Re: Set Frozenset?
R. David Murray wrote:
Lie Ryan lie.1...@gmail.com wrote:
Matt Nordhoff wrote:
Alan G Isaac wrote:
Hans Larsen schrieb:
How could I take an elemment from a set or a frozenset
On 3/8/2009 2:06 PM Diez B. Roggisch apparently wrote:
You iterate over them. If you only want one value, use
iter(the_set).next()
I recall a claim that
for result in myset: break
is the most efficient way to get one result.
Is this right? (It seems nearly the same.)
Alan Isaac
Checking Python 2.5 on Linux, your solution is much faster, but seeing
as they both come in under a microsecond, it hardly matters.
It's unexpected...
timeit.timeit('res=iter(myset).next()', 'myset=range(100)')
0.8894412399647
timeit.timeit('res=myset.next()', 'myset=range(100);
myset=iter(myset)')
0.4916552002516
timeit.timeit('for res in myset: break', 'myset=range(100)')
0.3293300797699
I'd never expect that for-loop assignment is even faster than a
precreated iter object (the second test)... but I don't think this
for-looping variable leaking behavior is guaranteed, isn't it?
My guess would be that what's controlling the timing here is
name lookup. Three in the first example, two in the second,
and one in the third.
You got it:
timeit.timeit('res=myset()', 'myset=range(100);
myset=iter(myset).next')
0.2646590399796
--
The following is a complete benchmark:
timeit.timeit('res=iter(myset).next()', 'myset=range(1000)',
number=1000)
8.514500200432
timeit.timeit('res=myset.next()', 'myset=range(1000);
myset=iter(myset)', number=1000)
4.550980280898
timeit.timeit('for res in myset: break', 'myset=range(1000)',
number=1000)
2.999421360421
timeit.timeit('res=myset()', 'myset=range(1000);
myset=iter(myset).next', number=1000)
2.222883241011
--
I also performed additional timing for overhead:
Local name lookup:
timeit.timeit('myset', 'myset=range(100)', number=1000)
1.108640079865
Global name lookup:
timeit.timeit('iter', number=1000)
1.814941079259
Attribute lookup:
timeit.timeit('myset.next', 'myset=range(1000);
myset=iter(myset)', number=1000)
3.301133397987
Combined multiple name lookup that troubled first test
timeit.timeit('iter(myset).next', 'myset=range(1000)',
number=1000)
6.559937480305
Creating iterables:
timeit.timeit('iter(myset)', 'myset=range(100)', number=1000)
4.25940671788
--
So adjusting the overheads:
Attribute lookup:
timeit.timeit('myset.next', 'myset=range(1000);
myset=iter(myset)', number=1000)
3.301133397987
The timing for Attribute also include a local name lookup (myset), so
the real attribute lookup time shold be:
3.301133397987 - 1.108640079865 = 2.192493318122
Creating iterables:
timeit.timeit('iter(myset)', 'myset=range(100)', number=1000)
4.25940671788
Creating iterable involve global name lookup, so the real time should be:
4.25940671788 - 1.814941079259 = 2.65638621
--
To summarize the adjusted overheads:
Local name lookup: 1.108640079865
Global name lookup: 1.814941079259
Attribute lookup: 2.192493318122
Creating iterables: 2.65638621
--
Back to the problem, now we'll be adjusting the timing of each codes:
'res=iter(myset).next()': 8.514500200432
Adjusting with the Combined multiple name lookup
8.514500200432 - 6.559937480305 = 1.954562720126
Another way to do the adjustment:
Adjusting global name lookup (iter):
8.514500200432 - 1.814941079259 = 6.699559121172
Adjusting iterable creation:
6.699559121172 - 2.65638621 = 4.255093482551
Adjusting attribute lookup:
4.255093482551 - 2.192493318122 = 2.062600164429
'res=myset.next()': 4.550980280898
Adjusting with |unadjusted| attribute lookup:
4.550980280898 - 3.301133397987 = 1.249846882911
Another way to do the adjustment:
Adjusting with local name lookup:
4.550980280898 - 1.108640079865 = 3.442340201033
Adjusting with attribute lookup:
3.442340201033 - 2.192493318122 = 1.249846882911
'for res in myset: break': 2.999421360421
Adjusting for local name lookup (myset):
2.999421360421 - 1.108640079865 = 1.890781280556
'res=myset()': 2.222883241011
Adjusting for local name lookup
2.222883241011 - 1.108640079865 = 1.114243161146
--
To summarize:
'res=iter(myset).next()': 1.954562720126 / 2.062600164429
'res=myset.next()': 1.249846882911 / 1.249846882911
'for res in myset: break': 1.890781280556
'res=myset()': 1.114243161146
--
To conclude, 'for res in myset: break' is actually not much faster than
'res=iter(myset).next()' except the former saves a lot of name lookup.
The problem with
Re: Set Frozenset?
Matt Nordhoff wrote:
Alan G Isaac wrote:
Hans Larsen schrieb:
How could I take an elemment from a set or a frozenset
On 3/8/2009 2:06 PM Diez B. Roggisch apparently wrote:
You iterate over them. If you only want one value, use
iter(the_set).next()
I recall a claim that
for result in myset: break
is the most efficient way to get one result.
Is this right? (It seems nearly the same.)
Alan Isaac
Checking Python 2.5 on Linux, your solution is much faster, but seeing
as they both come in under a microsecond, it hardly matters.
It's unexpected...
timeit.timeit('res=iter(myset).next()', 'myset=range(100)')
0.8894412399647
timeit.timeit('res=myset.next()', 'myset=range(100);
myset=iter(myset)')
0.4916552002516
timeit.timeit('for res in myset: break', 'myset=range(100)')
0.3293300797699
I'd never expect that for-loop assignment is even faster than a
precreated iter object (the second test)... but I don't think this
for-looping variable leaking behavior is guaranteed, isn't it?
Note: the second one exhausts the iter object.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Lie Ryan wrote: I recall a claim that for result in myset: break is the most efficient way to get one result. I'd never expect that for-loop assignment is even faster than a precreated iter object (the second test)... but I don't think this for-looping variable leaking behavior is guaranteed, isn't it? It is an intentional, documented feature: Names in the target list are not deleted when the loop is finished, but if the sequence is empty, it will not have been assigned to at all by the loop. -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Terry Reedy tjre...@udel.edu writes: I'd never expect that for-loop assignment is even faster than a precreated iter object (the second test)... but I don't think this for-looping variable leaking behavior is guaranteed, isn't it? It is an intentional, documented feature: ... I prefer thinking of it as a documented bug. It is fixed in 3.x. I usually avoid the [... for x in xiter] listcomp syntax in favor of list(... for x in xiter) just as an effort to be a bit less bug-prone. -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Paul Rubin wrote: Terry Reedy tjre...@udel.edu writes: I'd never expect that for-loop assignment is even faster than a precreated iter object (the second test)... but I don't think this for-looping variable leaking behavior is guaranteed, isn't it? It is an intentional, documented feature: ... I prefer thinking of it as a documented bug. It is fixed in 3.x. Nope to both. We are talking about for-loop statements. -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Lie Ryan lie.1...@gmail.com wrote:
Matt Nordhoff wrote:
Alan G Isaac wrote:
Hans Larsen schrieb:
How could I take an elemment from a set or a frozenset
On 3/8/2009 2:06 PM Diez B. Roggisch apparently wrote:
You iterate over them. If you only want one value, use
iter(the_set).next()
I recall a claim that
for result in myset: break
is the most efficient way to get one result.
Is this right? (It seems nearly the same.)
Alan Isaac
Checking Python 2.5 on Linux, your solution is much faster, but seeing
as they both come in under a microsecond, it hardly matters.
It's unexpected...
timeit.timeit('res=iter(myset).next()', 'myset=range(100)')
0.8894412399647
timeit.timeit('res=myset.next()', 'myset=range(100);
myset=iter(myset)')
0.4916552002516
timeit.timeit('for res in myset: break', 'myset=range(100)')
0.3293300797699
I'd never expect that for-loop assignment is even faster than a
precreated iter object (the second test)... but I don't think this
for-looping variable leaking behavior is guaranteed, isn't it?
My guess would be that what's controlling the timing here is
name lookup. Three in the first example, two in the second,
and one in the third.
--
R. David Murray http://www.bitdance.com
--
http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Hans Larsen schrieb: How could I take an elemment from a set or a frozenset On 3/8/2009 2:06 PM Diez B. Roggisch apparently wrote: You iterate over them. If you only want one value, use iter(the_set).next() I recall a claim that for result in myset: break is the most efficient way to get one result. Is this right? (It seems nearly the same.) Alan Isaac -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Alan G Isaac wrote: Hans Larsen schrieb: How could I take an elemment from a set or a frozenset On 3/8/2009 2:06 PM Diez B. Roggisch apparently wrote: You iterate over them. If you only want one value, use iter(the_set).next() I recall a claim that for result in myset: break is the most efficient way to get one result. Is this right? (It seems nearly the same.) Alan Isaac Checking Python 2.5 on Linux, your solution is much faster, but seeing as they both come in under a microsecond, it hardly matters. -- -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Hans Larsen schrieb: Could you help me ? How could I take an elemment from a set or a frozenset .-) ? From a string (unicode? Python3), or from a tuple,or from a list: Element by index or slice. From a dict: by key. But what concerning a set or frozenset! hope somebody can help! You iterate over them. If you only want one value, use iter(the_set).next() Diez -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Diez B. Roggisch wrote: Hans Larsen schrieb: Could you help me ? How could I take an elemment from a set or a frozenset .-) ? From a string (unicode? Python3), or from a tuple,or from a list: Element by index or slice. From a dict: by key. But what concerning a set or frozenset! hope somebody can help! You iterate over them. If you only want one value, use iter(the_set).next() or the_set.pop () TJG -- http://mail.python.org/mailman/listinfo/python-list
Re: Set Frozenset?
Tim Golden wrote: Diez B. Roggisch wrote: Hans Larsen schrieb: Could you help me ? How could I take an elemment from a set or a frozenset .-) ? From a string (unicode? Python3), or from a tuple,or from a list: Element by index or slice. From a dict: by key. But what concerning a set or frozenset! hope somebody can help! You iterate over them. If you only want one value, use iter(the_set).next() or the_set.pop () Which will, in addition, remove it from the set. (May not be what you want :) ). TJG -- http://mail.python.org/mailman/listinfo/python-list
