Question

I have wrote some very simple tests (I know, they are not ‘conclusive’, but they make me curious). I ran with optimization and all that jazz.

from time import time

alist = [ 2, 4, 6, 8, 10, 12, 24, 48, 64, 128 ]

def all_even( alist ):
    for val in alist:
        if not (val & 1) == 0:
            return False
    return True

def all_even_bad( alist ):
    result = False
    for val in alist:
        if not (val & 1) == 0:
            result = False
        else:
            result = True
    return result

def main():
    start = time()
    for i in range(1, 10000):
        all_even( alist )
    print('All even: {0}'.format(time() - start))

    start = time()
    for i in range(1, 10000):
        all_even_bad( alist )
    print('All even bad: {0}'.format(time() - start))


    start = time()
    for i in range(1, 10000):
        all( val & 1 == 0 for val in alist )
    print('All one: {0}'.format(time() - start))


if __name__ == '__main__':
    main()

I get results around:

> All even: 2.86299991608 
> All even bad: 3.71399998665 
> All one: 3.89900016785

It appears the built in function doesn’t bail out early?

Answer 1

all() definitely does exit early, I think the behavior difference is just the result of the overhead required to create the generator.

Here is some proof that all() does exit early:

In [8]: alist = [3] + [0] * 2**20    # alist bigger, with an early odd

In [10]: %timeit all_even(alist)
1000000 loops, best of 3: 309 ns per loop

In [11]: %timeit all_even_bad(alist)
10 loops, best of 3: 133 ms per loop

In [12]: %timeit all(val & 1 == 0 for val in alist)
1000000 loops, best of 3: 891 ns per loop

Note that even though all() is slower than all_even() here, it is still significantly faster than the version of the function that doesn’t exit early.