Django group by

# If you mean to do aggregation you can use the aggregation features of the ORM:
from django.db.models import Count
Members.objects.values('designation').annotate(dcount=Count('designation'))

# This results in a query similar to:
SELECT designation, COUNT(designation) AS dcount
FROM members GROUP BY designation

#and the output would be of the form
[{'designation': 'Salesman', 'dcount': 2}, 
 {'designation': 'Manager', 'dcount': 2}]

Source: stackoverflow.com

Similar codes

1 vote318 views2 answers
How to 404 custom page not found in django
Python Django
54 votes496 views2 answers
How to run django requirement.txt
Python Django
0 vote354 views1 answer
Django create view class
Python Django
2 votes362 views2 answers
Super in django manager
Python Django
0 vote394 views1 answer
Setting up a django prostgress project
Python Django
4 votes334 views2 answers
Arrayfield django example
Python Django
9 votes336 views2 answers
Django app
Python Django
0 vote359 views1 answer
Cannot cast type smallint to boolean django
Python Django
1 vote338 views1 answer
Django get model class from queryset
Python Django
0 vote347 views2 answers
Form.as_p in django
Python Django