Fix Python – Iterating over a numpy array


Asked By – Ram Rachum

Is there a less verbose alternative to this:

for x in xrange(array.shape[0]):
    for y in xrange(array.shape[1]):
        do_stuff(x, y)

I came up with this:

for x, y in itertools.product(map(xrange, array.shape)):
    do_stuff(x, y)

Which saves one indentation, but is still pretty ugly.

I’m hoping for something that looks like this pseudocode:

for x, y in array.indices:
    do_stuff(x, y)

Does anything like that exist?

Now we will see solution for issue: Iterating over a numpy array


I think you’re looking for the ndenumerate.

>>> a =numpy.array([[1,2],[3,4],[5,6]])
>>> for (x,y), value in numpy.ndenumerate(a):
...  print x,y
0 0
0 1
1 0
1 1
2 0
2 1

Regarding the performance. It is a bit slower than a list comprehension.

X = np.zeros((100, 100, 100))

%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop

%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop

If you are worried about the performance you could optimise a bit further by looking at the implementation of ndenumerate, which does 2 things, converting to an array and looping. If you know you have an array, you can call the .coords attribute of the flat iterator.

a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop

This question is answered By – SiggyF

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0