Django database Aggregate functions Cheat sheet

By: Varghese Chacko 1 year, 3 months ago

# Total number of books.
>>> Book.objects.count()
2452

# Total number of books with publisher=BaloneyPress
>>> Book.objects.filter(publisher__name='BaloneyPress').count()
73

# Average price across all books.
>>> from django.db.models import Avg
>>> Book.objects.all().aggregate(Avg('price'))
{'price__avg': 34.35}

# Max price across all books.
>>> from django.db.models import Max
>>> Book.objects.all().aggregate(Max('price'))
{'price__max': Decimal('81.20')}

# Difference between the highest priced book and the average price of all books.
>>> from django.db.models import FloatField
>>> Book.objects.aggregate(
...     price_diff=Max('price', output_field=FloatField()) - Avg('price'))
{'price_diff': 46.85}

# All the following queries involve traversing the Book<->Publisher
# foreign key relationship backwards.

# Each publisher, each with a count of books as a "num_books" attribute.
>>> from django.db.models import Count
>>> pubs = Publisher.objects.annotate(num_books=Count('book'))
>>> pubs
<QuerySet [<Publisher: BaloneyPress>, <Publisher: SalamiPress>, ...]>
>>> pubs[0].num_books
73

# Each publisher, with a separate count of books with a rating above and below 5
>>> from django.db.models import Q
>>> above_5 = Count('book', filter=Q(book__rating__gt=5))
>>> below_5 = Count('book', filter=Q(book__rating__lte=5))
>>> pubs = Publisher.objects.annotate(below_5=below_5).annotate(above_5=above_5)
>>> pubs[0].above_5
23
>>> pubs[0].below_5
12

# The top 5 publishers, in order by number of books.
>>> pubs = Publisher.objects.annotate(num_books=Count('book')).order_by('-num_books')[:5]
>>> pubs[0].num_books
1323