Iterating over arbitrary dimension of numpy.array
What you propose is quite fast, but the legibility can be improved with the clearer forms:
for i in range(c.shape[-1]):
print c[:,:,i]
or, better (faster, more general and more explicit):
for i in range(c.shape[-1]):
print c[...,i]
However, the first approach above appears to be about twice as slow as the swapaxes()
approach:
python -m timeit -s 'import numpy; c = numpy.arange(24).reshape(2,3,4)' \
'for r in c.swapaxes(2,0).swapaxes(1,2): u = r'
100000 loops, best of 3: 3.69 usec per loop
python -m timeit -s 'import numpy; c = numpy.arange(24).reshape(2,3,4)' \
'for i in range(c.shape[-1]): u = c[:,:,i]'
100000 loops, best of 3: 6.08 usec per loop
python -m timeit -s 'import numpy; c = numpy.arange(24).reshape(2,3,4)' \
'for r in numpy.rollaxis(c, 2): u = r'
100000 loops, best of 3: 6.46 usec per loop
I would guess that this is because swapaxes()
does not copy any data, and because the handling of c[:,:,i]
might be done through general code (that handles the case where :
is replaced by a more complicated slice).
Note however that the more explicit second solution c[...,i]
is both quite legible and quite fast:
python -m timeit -s 'import numpy; c = numpy.arange(24).reshape(2,3,4)' \
'for i in range(c.shape[-1]): u = c[...,i]'
100000 loops, best of 3: 4.74 usec per loop
I'd use the following:
c = numpy.arange(2 * 3 * 4)
c.shape = (2, 3, 4)
for r in numpy.rollaxis(c, 2):
print(r)
The function rollaxis creates a new view on the array. In this case it's moving axis 2 to the front, equivalent to the operation c.transpose(2, 0, 1)
.