Is There A Binary Or Operator In Python That Works On Arrays?

I have come from a matlab background to python and i am just wondering if there is a simple operator in python that will perform the following function: a = [1, 0, 0, 1, 0, 0] b =

Solution 1:

You could use a list comprehension. Use izip from itertools if you're using Python 2.

c = [x | y for x, y in zip(a, b)]

Alternatively, @georg pointed out in a comment that you can import the bitwise or operator and use it with map. This is only slightly faster than the list comprehension. map doesn't need wrapped with list() in Python 2.

importoperator
c = list(map(operator.or_, a, b))

Performance

List comprehension:

$ python -m timeit -s "a = [1, 0, 0, 1, 0, 0]; b = [0, 1, 0, 1, 0, 1]" \
> "[x | y for x, y in zip(a, b)]"1000000 loops, best of 3: 1.41 usec per loop

Map:

$ python -m timeit -s "a = [1, 0, 0, 1, 0, 0]; b = [0, 1, 0, 1, 0, 1]; \
> from operator import or_""list(map(or_, a, b))"1000000 loops, best of 3: 1.31 usec per loop

NumPy

$ python -m timeit -s "import numpy; a = [1, 0, 0, 1, 0, 0]; \
> b = [0, 1, 0, 1, 0, 1]""na = numpy.array(a); nb = numpy.array(b); na | nb"100000 loops, best of 3: 6.07 usec per loop

NumPy (where a and b have already been converted to numpy arrays):

$ python -m timeit -s "import numpy; a = numpy.array([1, 0, 0, 1, 0, 0]); \
> b = numpy.array([0, 1, 0, 1, 0, 1])""a | b"1000000 loops, best of 3: 1.1 usec per loop

Conclusion: Unless you need NumPy for other operations, it's not worth the conversion.

Solution 2:

If your data was in numpy arrays then yes this would work:

In [42]:

a = np.array([1, 0, 0, 1, 0, 0])
b = np.array([0, 1, 0, 1, 0, 1])
c = a|b
print(c)
[110101]
Out[42]:
[1, 1, 0, 1, 0, 1]

Solution 3:

map(lambda (a,b): a|b,zip(a,b))

Solution 4:

You can use numpy.bitwise_or

>>>import numpy>>>a = [1, 0, 0, 1, 0, 0]>>>b = [0, 1, 0, 1, 0, 1]>>>numpy.bitwise_or(a,b)
array([1, 1, 0, 1, 0, 1])

Solution 5:

c = [q|w for q,w inzip(a,b)]
print c
# [1, 1, 0, 1, 0, 1]