Question

I have two matrices. Both are filled with zeros and ones. One is a big one (3000 x 2000 elements), and the other is smaller ( 20 x 20 ) elements. I am doing something like:

newMatrix = (size of bigMatrix), filled with zeros
l = (a constant)

for y in xrange(0, len(bigMatrix[0])):
    for x in xrange(0, len(bigMatrix)):

        for b in xrange(0, len(smallMatrix[0])):
            for a in xrange(0, len(smallMatrix)):

                if (bigMatrix[x, y] == smallMatrix[x + a - l, y + b - l]):
                    newMatrix[x, y] = 1

Which is being painfully slow. Am I doing anything wrong? Is there a smart way to make this work faster?

edit: Basically I am, for each (x,y) in the big matrix, checking all the pixels of both big matrix and the small matrix around (x,y) to see if they are 1. If they are 1, then I set that value on newMatrix. I am doing a sort of collision detection.

Answer 1

I can think of a couple of optimisations there –
As you are using 4 nested python “for” statements, you are about as slow as you can be.

I can’t figure out exactly what you are looking for –
but for one thing, if your big matrix “1”s density is low, you can certainly use python’s “any” function on bigMtarix’s slices to quickly check if there are any set elements there — you could get a several-fold speed increase there:

step = len(smallMatrix[0])
for y in xrange(0, len(bigMatrix[0], step)):
    for x in xrange(0, len(bigMatrix), step):
        if not any(bigMatrix[x: x+step, y: y + step]):
            continue
        (...) 

At this point, if still need to interact on each element, you do another pair of indexes to walk each position inside the step – but I think you got the idea.

Apart from using inner Numeric operations like this “any” usage, you could certainly add some control flow code to break-off the (b,a) loop when the first matching pixel is found.
(Like, inserting a “break” statement inside your last “if” and another if..break pair for the “b” loop.

I really can’t figure out exactly what your intent is – so I can’t give you more specifc code.