bug with itertools.groupby?

Dave Angel davea at ieee.org
Wed Oct 7 04:07:41 CEST 2009


Kitlbast wrote:
> On Oct 7, 3:04 am, Raymond Hettinger <pyt... at rcn.com> wrote:
>   
>> On Oct 6, 4:06 pm, Kitlbast <vlad.shevche... at gmail.com> wrote:
>>
>>
>>
>>
>>
>>     
>>> Hi there,
>>>       
>>> the code below on Python 2.5.2:
>>>       
>>> from itertools import groupby
>>>       
>>> info_list =
>>>     {'profile': 'http://somesite.com/profile1', 'account': 61L},
>>>     {'profile': 'http://somesite.com/profile2', 'account': 64L},
>>>     {'profile': 'http://somesite.com/profile3', 'account': 61L},
>>> ]
>>>       
>>> grouped_by_account =roupby(info_list, lambda x: x['account'])
>>> for acc, iter_info_items in grouped_by_account:
>>>     print 'grouped acc: ', acc
>>>       
>>> gives output:
>>>       
>>> grouped acc:  61
>>> grouped acc:  64
>>> grouped acc:  61
>>>       
>>> am I doing something wrong?
>>>       
>> Try another variant of groupby() that doesn't require the data to be
>> sorted:
>>
>>    http://code.activestate.com/recipes/259173/
>>
>> Raymond
>>     
>
> I've checked few options of groupby() implementations
>
> 1. def groupby(_list, key_func):
>     res =}
>     for i in _list:
>         k =ey_func(i)
>         if k not in res:
>             res[k] =i]
>         else:
>             res[k].append(i)
>     return res
>
>
> 2. def groupby(_list, key_func):
>     res =}
>     [res.setdefault(key_func(i), []).append(i) for i in _list]
>     return res
>
>
> second one with setdefault works little bit slower then (1), although
> it use list comprehension
>
>   
Or option 3:   use defaultdict





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