Django equivalent of COUNT with GROUP BY


Question

I know Django 1.1 has some new aggregation methods. However I couldn't figure out equivalent of the following query:

SELECT player_type, COUNT(*) FROM players GROUP BY player_type;

Is it possible with Django 1.1's Model Query API or should I just use plain SQL?

1
31
9/5/2009 4:14:09 PM

Accepted Answer

If you are using Django 1.1 beta (trunk):

Player.objects.values('player_type').order_by().annotate(Count('player_type'))
  • values('player_type') - for inclusion only player_type field into GROUP BY clause.
  • order_by() - for exclusion possible default ordering that can cause not needed fields inclusion in SELECT and GROUP BY.
61
5/9/2009 7:40:12 AM

Django 1.1 does support aggregation methods like count. You can find the full documentation here.

To answer your question, you can use something along the lines of:

from django.db.models import Count
q = Player.objects.annotate(Count('games'))
print q[0]
print q[0].games__count

This will need slight tweaking depending on your actual model.

Edit: The above snippet generates aggregations on a per-object basis. If you want aggregation on a particular field in the model, you can use the values method:

from django.db.models import Count
q = Player.objects.values('playertype').annotate(Count('games')).order_by()
print q[0]
print q[0].games__count

order_by() is needed because fields that are in the default ordering are automatically selected even if they are not explicitly passed to values(). This call to order_by() clears any ordering and makes the query behave as expected.

Also, if you want to count the field that is used for grouping (equivalent to COUNT(*)), you can use:

from django.db.models import Count
q = Player.objects.values('playertype').annotate(Count('playertype')).order_by()
print q[0]
print q[0].playertype__count

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