numpy.all with integer arguments returns an integer

Question

Why do this happen?

>>> map(numpy.all, range(-2, 3))
[-2, -1, 0, 1, 2]

Is it intentional or is the integer just falling through a crack?

Does it have to do with:

>>> map(numpy.all, [False, True])
[False, True]

I'm running Numpy 1.8.0.dev-74b08b3 and Python 2.7.3

Answer 1

Using map(numpy.all, range(-2,3)) is actually creating a list with:

[numpy.all(-2), numpy.all(-1), numpy.all(0), numpy.all(1), numpy.all(2)]

giving

[-2, -1, 0, 1, 2]

If you did map(lambda x: numpy.all([x]), range(-2,3)), it would do:

[numpy.all([-2]), numpy.all([-1]), numpy.all([0]), numpy.all([1]), numpy.all([2])]

giving

[True, True, False, True, True]

As posted by @Mark Dickinson, there is a known issue with numpy.all in which it returns the input value instead of True or False for some inputs. In your second example map(numpy.all, [False, True]) does exactly as before, returning the input value.