Re: sorting QuerySets

[Ned Batchelder]

Don't forget that sort() takes a key argument to sort by an arbitrary function of the elements, so there's no need to build extra datastructures: object_list = Word.objects.all().order_by('-product_count')[:100] object_list.sort(key=lambda w: w.title) --Ned. DavidA wrote: Tom Smith wrote: With about a million records I can't imagine doing it any other way... I want to find my top 100 words (by product_count) then sort them alphabetically.... to make a tag cloud... thanks... If you are only dealing with 100 records after the query, then its not a big deal to just sort them by some property in memory. Something like: # get the top 100 objects by the first criteria object_list = Word.objects.all().order_by('-product_count')[:100] # create a dict of the second criteria -> index of the current list index = dict([(word.value, i) for i, word in enumerate(object_list)]) # get a new list of objects ordered by the second criteria sorted_object_list = [object_list[index[key]] for key in sorted(index.keys())] I think I got all that syntax right... -Dave . -- Ned Batchelder, http://nedbatchelder.com --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users -~----------~----~----~----~------~----~------~--~---