,0,0] = a[:-1,:-1]
etc.

(or use hstack and vstack to the same effect).

But a nice thing (or maybe not so nice) about the strided approach is that it doesn't involve copy any data of a - it is just a view.

As to splitting the job into pieces, I can imagine using slices of C (on the 1st 2 dimensions), e.g.

 C[0:100,0:100,:,:].sum(axis=(2,3))

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