numpy.bitwise_and.reduce behaving unexpectedly?

2019-02-19 10:49发布

问题:

The ufunc.reduce for numpy.bitwise_and.reduce does not appear to behave properly... am I misusing it?

>>> import numpy as np
>>> x = [0x211f,0x1013,0x1111]
>>> np.bitwise_or.accumulate(x)
array([ 8479, 12575, 12575])
>>> np.bitwise_and.accumulate(x)
array([8479,   19,   17])
>>> '%04x' % np.bitwise_or.reduce(x)
'311f'
>>> '%04x' % np.bitwise_and.reduce(x)
'0001'

The result of reduce() should be the last value of accumulate() and it's not. What am I missing here?

For the moment, I can work around by using DeMorgan's identity (swapping OR and AND, and inverting input and output):

>>> ~np.bitwise_or.reduce(np.invert(x))
17

回答1:

According to the documentation you provided, ufunc.reduce uses op.identity as an initial value.

numpy.bitwise_and.identity is 1, not 0xffffffff.... nor -1.

>>> np.bitwise_and.identity
1

So numpy.bitwise_and.reduce([0x211f,0x1013,0x1111]) is equivalent to:

>>> np.bitwise_and(np.bitwise_and(np.bitwise_and(1, 0x211f), 0x1013), 0x1111)
1
>>> 1 & 0x211f & 0x1013 & 0x1111
1

>>> -1 & 0x211f & 0x1013 & 0x1111
17

There seems to be no way to specify the initial value according to the documentation. (unlike Python builtin function reduce)