Python: rewrite a looping numpy math function to run on GPU

Question

Can someone help me rewrite this one function (the doTheMath function) to do the calculations on the GPU? I used a few good days now trying to get my head

Answer 1

Tweak #1

Its usually advised to vectorize things when working with NumPy arrays. But with very large arrays, I think you are out of options there. So, to boost performance, a minor tweak is possible to optimize on the last step of summing.

We could replace the step that makes an array of 1s and 0s and does summing :

np.where(((abcd <= data2a) & (abcd >= data2b)), 1, 0).sum()

with np.count_nonzero that works efficiently to count True values in a boolean array, instead of converting to 1s and 0s -

np.count_nonzero((abcd <= data2a) & (abcd >= data2b))

Runtime test -

In [45]: abcd = np.random.randint(11,99,(10000))

In [46]: data2a = np.random.randint(11,99,(10000))

In [47]: data2b = np.random.randint(11,99,(10000))

In [48]: %timeit np.where(((abcd <= data2a) & (abcd >= data2b)), 1, 0).sum()
10000 loops, best of 3: 81.8 µs per loop

In [49]: %timeit np.count_nonzero((abcd <= data2a) & (abcd >= data2b))
10000 loops, best of 3: 28.8 µs per loop

Tweak #2

Use a pre-computed reciprocal when dealing with cases that undergo implicit broadcasting. Some more info here. Thus, store reciprocal of dif and use that instead at the step :

((((A  - Cmin) / dif) + ((B  - Cmin) / dif) + ...

Sample test -

In [52]: A = np.random.rand(10000)

In [53]: dif = 0.5

In [54]: %timeit A/dif
10000 loops, best of 3: 25.8 µs per loop

In [55]: %timeit A*(1.0/dif)
100000 loops, best of 3: 7.94 µs per loop

You have four places using division by dif. So, hopefully this would bring out noticeable boost there too!