Is divmod() faster than using the % and // operators?

To measure is to know (all timings on a Macbook Pro 2.8Ghz i7):

>>> import sys, timeit>>> sys.version_infosys.version_info(major=2, minor=7, micro=12, releaselevel='final', serial=0)>>> timeit.timeit('divmod(n, d)', 'n, d = 42, 7')0.1473848819732666>>> timeit.timeit('n // d, n % d', 'n, d = 42, 7')0.10324406623840332

The divmod() function is at a disadvantage here because you need to look up the global each time. Binding it to a local (all variables in a timeit time trial are local) improves performance a little:

>>> timeit.timeit('dm(n, d)', 'n, d = 42, 7; dm = divmod')0.13460898399353027

but the operators still win because they don't have to preserve the current frame while a function call to divmod() is executed:

>>> import dis>>> dis.dis(compile('divmod(n, d)', '', 'exec'))  1           0 LOAD_NAME                0 (divmod)              3 LOAD_NAME                1 (n)              6 LOAD_NAME                2 (d)              9 CALL_FUNCTION            2             12 POP_TOP                          13 LOAD_CONST               0 (None)             16 RETURN_VALUE        >>> dis.dis(compile('(n // d, n % d)', '', 'exec'))  1           0 LOAD_NAME                0 (n)              3 LOAD_NAME                1 (d)              6 BINARY_FLOOR_DIVIDE               7 LOAD_NAME                0 (n)             10 LOAD_NAME                1 (d)             13 BINARY_MODULO                    14 BUILD_TUPLE              2             17 POP_TOP                          18 LOAD_CONST               0 (None)             21 RETURN_VALUE        

The // and % variant uses more opcodes, but the CALL_FUNCTION bytecode is a bear, performance wise.

In PyPy, for small integers there isn't really much of a difference; the small speed advantage the opcodes have melts away under the sheer speed of C integer arithmetic:

>>>> import platform, sys, timeit>>>> platform.python_implementation(), sys.version_info('PyPy', (major=2, minor=7, micro=10, releaselevel='final', serial=42))>>>> timeit.timeit('divmod(n, d)', 'n, d = 42, 7', number=10**9)0.5659301280975342>>>> timeit.timeit('n // d, n % d', 'n, d = 42, 7', number=10**9)0.5471200942993164

(I had to crank the number of repetitions up to 1 billion to show how small the difference really is, PyPy is blazingly fast here).

However, when the numbers get large, divmod() wins by a country mile:

>>>> timeit.timeit('divmod(n, d)', 'n, d = 2**74207281 - 1, 26', number=100)17.620037078857422>>>> timeit.timeit('n // d, n % d', 'n, d = 2**74207281 - 1, 26', number=100)34.44323515892029

(I now had to tune down the number of repetitions by a factor of 10 compared to hobbs' numbers, just to get a result in a reasonable amount of time).

This is because PyPy no longer can unbox those integers as C integers; you can see the striking difference in timings between using sys.maxint and sys.maxint + 1:

>>>> timeit.timeit('divmod(n, d)', 'import sys; n, d = sys.maxint, 26', number=10**7)0.008622884750366211>>>> timeit.timeit('n // d, n % d', 'import sys; n, d = sys.maxint, 26', number=10**7)0.007693052291870117>>>> timeit.timeit('divmod(n, d)', 'import sys; n, d = sys.maxint + 1, 26', number=10**7)0.8396248817443848>>>> timeit.timeit('n // d, n % d', 'import sys; n, d = sys.maxint + 1, 26', number=10**7)1.0117690563201904

Martijn's answer is correct if you're using "small" native integers, where arithmetic operations are very fast compared to function calls. However, with bigints, it's a whole different story:

>>> import timeit>>> timeit.timeit('divmod(n, d)', 'n, d = 2**74207281 - 1, 26', number=1000)24.22666597366333>>> timeit.timeit('n // d, n % d', 'n, d = 2**74207281 - 1, 26', number=1000)49.517399072647095

when dividing a 22-million-digit number, divmod is almost exactly twice as fast as doing the division and modulus separately, as you might expect.

On my machine, the crossover occurs somewhere around 2^63, but don't take my word for it. As Martijn says, measure! When performance really matters, don't assume that what held true in one place will still be true in another.