Re: Best way to extract an item from a set of len 1
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]' > 10 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]' > 10 loops, best of 3: 1.48 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' '[x for x in s]' > 100 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()' > 10 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)' > 10 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()' > 10 loops, best of 3: 2.21 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=list(s).pop()' > 10 loops, best of 3: 3.2 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' 'x=tuple(s)[0]' > 10 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' > 100 loops, best of 3: 0.664 usec per loop > > > Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Alex Martelli wrote: > Rene Pijlman <[EMAIL PROTECTED]> wrote: >> Peter Otten: >> s = set(["one-and-only"]) >> item, = s >... >> >The comma may easily be missed, though. >> >> You could write: >> >> (item,) = s >> >> But I'm not sure if this introduces additional overhead. > > Naah...: > > helen:~ alex$ python -mtimeit -s's=set([23])' 'x,=s' > 100 loops, best of 3: 0.689 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' '(x,)=s' > 100 loops, best of 3: 0.652 usec per loop > helen:~ alex$ python -mtimeit -s's=set([23])' '[x]=s' > 100 loops, best of 3: 0.651 usec per loop > > ...much of a muchness. And that is no coincidence. All three variants are compiled to the same bytecode: >>> import dis >>> def a(): x, = s ... >>> def b(): (x,) = s ... >>> def c(): [x] = s ... >>> dis.dis(a) 1 0 LOAD_GLOBAL 0 (s) 3 UNPACK_SEQUENCE 1 6 STORE_FAST 0 (x) 9 LOAD_CONST 0 (None) 12 RETURN_VALUE >>> dis.dis(b) 1 0 LOAD_GLOBAL 0 (s) 3 UNPACK_SEQUENCE 1 6 STORE_FAST 0 (x) 9 LOAD_CONST 0 (None) 12 RETURN_VALUE >>> dis.dis(c) 1 0 LOAD_GLOBAL 0 (s) 3 UNPACK_SEQUENCE 1 6 STORE_FAST 0 (x) 9 LOAD_CONST 0 (None) 12 RETURN_VALUE Peter -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Rene Pijlman <[EMAIL PROTECTED]> wrote: > Peter Otten: > s = set(["one-and-only"]) > item, = s ... > >The comma may easily be missed, though. > > You could write: > > (item,) = s > > But I'm not sure if this introduces additional overhead. Naah...: helen:~ alex$ python -mtimeit -s's=set([23])' 'x,=s' 100 loops, best of 3: 0.689 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' '(x,)=s' 100 loops, best of 3: 0.652 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' '[x]=s' 100 loops, best of 3: 0.651 usec per loop ...much of a muchness. Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
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]' 10 loops, best of 3: 2.55 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' 'x=[x for x in s][0]' 10 loops, best of 3: 1.48 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' '[x for x in s]' 100 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()' 10 loops, best of 3: 1.71 usec per loop > 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)' 10 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()' 10 loops, best of 3: 2.21 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' 'x=list(s).pop()' 10 loops, best of 3: 3.2 usec per loop helen:~ alex$ python -mtimeit -s's=set([23])' 'x=tuple(s)[0]' 10 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' 100 loops, best of 3: 0.664 usec per loop Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Peter Otten wrote:
> > When you have a set, known to be of length one, is there a "best"
> > ("most pythonic") way to retrieve that one item?
>
> >>> s = set(["one-and-only"])
> >>> item, = s
> >>> item
> 'one-and-only'
>
> This works for any iterable and guarantees that it contains exactly one
> item. The comma may easily be missed, though.
you can make this a bit more obvious:
>>> [item] = s
this is almost twice as fast as the fastest alternative from my previous
post.
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Re: Best way to extract an item from a set of len 1
Peter Otten: s = set(["one-and-only"]) item, = s item >'one-and-only' > >This works for any iterable and guarantees that it contains exactly one >item. Nice! >The comma may easily be missed, though. You could write: (item,) = s But I'm not sure if this introduces additional overhead. -- René Pijlman Wat wil jij worden? http://www.carrieretijger.nl -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Tim Chase <[EMAIL PROTECTED]> wrote: ... > To get the item, i had to resort to methods that feel less than > the elegance I've come to expect from python: > > >>> item = [x for x in s][0] A shorter, clearer expression of the same idea: item = list(s)[0] or item = list(s).pop() > or the more convoluted two-step > > >>> item = s.pop() > >>> s.add(item) which in turn suggests item = set(s).pop() Similar ideas include iter(s).next() and s.copy().pop(). Basically: s has no way to get the item non-destructively, so, either make a copy (and use the destructive-get 'pop' on the copy) or build from s a type which DOES have ways to get the item (iterator, list, etc) be they destructive or not. As for speed, measuring is the only way, and timeit is your friend. As for elegance, the most concise readable form is "set(s).pop()" and that's what I would use. Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Tim Chase:
>When you have a set, known to be of length one, is there a "best"
>("most pythonic") way to retrieve that one item?
e = s.copy().pop() #:-)
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René Pijlman
Wat wil jij worden? http://www.carrieretijger.nl
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Re: Best way to extract an item from a set of len 1
That's cute. :-) Fuzzyman -- http://mail.python.org/mailman/listinfo/python-list
Re: Best way to extract an item from a set of len 1
Tim Chase wrote:
> When you have a set, known to be of length one, is there a "best"
> ("most pythonic") way to retrieve that one item?
>>> s = set(["one-and-only"])
>>> item, = s
>>> item
'one-and-only'
This works for any iterable and guarantees that it contains exactly one
item. The comma may easily be missed, though.
Peter
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Re: Best way to extract an item from a set of len 1
Tim Chase wrote: > I suppose I was looking for something like > > >>> item = s.aslist()[0] > > which feels a little more pythonic (IMHO). Is one solution > preferred for speed over others (as this is happening in a fairly > deeply nested loop)? 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. here's a faster variant: item = iter(s).next() but at least on my machine, your two-step solution item = s.pop(); s.add(item) seems to be even faster. -- http://mail.python.org/mailman/listinfo/python-list
Best way to extract an item from a set of len 1
When you have a set, known to be of length one, is there a "best"
("most pythonic") way to retrieve that one item?
# given that I've got Python2.3.[45] on hand,
# hack the following two lines to get a "set" object
>>> import sets
>>> set = sets.Set
>>> s = set(['test'])
>>> len(s)
1
>>> s[0]
Traceback (most recent call last):
File "", line 1, in ?
TypeError: unindexable object
(which is kinda expected, given that it's unordered...an index
doesn't make much sense)
To get the item, i had to resort to methods that feel less than
the elegance I've come to expect from python:
>>> item = [x for x in s][0]
or the more convoluted two-step
>>> item = s.pop()
>>> s.add(item)
or even worse, intruding into private members
>>> item = s._data.keys()[0]
Is any of these more "pythonic" than the others? Is there a more
elegant 2.3.x solution? If one upgrades to 2.4+, is there
something even more elegant? I suppose I was looking for
something like
>>> item = s.aslist()[0]
which feels a little more pythonic (IMHO). Is one solution
preferred for speed over others (as this is happening in a fairly
deeply nested loop)?
Any tips, preferences, input, suggestions, pointers to obvious
things I've missed, or the like?
Thanks,
-tkc
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