On Thu, Nov 5, 2009 at 4:09 PM, Alexander Belopolsky
<alexander.belopolsky at gmail.com> wrote:
> On Thu, Nov 5, 2009 at 3:43 PM, Chris Bergstresser <chris at 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
RetroSearch is an open source project built by @garambo | Open a GitHub Issue
Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo
HTML:
3.2
| Encoding:
UTF-8
| Version:
0.7.4