django-aggregate-if 0.3.1

Django Aggregate If: Condition aggregates for Django

Aggregate-if adds conditional aggregates to Django.

Conditional aggregates can help you reduce the ammount of queries to obtain aggregated information, like statistics for example.

Imagine you have a model Offer like this one:

class Offer(models.Model):
    sponsor = models.ForeignKey(User)
    price = models.DecimalField(max_digits=9, decimal_places=2)
    status = models.CharField(max_length=30)
    expire_at = models.DateField(null=True, blank=True)
    created_at = models.DateTimeField(auto_now_add=True)
    updated_at = models.DateTimeField(auto_now=True)

    OPEN = "OPEN"
    REVOKED = "REVOKED"
    PAID = "PAID"

Let’s say you want to know:

1. How many offers exists in total;

2. How many of them are OPEN, REVOKED or PAID;

3. How much money was offered in total;

4. How much money is in OPEN, REVOKED and PAID offers;

To get these informations, you could query:

from django.db.models import Count, Sum

Offer.objects.count()
Offer.objects.filter(status=Offer.OPEN).aggregate(Count('pk'))
Offer.objects.filter(status=Offer.REVOKED).aggregate(Count('pk'))
Offer.objects.filter(status=Offer.PAID).aggregate(Count('pk'))
Offer.objects.aggregate(Sum('price'))
Offer.objects.filter(status=Offer.OPEN).aggregate(Sum('price'))
Offer.objects.filter(status=Offer.REVOKED).aggregate(Sum('price'))
Offer.objects.filter(status=Offer.PAID).aggregate(Sum('price'))

In this case, 8 queries were needed to retrieve the desired information.

With conditional aggregates you can get it all with only 1 query:

from django.db.models import Q
from aggregate_if import Count, Sum

Offer.objects.aggregate(
    Count('pk'),
    Count('pk', only=Q(status=Offer.OPEN)),
    Count('pk', only=Q(status=Offer.REVOKED)),
    Count('pk', only=Q(status=Offer.PAID)),
    Sum('price'),
    Sum('price', only=Q(status=Offer.OPEN)),
    Sum('price'), only=Q(status=Offer.REVOKED)),
    Sum('price'), Q(status=Offer.PAID)),
)

$ pip install django-aggregate-if

License

The MIT License.