Counting Number of Element in a smart way

[bowlofred]

If the only objection is writing out the triplets, you could do the following. It runs the same commands, just does it in a loop.

from itertool import product
triplets = tuple(product((0, 1), repeat = 3))
sums = {}
for triple in triplets:
    sums[triple] = sum(np.all((results-np.array(triple))==0, axis=1))

The sums will be in a dict sums rather than as separate variables. But you can pull them out, or you could change how the keys are named.

However, this doesn't change the fundamental calls made. Perhaps someone else can see if there's a better way to have numpy do the sum more directly.