Alex Martelli a écrit : > Fredrik Lundh <[EMAIL PROTECTED]> wrote: > ... > >>the obvious solution is >> >> item = list(s)[0] >> >>but that seems to be nearly twice as slow as [x for x in s][0] >>under 2.4. hmm. > > > Funny, and true on my laptop too: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=list(s)[0]' > 100000 loops, best of 3: 2.55 usec per loop
That's probably because of the name lookup needed for "list" > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=[x for x in s][0]' > 100000 loops, best of 3: 1.48 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' '[x for x in s]' > 1000000 loops, best of 3: 1.36 usec per loop > > Exploiting the design defect whereby a LC leaves variables bound can > shave another few percents off it, as shown. > > >>here's a faster variant: >> >> item = iter(s).next() > > > Not all that fast here: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=iter(s).next()' > 100000 loops, best of 3: 1.71 usec per loop That's probably because of the 2 name lookups needed :) "iter" and "next" >>but at least on my machine, your two-step solution >> >> item = s.pop(); s.add(item) >>seems to be even faster. > > > Not really, here: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=s.pop();s.add(x)' > 100000 loops, best of 3: 1.49 usec per loop > > No joy from several variations on transform-and-pop: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=set(s).pop()' > 100000 loops, best of 3: 2.21 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=list(s).pop()' > 100000 loops, best of 3: 3.2 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=tuple(s)[0]' > 100000 loops, best of 3: 1.79 usec per loop > > > Fastest I've found is unpacking-assignment: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x,=s' > 1000000 loops, best of 3: 0.664 usec per loop > > > Alex -- http://mail.python.org/mailman/listinfo/python-list
