On 06/08/2014 06:34, Gayathri J wrote:
Dear Peter
Below is the code I tried to check if itertools.product() was faster
than normal nested loops...
they arent! arent they supposed to be...or am i making a mistake? any idea?
*
*
*############################################################
*
*# -*- coding: utf-8 -*-
*
*import numpy as np*
*import time*
*from itertools import product,repeat*
*def main():*
* # N - size of grid*
* # nvel - number of velocities*
* # times - number of times to run the functions*
* N=256*
* times=3*
* f=np.random.rand(N,N,N)*
**
* # using loops*
* print "normal nested loop"*
* python_dot_loop1(f,times,N)*
*
*
* print "nested loop using itertools.product()"*
* python_dot_loop2(f,times,N)*
*
*
*def python_dot_loop1(f,times,N):*
* for t in range(times):*
* t1=time.time()*
* for i in range(N):*
* for j in range(N):*
* for k in range(N):*
* f[i,j,k] = 0.0*
* print "python dot loop " + str(time.time()-t1)*
**
*def python_dot_loop2(f,times,N):*
* rangeN=range(N)*
* for t in range(times):*
* t1=time.time()*
* for i,j,k in product(rangeN,repeat=3):*
* f[i,j,k]=0.0*
* print "python dot loop " + str(time.time()-t1)*
*
*
*
*
*if __name__=='__main__':*
* main()*
*############################################################*
Who cares, well not I for one? Give me slower but accurate code over
faster but inaccurate code any day of the week?
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
--
https://mail.python.org/mailman/listinfo/python-list
