Iterating over arbitrary dimension of numpy.array in Python Programming Language?

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Answer(s) available: 1

Savitri Dora

, Agricultural Sciences Professor, Retention

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What you propose is quite fast, but the legibility can be improved with the clearer forms:

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for i in range(c.shape[-1]):"
    print c[:,:,i]"
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or, better (faster, more general and more explicit):

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for i in range(c.shape[-1]):"
    print c[...,i]"
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However, the first approach above appears to be about twice as slow as the swapaxes() approach:

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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"
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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"
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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"
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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).

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Note however that the more explicit second solution c[...,i] is both quite legible and quite fast:

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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"
"

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