On Thu, Nov 5, 2009 at 4:09 PM, Alexander Belopolsky
<alexander.belopol...@gmail.com> wrote:
> On Thu, Nov 5, 2009 at 3:43 PM, Chris Bergstresser <ch...@subtlety.com> wrote:
>> .. and "x = iter(s).next()" raises a StopIteration
>> exception.
>
> And that's why the documented recipe should probably recommend
> next(iter(s), default) instead. Especially because iter(s).next() is
> not even valid code in 3.0.
This seems reasonably legible to you? Strikes me as coding by
incantation. Also, while I've heard people say that the naive
approach is slower, I'm not getting that result. Here's my test:
>>> smrt = timeit.Timer("next(iter(s))", "s=set(range(100))")
>>> smrt.repeat(10)
[1.2845709323883057, 0.60247397422790527, 0.59621405601501465,
0.59133195877075195, 0.58387589454650879, 0.56839084625244141,
0.56839680671691895, 0.56877803802490234, 0.56905913352966309,
0.56846404075622559]
>>> naive = timeit.Timer("x=s.pop();s.add(x)", "s=set(range(100))")
>>> naive.repeat(10)
[0.93139314651489258, 0.53566789627075195, 0.53674602508544922,
0.53608798980712891, 0.53634309768676758, 0.53557991981506348,
0.53578495979309082, 0.53666114807128906, 0.53576493263244629,
0.53491711616516113]
Perhaps my test is flawed in some way?
Geremy Condra
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