Python is darn fast (was: How fast is Python)
Michele Simionato
mis6 at pitt.edu
Sun Aug 24 10:38:49 EDT 2003
I finally came to the conclusion that the exceeding good performance
of Psyco was due to the fact that the function was called a million
times with the *same* argument. Evidently Psyco is smart enough to
notice that. Changing the argument at each call
(erfc(0.456) -> i/1000000.0) slows down Python+Psyco at 1/4 of C speed.
Psyco improves Python performance by an order of magnitude, but still it
is not enough :-(
I was too optimistic!
Here I my numbers for Python 2.3, Psyco 1.0, Red Hat Linux 7.3,
Pentium II 366 MHz:
$ time p23 erf.py
real 0m3.245s
user 0m3.164s
sys 0m0.037s
This is more than four times slower than optimized C:
$ gcc erf.c -lm -O3
$ time ./a.out
real 0m0.742s
user 0m0.725s
sys 0m0.002s
Here is the situation for pure Python
$time p23 erf.jy
real 0m27.470s
user 0m27.162s
sys 0m0.023s
and, just for fun, here is Jython performance:
$ time jython erf.jy
real 0m44.395s
user 0m42.602s
sys 0m0.389s
----------------------------------------------------------------------
$ 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(i/1000000.0)
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(i/1000000.0);
}
return 0;
}
Michele Simionato, Ph. D.
MicheleSimionato at libero.it
http://www.phyast.pitt.edu/~micheles/
---- Currently looking for a job ----
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