how to sort by a computed value in django

Question

Hey I want to sort objects based on a computed value in django... how do I do it?

Here is an example User profile model based on stack overflow that explains my predicament:

class Profile(models.Model):
    user = models.ForeignKey(User)

    def get_reputation():
        ... 
        return reputation
    reputation = property(get_reputation)

So, say I want to sort users by reputation. How do I do that? I know you can't just do this:

Profile.objects.order_by("-reputation")

Thanks for your help everyone :)

Answer 1

If you need to do the sorting in the database (because you have lots of records, and need to e.g. paginate them), the only real option is to turn reputation into a denormalized field (e.g. updated in an overridden save() method on the model).

Answer 2

Since your calculation code exists only within Python, you have to perform the sorting in Python as well:

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

Answer 3

As of Django-1.8, it is possible to sort a QuerySet using query expressions as long as the computed value can be rendered in SQL. You can also use annotations to the model as the order_by term.

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

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

Links:

• Aggregation - order_by
• Making Queries - filters can reference fields on a model
• Query Expressions - Built in Expressions

See also this SO Q&A: Order a QuerySet by aggregate field value