Django Aggregation: Summation of Multiplication of

Question

I have a model some thing like this

class Task(models.Model):
   progress = models.PositiveIntegerField()
   estimated_days = models.PositiveIntegerField()

Now I would like to do a calculation Sum(progress * estimated_days) on the database level. Using Django Aggregation I can have the sum for each field but not the summation of multiplication of fields.

Answer 1

With Django 1.8 and above you can now pass an expression to your aggregate:

 from django.db.models import F

 Task.objects.aggregate(total=Sum(F('progress') * F('estimated_days')))['total']

Constants are also available, and everything is combinable:

 from django.db.models import Value

 Task.objects.aggregate(total=Sum('progress') / Value(10))['total']

Answer 2

Do you have several options:

1. Raw query
2. Emulbreh's undocumented approach
3. Create a third field progress_X_estimated_days and update it in save overwrited method. Then do aggregation through this new field.

Overwriting:

class Task(models.Model):
   progress = models.PositiveIntegerField()
   estimated_days = models.PositiveIntegerField()
   progress_X_estimated_days = models.PositiveIntegerField(editable=False)

   def save(self, *args, **kwargs):
      progress_X_estimated_days = self.progress * self.estimated_days
      super(Task, self).save(*args, **kwargs)

Answer 3

Update: for Django >= 1.8 please follow the answer provided by @kmmbvnr

it's possible using Django ORM:

here's what you should do:

from django.db.models import Sum

total = ( Task.objects
            .filter(your-filter-here)
            .aggregate(
                total=Sum('progress', field="progress*estimated_days")
             )['total']
         )

Note: if the two fields are of different types, say integer & float, the type you want to return should be passed as the first parameter of Sum

It's a late answer, but I guess it'll help someone looking for the same.

Answer 4

The solution depends on Django version.

• django < 1.8

  from django.db.models import Sum
  MyModel.objects.filter(<filters>).aggregate(Sum('field1', field="field1*field2"))
  

• django >= 1.8

  from django.db.models import Sum, F
  MyModel.objects.filter(<filters>).aggregate(Sum(F('field1')*F('field2')))