How to sort by computed value in django

Question

Hi, I want to sort objects based on computed value in django ... how to do this?

Here is an example of a stack overflow model of a user profile that explains my predicament:

class Profile(models.Model): user = models.ForeignKey(User) def get_reputation(): ... return reputation reputation = property(get_reputation) 

So, let's say I want to sort users by reputation. How should I do it? I know that you cannot just do this:

 Profile.objects.order_by("-reputation") 

Thanks for all your help :)

Answer 1

If you need to sort in the database (because you have many records and need to paginate them, for example), the only real option is to turn the reputation into a denormalized field (for example, updated in the overridden save() method on the model).

Answer 2

Since your calculation code only exists inside Python, you also need to do the sorting in Python:

 sorted (Profile.objects.all (), key = lambda p: p.reputation) 

Answer 3

As of Django-1.8, you can sort a QuerySet using a query expression expression , if the calculated value can be displayed in SQL, you can also use annotations for the model as an order_by term.

 from django.db.models import F Profile.objects.annotate(reputation=(F('<field>') + ...)).order_by("-reputation") 

Basic operations such as + , - , * , / and ** can be used with F() . In addition, there are several other functions such as Count , Avg , Max , ...

References:

• Aggregation - order_by
• Query execution - filters can refer to fields on the model
• Query Expressions - Inline Expressions

See also SO Q & A: QuerySet Order by Aggregation Field Value