On Fri, Dec 11, 2009 at 2:43 PM, MRAB <pyt...@mrabarnett.plus.com> wrote:
> John Arbash Meinel wrote:
>>
>> Roy Hyunjin Han wrote:
>>>
>>> While debugging a network algorithm in Python 2.6.2, I encountered
>>> some strange behavior and was wondering whether it has to do with some
>>> sort of code optimization that Python does behind the scenes.
>>>
>>>
>>> ************
>>> After initialization: defaultdict(<type 'set'>, {1: set([1])})
>>> Popping and updating in two steps: defaultdict(<type 'set'>, {1:
>>> set([1])})
>>> ************
>>> After initialization: defaultdict(<type 'set'>, {1: set([1])})
>>> Popping and updating in one step: defaultdict(<type 'set'>, {})
>>>
>>>
>>> import collections
>>> print '************'
>>> x = collections.defaultdict(set)
>>> x[1].update([1])
>>> print 'After initialization: %s' % x
>>> items = x.pop(1)
>>> x[1].update(items)
>>> print 'Popping and updating in two steps: %s' % x
>>> print '************'
>>> y = collections.defaultdict(set)
>>> y[1].update([1])
>>> print 'After initialization: %s' % y
>>> y[1].update(y.pop(1))
>>> print 'Popping and updating in one step: %s' % y
>>>
>>
>> y[1].update(y.pop(1))
>>
>> is going to be evaluating y[1] before it evaluates y.pop(1).
>> Which means that it has the original set returned, which is then removed
>> by y.pop, and updated.
>>
>> You probably get the same behavior without using a defaultdict:
>> y.setdefault(1, set()).update(y.pop(1))
>> ^^^^^^^^^^^^^^^^^^^^^^- evaluated first
>>
>>
>> Oh and I should probably give the standard: "This list is for the
>> development *of* python, not development *with* python."
>>
> To me the question was whether Python was behaving correctly; the
> behaviour was more subtle than the legendary mutable default argument.
Thanks, John and MRAB. I was pointing it out on this list because I
felt like it was counterintuitive and that the result should be the
same whether the developer decides to do it in two steps or in one
step.
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