In Python, what is a good way to round towards zero in integer division?

Do floating point division then convert to an int. No extra modules needed.

Python 3:

>>> int(-1 / 2)0>>> int(-3 / 2)-1>>> int(1 / 2)0>>> int(3 / 2)1

Python 2:

>>> int(float(-1) / 2)0>>> int(float(-3) / 2)-1>>> int(float(1) / 2)0>>> int(float(3) / 2)1

Python's default division of integers is return the floor (towards negative infinity) with no ability to change that. You can read the BDFL's reason why.

To do 'round up' division, you would use:

>>> a=1>>> b=2>>> (a+(-a%b))//b1>>> a,b=-1,2>>> (a+(-a%b))//b0

To do truncation towards zero, and maintain integer division, you use (a+(-a%b))//b if either a or b are negative and the default division if both are positive.

This will do integer division and always round towards zero:

>>> a=1>>> b=2>>> a//b if a*b>0 else (a+(-a%b))//b0>>> a=-1>>> b=2>>> a//b if a*b>0 else (a+(-a%b))//b0>>> a,b=-3,2>>> a//b if a*b>0 else (a+(-a%b))//b-1>>> a,b=3,2>>> a//b if a*b>0 else (a+(-a%b))//b1

footnote

Interestingly enough, C99 declares that round towards zero is the default:

#include <stdio.h>int main(int argc, const char * argv[]){    int a=-3;    int b=2;    printf("a=%d, b=%d, a/b=%d\n",a,b,a/b);    a=3;    printf("a=%d, b=%d, a/b=%d\n",a,b,a/b);    return 0;}

Prints:

a=-3, b=2, a/b=-1a=3, b=2, a/b=1

For what it's worth, my own favourite solution is this one. Integer arithmetic only, a single division, and everything else linear time:

def integer_divide_towards_zero(a, b):    return -(-a // b) if a < 0 else a // b

That assumes that b is positive, but in most of the applications I've seen that's true. If you need to deal with negative b too, then the function becomes marginally more complicated:

def integer_divide_towards_zero(a, b):    return -(-a // b) if (a < 0) ^ (b < 0) else a // b

Some sample outputs:

>>> integer_divide_towards_zero(11, 3)3>>> integer_divide_towards_zero(-11, 3)-3>>> integer_divide_towards_zero(6, 3)2>>> integer_divide_towards_zero(-6, 3)-2>>> integer_divide_towards_zero(11, -3)-3>>> integer_divide_towards_zero(-11, -3)3