Fastest way to check if a number is divisible by another in python

问题

So I’ve been doing something with primes in python, and I’m currently using this

def isDivisible(number,divisor):
    if number % divisor == 0:
        return True
    return False

to check if a number is divisible by the divisor. So I was wondering if there was a faster way to do this?

回答1:

What about:

return (number % divisor == 0)

回答2:

A speed test shows that checking not() is faster than a != 0 solution:

%%timeit 
not(8 % 3)
# 19 ns ± 0.925 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%%timeit 
8 % 3 != 0
# 27.1 ns ± 0.929 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

回答3:

I doubt there is a "faster" way of checking this. And it seems pretty simple. However, I would write your function as:

def isDivisible(number, divisor):
    return number % divisor == 0

回答4:

maybe you can use lambda:

isDivisible = lambda x,y: x%y==0

isDivisible(4,2)

output:

True

回答5:

Not faster, but note that number % divisor == 0 already returns a boolean. So you colud simply do

is_divisible = lambda number, divisor: number % divisor == 0

to define your funtion. This however is still the same method you are using. Could be marginlly faster, I havn't tested.

来源：https://stackoverflow.com/questions/52773914/fastest-way-to-check-if-a-number-is-divisible-by-another-in-python