Python is darn fast (was: How fast is Python)

[Michele Simionato]

I posted this few weeks ago (remember the C Sharp thread?) but it went
unnoticed on the large mass of posts, so let me retry. Here I get Python+
Psyco twice as fast as optimized C, so I would like to now if something
is wrong on my old laptop and if anybody can reproduce my results.
Here are I my numbers for calling the error function a million times
(Python 2.3, Psyco 1.0, Red Hat Linux 7.3, Pentium II 366 MHz):

$ time p23 erf.py
real    0m0.614s
user    0m0.551s
sys     0m0.029s

This is twice as fast as optimized C:

$ gcc erf.c -lm -o3
$ time ./a.out
real    0m1.125s
user    0m1.086s
sys     0m0.006s
 
Here is the situation for pure Python

$time p23 erf.jy
real    0m25.761s
user    0m25.012s
sys     0m0.049s

and, just for fun, here is Jython performance:

$ time jython erf.jy
real    0m42.979s
user    0m41.430s
sys     0m0.361s

The source code follows (copied from Alex Martelli's post):

----------------------------------------------------------------------

$ cat erf.py
import math
import psyco
psyco.full()

def erfc(x):
    exp = math.exp

    p  =  0.3275911
    a1 =  0.254829592
    a2 = -0.284496736
    a3 =  1.421413741
    a4 = -1.453152027
    a5 =  1.061405429

    t = 1.0 / (1.0 + p*x)
    erfcx = ( (a1 + (a2 + (a3 +
                          (a4 + a5*t)*t)*t)*t)*t ) * exp(-x*x)
    return erfcx

def main():
    erg = 0.0

    for i in xrange(1000000):
        erg += erfc(0.456)

if __name__ == '__main__':
    main()

--------------------------------------------------------------------------

# python/jython version = same without "import psyco; psyco.full()"

--------------------------------------------------------------------------

$cat erf.c
#include <stdio.h>
#include <math.h>

double erfc( double x )
{
    double p, a1, a2, a3, a4, a5;
    double t, erfcx;

    p  =  0.3275911;
    a1 =  0.254829592;
    a2 = -0.284496736;
    a3 =  1.421413741;
    a4 = -1.453152027;
    a5 =  1.061405429;

    t = 1.0 / (1.0 + p*x);
    erfcx = ( (a1 + (a2 + (a3 +
     (a4 + a5*t)*t)*t)*t)*t ) * exp(-x*x);

    return erfcx;
}

int main()
{
    double erg=0.0;
    int i;

    for(i=0; i<1000000; i++)
    {
       erg = erg + erfc(0.456);
    }

    return 0;
}

Michele Simionato, Ph. D.
MicheleSimionato at libero.it
http://www.phyast.pitt.edu/~micheles
--- Currently looking for a job ---