Message 65737 - Python tracker (original) (raw)

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Alexander Belopolsky wrote:

Alexander Belopolsky <belopolsky@users.sourceforge.net> added the comment:

This has nothing to do with set.update, the difference is due to the time to setup the generator:

$ python -m timeit -s 'x = set(range(10000)); y = []' 'x.update(y)' 1000000 loops, best of 3: 0.38 usec per loop $ python -m timeit -s 'x = set(range(10000)); y = (i for i in [])' 'x.update(y)' 1000000 loops, best of 3: 0.335 usec per loop

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nosy: +belopolsky

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Tracker <report@bugs.python.org> <http://bugs.python.org/issue2672>

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That is true, though if I just force a generator overhead:

% python -m timeit -s 'x = set(range(10000)); y = []' 'x.update(y)' 1000000 loops, best of 3: 0.204 usec per loop % python -m timeit -s 'x = set(range(10000)); y = (i for i in [])' 'x.update(y)' 10000000 loops, best of 3: 0.173 usec per loop % python -m timeit -s 'x = set(range(10000)); l = []' 'x.update(i for i in l)' 1000000 loops, best of 3: 0.662 usec per loop python -m timeit -s 'x = set(range(10000)); l = []; y = (i for i in l)' '(i for i in l); x.update(y)' 1000000 loops, best of 3: 1.87 usec per loop

So if you compare consuming a generator multiple times to creating it each time, it is 0.662 usec - 0.173 usec = 0.489 usec to create a generator.

So why does: "(i for i in l); x.update(y)" take an additional 1.208 usec.

(I'm certainly willing to believe that set.update() is generator/list agnostic, but something weird is still happening.)

John =:->

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