How to convert a numpy.array to binary?

Question

I'm using the following function with Python2.7:

def array2int(pixels):
    out = 0
    for bit in pixels:
        out = (out << 1) | bit
    return out

Which usually works, but if I pass

v=np.array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
        1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
array2int(v.astype(int))

it returns -262145.

@PaulPanzer ahah, well in that case you can use this modification: int("".join(map(lambda x: str(int(x)), v.astype(bool))), 2) — pault
– pault, Commented Dec 1, 2017 at 18:00

Paul Panzer · Accepted Answer · 2017-12-01 18:41:53Z

3

Unlike Python, numpy by default uses fixed size integers. Those can overflow:

1<<65
# 36893488147419103232 # correct
npone = np.ones([1], dtype=int)[0]
npone
# 1
type(npone)
# <class 'numpy.int64'>
npone<<65
# 2 # wrong

When adding or bitwise oring or whatever a python int and a numpy int, numpy typically wins and the result will be a numpy int:

out = 1
type(out)
# <class 'int'>
out = (out << 1) | npone
type(out)
# <class 'numpy.int64'>

To prevent that from happening in your function you can explicitly cast bit to a proper int:

        out = (out << 1) | int(bit)

answered Dec 1, 2017 at 18:41

Paul Panzer

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