On 2005-12-07, Bengt Richter <[EMAIL PROTECTED]> wrote: >>In my experience, time.time() on Linux has a resolution of >>about 1us. The delta I get when I do >> >> print time.time()-time.time() >> >>is usually about 2-3us, but some of that is probably due to the >>overhead involved. >> > Try > > >>> import time > >>> t=time.time; c=time.clock > >>> min(filter(None,(-float.__sub__(c(),c()) for x in xrange(10000)) ))*1e3 > 0.0058666657878347905 > >>> min(filter(None,(-float.__sub__(t(),t()) for x in xrange(10000)) ))*1e3 > 9.9999904632568359
>>> import time >>> t=time.time; c=time.clock >>> min(filter(None,(-float.__sub__(c(),c()) for x in xrange(10000)) ))*1e3 10.000000000000002 >>> min(filter(None,(-float.__sub__(t(),t()) for x in xrange(10000)) ))*1e3 0.00095367431640625 Yup. That has less overhead than my original example because you've avoided the extra name lookup: >>> for f in range(10): ... print t()-t() ... -4.05311584473e-06 -1.90734863281e-06 -1.90734863281e-06 -2.14576721191e-06 -2.86102294922e-06 -1.90734863281e-06 -2.14576721191e-06 -2.14576721191e-06 -9.53674316406e-07 -1.90734863281e-06 The min delta seen is 0.95us. I'm guessing thats function/system call overhead and not timer resolution. -- Grant Edwards grante Yow! I HAVE a towel. at visi.com -- http://mail.python.org/mailman/listinfo/python-list