def most_commented(self, num=5, free=True): """ Returns the ``num`` objects with the highest comment counts, in order. Pass ``free=False`` if you're using the registered comment model (Comment) instead of the anonymous comment model (FreeComment). """ from django.db import connection from django.contrib.comments import models as comment_models from django.contrib.contenttypes.models import ContentType if free: comment_opts = comment_models.FreeComment._meta else: comment_opts = comment_models.Comment._meta ctype = ContentType.objects.get_for_model(self.model) query = """SELECT object_id, COUNT(*) AS score FROM %s WHERE content_type_id = %%s AND is_public = 1 GROUP BY object_id ORDER BY score DESC""" % comment_opts.db_table cursor = connection.cursor() cursor.execute(query, [ctype.id]) object_ids = [row[0] for row in cursor.fetchall()[:num]] # Use ``in_bulk`` here instead of an ``id__in`` filter, because ``id__in`` # would clobber the ordering. object_dict = self.in_bulk(object_ids) return [object_dict[object_id] for object_id in object_ids]