> 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.
>
> if __name__=='__main__':
> from timeit import Timer
> for y in range(5):
> t = Timer("proper_divisors_sum1(500000)", "from __main__
> import proper_divisors_sum")
> print round(t.timeit(number=10000),3)
>
> Thanks,
>
> Dick
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