Numpy `logical_or` for more than two arguments

Question

Numpy\'s logical_or function takes no more than two arrays to compare. How can I find the union of more than two arrays? (The same question could be asked wit

Answer 1

As boolean algebras are both commutative and associative by definition, the following statements or equivalent for boolean values of a, b and c.

a or b or c

(a or b) or c

a or (b or c)

(b or a) or c

So if you have a "logical_or" which is dyadic and you need to pass it three arguments (a, b, and c), you can call

logical_or(logical_or(a, b), c)

logical_or(a, logical_or(b, c))

logical_or(c, logical_or(b, a))

or whatever permutation you like.

---

Back to python, if you want to test whether a condition (yielded by a function test that takes a testee and returns a boolean value) applies to a or b or c or any element of list L, you normally use

any(test(x) for x in L)

Answer 2

I use this workaround which can be extended to n arrays:

>>> a = np.array([False, True, False, False])
>>> b = np.array([True, False, False, False])
>>> c = np.array([False, False, False, True])
>>> d = (a + b + c > 0) # That's an "or" between multiple arrays
>>> d
array([ True,  True, False,  True], dtype=bool)

Answer 3

using the sum function:

a = np.array([True, False, True])
b = array([ False, False,  True])
c = np.vstack([a,b,b])

Out[172]: 
array([[ True, False,  True],
   [False, False,  True],
   [False, False,  True]], dtype=bool)

np.sum(c,axis=0)>0
Out[173]: array([ True, False,  True], dtype=bool)

Answer 4

Building on abarnert's answer for n-dimensional case:

TL;DR: np.logical_or.reduce(np.array(list))

Answer 5

I've tried the following three different methods to get the logical_and of a list l of k arrays of size n:

1. Using a recursive numpy.logical_and (see below)
2. Using numpy.logical_and.reduce(l)
3. Using numpy.vstack(l).all(axis=0)

Then I did the same for the logical_or function. Surprisingly enough, the recursive method is the fastest one.

import numpy
import perfplot

def and_recursive(*l):
    if len(l) == 1:
        return l[0].astype(bool)
    elif len(l) == 2:
        return numpy.logical_and(l[0],l[1])
    elif len(l) > 2:
        return and_recursive(and_recursive(*l[:2]),and_recursive(*l[2:]))

def or_recursive(*l):
    if len(l) == 1:
        return l[0].astype(bool)
    elif len(l) == 2:
        return numpy.logical_or(l[0],l[1])
    elif len(l) > 2:
        return or_recursive(or_recursive(*l[:2]),or_recursive(*l[2:]))

def and_reduce(*l):
    return numpy.logical_and.reduce(l)

def or_reduce(*l):
    return numpy.logical_or.reduce(l)

def and_stack(*l):
    return numpy.vstack(l).all(axis=0)

def or_stack(*l):
    return numpy.vstack(l).any(axis=0)

k = 10 # number of arrays to be combined

perfplot.plot(
    setup=lambda n: [numpy.random.choice(a=[False, True], size=n) for j in range(k)],
    kernels=[
        lambda l: and_recursive(*l),
        lambda l: and_reduce(*l),
        lambda l: and_stack(*l),
        lambda l: or_recursive(*l),
        lambda l: or_reduce(*l),
        lambda l: or_stack(*l),
    ],
    labels = ['and_recursive', 'and_reduce', 'and_stack', 'or_recursive', 'or_reduce', 'or_stack'],
    n_range=[2 ** j for j in range(20)],
    logx=True,
    logy=True,
    xlabel="len(a)",
    equality_check=None
)

Here below the performances for k = 4.

And here below the performances for k = 10.

It seems that there is an approximately constant time overhead also for higher n.

Answer 6

In case someone still need this - Say you have three Boolean arrays a, b, c with the same shape, this gives and element-wise:

a * b * c

this gives or:

a + b + c

Is this what you want? Stacking a lot of logical_and or logical_or is not practical.