[Tutor] List comprehension question
Richard D. Moores
rdmoores at gmail.com
Fri Nov 12 15:41:48 CET 2010
On Fri, Nov 12, 2010 at 05:15, David Hutto <smokefloat at 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.
> 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?
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