bug with itertools.groupby?
Dave Angel
davea at ieee.org
Tue Oct 6 22:07:41 EDT 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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