[Python-Dev] Retrieve an arbitrary element from a set without removing it (original) (raw)

Willi Richert w.richert at gmx.net
Sat Oct 24 17:41:18 CEST 2009

Hi,

someone on this list mentioned that much of the s.get() time is spend on the name lookup for get(). That is indeed the case:

=================== from timeit import *

stats = ["for i in xrange(1000): iter(s).next() ", "for i in xrange(1000): \n\tfor x in s: \n\t\tbreak", "for i in xrange(1000): s.add(s.pop()) ", "for i in xrange(1000): s.get() ", "g=s.get;\nfor i in xrange(1000): g() "]

for stat in stats: t = Timer(stat, setup="s=set(range(1000))") print "Time for %s:\t %f"%(stat, t.timeit(number=1000))

Time for for i in xrange(1000): iter(s).next() : 0.448227 Time for for i in xrange(1000): for x in s: break: 0.141669 Time for for i in xrange(1000): s.add(s.pop()) : 0.348055 Time for for i in xrange(1000): s.get() : 0.148580 Time for g=s.get; for i in xrange(1000): g() : 0.080563

So, now set.get() is indeed the fastest and preferable solution if you need massive amounts of retrieving elements from a set without removing them.

wr

Am Freitag, 23. Oktober 2009 22:53:24 schrieb Willi Richert:

Hi,

surprised about the performance of for/break provided by Vitor, I did some more testing. It revealed that indeed we can forget the get() (which was implemented as a stripped down pop()): from timeit import * stats = ["for i in xrange(1000): iter(s).next() ", "for i in xrange(1000): \n\tfor x in s: \n\t\tbreak ", "for i in xrange(1000): s.add(s.pop()) ", "for i in xrange(1000): s.get() "] for stat in stats: t = Timer(stat, setup="s=set(range(100))") try: print "Time for %s:\t %f"%(stat, t.timeit(number=1000)) except: t.printexc()

$ ./testget.py Time for for i in xrange(1000): iter(s).next() : 0.433080 Time for for i in xrange(1000): for x in s: break : 0.148695 Time for for i in xrange(1000): s.add(s.pop()) : 0.317418 Time for for i in xrange(1000): s.get() : 0.146673 In some tests, for/break was even slightly faster then get(). As always, intuition regarding performance bottlenecks is flawed ;-) Anyway, thanks for all the helpful comments, especially to Stefan for the http://comments.gmane.org/gmane.comp.python.ideas/5606 link. Regards, wr Am Freitag, 23. Oktober 2009 19:25:48 schrieb John Arbash Meinel: > Vitor Bosshard wrote: > > 2009/10/23 Willi Richert <w.richert at gmx.net>: > >> Hi, > >> > >> recently I wrote an algorithm, in which very often I had to get an > >> arbitrary element from a set without removing it. > >> > >> Three possibilities came to mind: > >> > >> 1. > >> x = someset.pop() > >> someset.add(x) > >> > >> 2. > >> for x in someset: > >> break > >> > >> 3. > >> x = iter(someset).next() > >> > >> > >> Of course, the third should be the fastest. It nevertheless goes > >> through all the iterator creation stuff, which costs some time. I > >> wondered, why the builtin set does not provide a more direct and > >> efficient way for retrieving some element without removing it. Is > >> there any reason for this? > >> > >> I imagine something like > >> > >> x = someset.get() > > > > I see this as being useful for frozensets as well, where you can't get > > an arbitrary element easily due to the obvious lack of .pop(). I ran > > into this recently, when I had a frozenset that I knew had 1 element > > (it was the difference between 2 other sets), but couldn't get to that > > element easily (get the pun?) > > So in my testing (2) was actually the fastest. I assumed because .next() > was a function call overhead, while: > for x in someset: > break > > Was evaluated inline. It probably still has to call PyObjectGetIter, > however it doesn't have to create a stack frame for it. > > This is what "timeit" tells me. All runs are of the form: > python -m timeit -s "s = set([10])" ... > > 0.101us "for x in s: break; x" > 0.130us "for x in s: pass; x" > 0.234us -s "n = next; i = iter" "x = n(i(s)); x" > 0.248us "x = next(iter(s)); x" > 0.341us "x = iter(s).next(); x" > > So 'for x in s: break' is about 2x faster than next(iter(s)) and 3x > faster than (iter(s).next()). > I was pretty surprised that it was 30% faster than "for x in s: pass". I > assume it has something to do with a potential "else:" statement? > > Note that all of these are significantly < 1us. So this only matters if_ _> it is something you are doing often. > > I don't know your specific timings, but I would guess that: > for x in s: break > > Is actually going to be faster than your > s.get() > > Primarily because s.get() requires an attribute lookup. I would think > your version might be faster for: > stat2 = "g = s.get; for i in xrange(100): g()" > > However, that is still a function call, which may be treated differently > by the interpreter than the for:break loop. I certainly suggest you try > it and compare. > > John > =:->

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