I use itertools.product to generate all possible variations of 4 elements of length 13. The 4 and 13 can be arbitrary, but as it is, I get 4^13 results, which is a lot. I need the result as a Numpy array and currently do the following:
c = it.product([1,-1,np.complex(0,1), np.complex(0,-1)], repeat=length)
sendbuf = np.array(list(c))
With some simple profiling code shoved in between, it looks like the first line is pretty much instantaneous, whereas the conversion to a list and then Numpy array takes about 3 hours. Is there a way to make this quicker? It's probably something really obvious that I am overlooking.
Thanks!
The NumPy equivalent of itertools.product()
is numpy.indices()
, but it will only get you the product of ranges of the form 0,...,k-1:
numpy.rollaxis(numpy.indices((2, 3, 3)), 0, 4)
array([[[[0, 0, 0],
[0, 0, 1],
[0, 0, 2]],
[[0, 1, 0],
[0, 1, 1],
[0, 1, 2]],
[[0, 2, 0],
[0, 2, 1],
[0, 2, 2]]],
[[[1, 0, 0],
[1, 0, 1],
[1, 0, 2]],
[[1, 1, 0],
[1, 1, 1],
[1, 1, 2]],
[[1, 2, 0],
[1, 2, 1],
[1, 2, 2]]]])
For your special case, you can use
a = numpy.indices((4,)*13)
b = 1j ** numpy.rollaxis(a, 0, 14)
(This won't run on a 32 bit system, because the array is to large. Extrapolating from the size I can test, it should run in less than a minute though.)
EIDT: Just to mention it: the call to numpy.rollaxis()
is more or less cosmetical, to get the same output as itertools.product()
. If you don't care about the order of the indices, you can just omit it (but it is cheap anyway as long as you don't have any follow-up operations that would transform your array into a contiguous array.)
EDIT2: To get the exact analogue of
numpy.array(list(itertools.product(some_list, repeat=some_length)))
you can use
numpy.array(some_list)[numpy.rollaxis(
numpy.indices((len(some_list),) * some_length), 0, some_length + 1)
.reshape(-1, some_length)]
This got completely unreadable -- just tell me whether I should explain it any further :)