On Fri, Nov 12, 2010 at 05:15, David Hutto <smokefl...@gmail.com> wrote: >> repeatedly, returning a list of results. ... >> >> I'm sorry, Steven, but I could you revise this code to use repeat=5 >> instead of the for loop? I can't see how to do it. > >>>> help(timeit.Timer > > repeat(self, repeat=3, number=1000000) > | Call timeit() a few times. > | > | This is a convenience function that calls the timeit() > | repeatedly, returning a list of results. The first argument > | specifies how many times to call timeit(), defaulting to 3; > | the second argument specifies the timer argument, defaulting > | to one million. > | > | Note: it's tempting to calculate mean and standard deviation > | from the result vector and report these. However, this is not > | very useful. In a typical case, the lowest value gives a > | lower bound for how fast your machine can run the given code > | snippet; higher values in the result vector are typically not > | caused by variability in Python's speed, but by other > | processes interfering with your timing accuracy. So the min() > | of the result is probably the only number you should be > | interested in. After that, you should look at the entire > | vector and apply common sense rather than statistics.
Look, I've already shown I know where the docs are, and have read them. I don't understand how to use repeat() with my code. Wasn't that clear to you? _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
