How to judge a int number odd or even? (the binary way)

Question

I wanna use basic knowledge to improve the efficiency of code. I know that in binary system. when the last digit of number is 1,this is a odd number.and 0 is even number. How to use this way to judge a int number in python? Is that python give any build-in method to do it?

Answer 1

AND it with 1:

  0000101001000101
  0000000000000001
&
__________________
  0000000000000001

If you get 1, the number is odd. If you get 0, the number is even. While this works, I would use the modulo operator instead:

>>> 8888 % 2
0
>>> 8881 % 2
1

It's faster and more straightforward to read:

In [5]: numbers = [random.randint(1, 1000000) for n in range(100000)]

In [6]: %timeit [n & 1 == 0 for n in numbers]
11 ms ± 390 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [7]: %timeit [n % 2 == 0 for n in numbers]
8.05 ms ± 244 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

You can & the number and 1, if you get 0 then the number is even, 1 means the number is odd.

>>> 2 & 1
0
>>> 3 & 1
1