Can this Python logical expression be replaced with a Numpy one?

• & works for scalars.

• np.logicaland.reduce()

• np.all() with a list []

# Source - https://stackoverflow.com/a/79919211
Posted by Mohammad Othman
Retrieved 2026-04-02, License - CC BY-SA 4.0
import numpy as np
import time
N, M, K = 10000, 10000, 1000000
a = np.random.randint(2 * K, size=(N, M))
element = 9999
def methodand():
    return (
        np.any(a[0, :] == element)
        and np.any(a[-1, :] == element)
        and np.any(a[:, 0] == element)
        and np.any(a[:, -1] == element)
    )
def methodbitwise():
    return (
        np.any(a[0, :] == element)
        & np.any(a[-1, :] == element)
        & np.any(a[:, 0] == element)
        & np.any(a[:, -1] == element)
    )
def methodlogicalreduce():
    return np.logicaland.reduce(
        [
            np.any(a[0, :] == element),
            np.any(a[-1, :] == element),
            np.any(a[:, 0] == element),
            np.any(a[:, -1] == element),
        ]
    )
def methodnpall():
    return np.all(
        [
            np.any(a[0, :] == element),
            np.any(a[-1, :] == element),
            np.any(a[:, 0] == element),
            np.any(a[:, -1] == element),
        ]
    )
iterations = 10000
methods = [
    ('Python "and"', methodand),
    ('Bitwise "&"', methodbitwise),
    ("np.logicaland.reduce()", methodlogicalreduce),
    ("np.all() with list", methodnpall),
]
for name, func in methods:
    start = time.perfcounter()
    for  in range(iterations):
        result = func()
    elapsed = time.perfcounter() - start
    print(f"{name:

I don't think you can replace all of the ands with numpy operations, but you can replace two of them:

(a[::N-1, :] == K).any(axis=1).all() and (a[:, ::M-1] == K).any(axis=0).all()

It is faster still (for the array size given) to just compute a==K for the whole array:

k = a == K
k[::N-1, :].any(axis=1).all() and k[:, ::M-1].any(axis=0).all()

Timings are: 8.7us for the original in the question, 6.2us for the first way and 4.8us for the second, for M, N, K = 31, 31, 5.

I am aware that this wasn't really the question, but it bugs me a bit.

Checking all items in the bordering vectors for the purpose of knowing if any match, is overkill. If we only want to know whether there is one match, we can stop checking the vector as soon as we found one. To do this, we can implement a small function.

def containsk(arr, K):
    for a in arr:
        if a == K:
            return True
    return False

np.any(a[0, :]==K) & np.any(a[-1, :]==K) & np.any(a[:, 0]==K) & np.any(a[:, -1]==K)

containsk(a[0, :],K) & containsk(a[-1, :],K) & containsk(a[:, 0], K) & contains_k(a[:, -1],K)

needs only 0.1 second to complete.