["python","django","django-queryset","greatest-n-per-group"]
Django Query That Get Most Recent Objects From Different Categories

I have two models A and B. All B objects have a foreign key to an A object. Given a set of A objects, is there anyway to use the ORM to get a set of B objects containing the most recent object created for each A object. Here's an simplified example: class Bakery(models.Model): town = models.CharField(max_length=255) class Cake(models.Model): bakery = models.ForeignKey(Bakery, on_delete=models.CASCADE) baked_at = models.DateTimeField() So I'm looking for a query that returns the most recent cake baked in each bakery in Anytown, USA.
As far as I know, there is no one-step way of doing this in Django ORM, but you can split it into two queries: from django.db.models import Max bakeries = Bakery.objects.annotate( hottest_cake_baked_at=Max('cake__baked_at') ) hottest_cakes = Cake.objects.filter( baked_at__in=[b.hottest_cake_baked_at for b in bakeries] ) If id's of cakes are progressing along with bake_at timestamps, you can simplify and disambiguate the above code (in case two cakes arrives at the same time you can get both of them): from django.db.models import Max hottest_cake_ids = Bakery.objects.annotate( hottest_cake_id=Max('cake__id') ).values_list('hottest_cak‌​e_id', flat=True) hottest_cakes = Cake.objects.filter(id__in=hottest_cake_ids) BTW credits for this goes to Daniel Roseman, who once answered similar question of mine: http://groups.google.pl/group/django-users/browse_thread/thread/3b3cd4cbad478d34/3e4c87f336696054?hl=pl&q= If the above method is too slow, then I know also second method - you can write custom SQL producing only those Cakes, that are hottest in relevant Bakeries, define it as database VIEW, and then write unmanaged Django model for it. It's also mentioned in the above django-users thread. Direct link to the original concept is here: http://web.archive.org/web/20130203180037/http://wolfram.kriesing.de/blog/index.php/2007/django-nice-and-critical-article#comment-48425 Hope this helps.