On Mon, Jul 11, 2011 at 3:06 PM, Cal Leeming [Simplicity Media Ltd] <
cal.leem...@simplicitymedialtd.co.uk> wrote:
>
>
> On Mon, Jul 11, 2011 at 3:04 PM, Cal Leeming [Simplicity Media Ltd] <
> cal.leem...@simplicitymedialtd.co.uk> wrote:
>
>>
>>
>> On Mon, Jul 11, 2011 at 2:57 PM, Michel30 <forerunn...@gmail.com> wrote:
>>
>>> Hi all,
>>>
>>> I have a basic search function that uses Q objects.
>>> After profiling it I found that the actual (mysql) database query
>>> finishes in fractions of seconds but the iterating after this can take
>>> up to 50 seconds per 10.000 results.
>>>
>>> I have been trying to speed it up but I have had not much results..
>>>
>>> My query is this one:
>>>
>>> found_entries = model.objects.filter((Q-objects),
>>> obsolete=0).order_by('-version','docid')
>>>
>>> So far so good, but then I need a dictionary to retrieve only unique
>>> 'documentid's'.
>>>
>>
>> You could do:
>>
>> # grab all results
>> _res =
>> model.objects.filter((Q-objects),obsolete=0).order_by('-version','docid').values()
>> # re-map them into (id, obj)
>> _res = map(lambda x: [x.docid, x], _res)
>>
> Sorry, small adjustment:
# re-map them into (id, obj), model(**x) will remap the values into new ORM
objects
_res = map(lambda x: [x.get('docid'), model(**x)], _res)
# wrap in a dict(), which uses index position 0 as the key, and index
>> position 1 as the value
>> _res = dict(_res)
>>
>
> Just tested the same principle, and it seems to work fine. It uses last
> object found as the final choice if dups are found.
>
> >>> _res = [ [1,2], [1,3], [2,4], [2,5] ]
> >>> map(lambda x: [x[0], x[1]], _res)
> [[1, 2], [1, 3], [2, 4], [2, 5]]
> >>> map(lambda x: [x[0], x[1]], _res)
> [[1, 2], [1, 3], [2, 4], [2, 5]]
> >>> dict(map(lambda x: [x[0], x[1]], _res))
> {1: 3, 2: 5}
> >>>
>
>
>
>
>>
>> Let me know if this give you the results you need.
>>
>>
>>> rev_dict = {}
>>>
>>> This is the part that hurts:
>>>
>>> for d in found_entries:
>>> rev_dict[d.documentid] = d
>>>
>>> And then some more sorting and filtering:
>>>
>>> filtered_entries = rev_dict.values()
>>> filtered_entries.sort(key=lambda d: d.revisiondate, reverse=True)
>>>
>>> Does anyone have some better ideas to achieve this?
>>>
>>> Thanks
>>>
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>>
>
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