Django snippets

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]