On Thu, 29 Jul 2010 08:45:23 -0400 Joe Riopel <goo...@gmail.com> wrote: > On Thu, Jul 29, 2010 at 8:34 AM, Mahmood Naderan <nt_mahm...@yahoo.com> wrote: > > the output should be 7600 (s) for example. What is the best and easiest way > > to do that? > > Take a look at time.clock()
I don't know if that's what he wants. The clock() method returns processor time, not wall time. Python 2.6.5 (r265:79063, Jul 8 2010, 16:01:18) [GCC 4.1.3 20080704 prerelease (NetBSD nb2 20081120)] on netbsd5 Type "help", "copyright", "credits" or "license" for more information. >>> from time import time, clock, sleep >>> t = time() >>> print time() - t, clock() 0.000596046447754 0.03 >>> sleep(3) >>> print time() - t, clock() 3.03474903107 0.03 >>> x = open("BIGFILE").read() >>> print time() - t, clock() 10.2008538246 1.42 -- D'Arcy J.M. Cain <da...@druid.net> | Democracy is three wolves http://www.druid.net/darcy/ | and a sheep voting on +1 416 425 1212 (DoD#0082) (eNTP) | what's for dinner. -- http://mail.python.org/mailman/listinfo/python-list