CODEWITHSUNDEEP
pythonarrayslistnumpybit-manipulation
havok2063
havok2063

Reputation: 567

bitwise operations between elements in a list

I have a of list of bitwise elements, e.g. [1,1,1], and I want to do a bitwise OR operation between every element in the list. So, e.g.

for [1,1,1] do

1 | 1 | 1 = 1

or for [1,17,1] do

1 | 17 | 1 = 17

How can I do this without looping? Numpy's bitwise_or only seems to work on 2 arrays. Is there a bitwise & or | that works on every element, similar to sum, or np.mean? Thanks.

Upvotes: 17

Views: 16984

Answers (4)

Unmitigated
Unmitigated

Reputation: 89204

In Python 3, reduce is no longer a built-in function. functools.reduce can be used in conjunction with operator.or_ instead.

from functools import reduce
import operator
l = [1, 17, 1]
result = reduce(operator.or_, l)
print(result) # 17

Alternatively, the following solution can be used just for ints:

from functools import reduce
l = [1, 17, 1]
result = reduce(int.__or__, l)

Upvotes: 0

Christian Borgelt
Christian Borgelt

Reputation: 119

Without importing anything, neither numpy nor operator.ior, as suggested in the other answers:

a = [1,17,1]
reduce(lambda x,y: x | y, a)

Edit: However, when I benchmarked different options, this was faster:

a = [1,17,1]; b = 0
for x in a: b |= x

This second option also has the advantage that it works in Python 3, from which reduce has been eliminated (although it can still be imported from functools).

Upvotes: 3

tcarobruce
tcarobruce

Reputation: 3838

This works for numpy reduce:

>>> ar = numpy.array([1,17,1])
>>> numpy.bitwise_or.reduce(ar)
17

Upvotes: 13

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 250891

You can use reduce with operator.ior:

>>> from operator import ior
>>> lst = [1, 17, 1]
>>> reduce(ior, lst)
17

And as suggested by @DSM in comments the numpy equivalent will be:

>>> import numpy as np
>>> arr = np.array(lst)
>>> np.bitwise_or.reduce(arr)
17

Upvotes: 32

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