Why is x**4.0
faster than x**4
in Python 3*?
Python 3 int
objects are a full fledged object designed to support an arbitrary size; due to that fact, they are handled as such on the C level (see how all variables are declared as PyLongObject *
type in long_pow
). This also makes their exponentiation a lot more trickier and tedious since you need to play around with the ob_digit
array it uses to represent its value to perform it. (Source for the brave. — See: Understanding memory allocation for large integers in Python for more on PyLongObject
s.)
Python float
objects, on the contrary, can be transformed to a C double
type (by using PyFloat_AsDouble
) and operations can be performed using those native types. This is great because, after checking for relevant edge-cases, it allows Python to use the platforms’ pow
(C’s pow
, that is) to handle the actual exponentiation:
/* Now iv and iw are finite, iw is nonzero, and iv is
* positive and not equal to 1.0. We finally allow
* the platform pow to step in and do the rest.
*/
errno = 0;
PyFPE_START_PROTECT("pow", return NULL)
ix = pow(iv, iw);
where iv
and iw
are our original PyFloatObject
s as C double
s.
For what it’s worth: Python 2.7.13
for me is a factor 2~3
faster, and shows the inverse behaviour.
The previous fact also explains the discrepancy between Python 2 and 3 so, I thought I’d address this comment too because it is interesting.
In Python 2, you’re using the old int
object that differs from the int
object in Python 3 (all int
objects in 3.x are of PyLongObject
type). In Python 2, there’s a distinction that depends on the value of the object (or, if you use the suffix L/l
):
# Python 2
type(30) # <type 'int'>
type(30L) # <type 'long'>
The <type 'int'>
you see here does the same thing float
s do, it gets safely converted into a C long
when exponentiation is performed on it (The int_pow
also hints the compiler to put ’em in a register if it can do so, so that could make a difference):
static PyObject *
int_pow(PyIntObject *v, PyIntObject *w, PyIntObject *z)
{
register long iv, iw, iz=0, ix, temp, prev;
/* Snipped for brevity */
this allows for a good speed gain.
To see how sluggish <type 'long'>
s are in comparison to <type 'int'>
s, if you wrapped the x
name in a long
call in Python 2 (essentially forcing it to use long_pow
as in Python 3), the speed gain disappears:
# <type 'int'>
(python2) ➜ python -m timeit "for x in range(1000):" " x**2"
10000 loops, best of 3: 116 usec per loop
# <type 'long'>
(python2) ➜ python -m timeit "for x in range(1000):" " long(x)**2"
100 loops, best of 3: 2.12 msec per loop
Take note that, though the one snippet transforms the int
to long
while the other does not (as pointed out by @pydsinger), this cast is not the contributing force behind the slowdown. The implementation of long_pow
is. (Time the statements solely with long(x)
to see).
[…] it doesn’t happen outside of the loop. […] Any idea about that?
This is CPython’s peephole optimizer folding the constants for you. You get the same exact timings either case since there’s no actual computation to find the result of the exponentiation, only loading of values:
dis.dis(compile('4 ** 4', '', 'exec'))
1 0 LOAD_CONST 2 (256)
3 POP_TOP
4 LOAD_CONST 1 (None)
7 RETURN_VALUE
Identical byte-code is generated for '4 ** 4.'
with the only difference being that the LOAD_CONST
loads the float 256.0
instead of the int 256
:
dis.dis(compile('4 ** 4.', '', 'exec'))
1 0 LOAD_CONST 3 (256.0)
2 POP_TOP
4 LOAD_CONST 2 (None)
6 RETURN_VALUE
So the times are identical.
*All of the above apply solely for CPython, the reference implementation of Python. Other implementations might perform differently.