Django: remove duplicates (group by) from queryset by related model field

Question

I have a Queryset with a couple of records, and I wan't to remove duplicates using the related model field. For example:

class User(models.Model):
    group = models.ForeignKey('Group')
    ...

class Address(models.Model):
    ...
    models.ForeignKey('User')

addresses = Address.objects.filter(user__group__id=1).order_by('-id')

This returns a QuerySet of Address records, and I want to group by the User ID.

• I can't use .annotate because I need all fields from Address, and the relationship between Address and User
• I can't use .distinct() because it doesn't work, since all addresses are distinct, and I want distinct user addresses.

I could:

addresses = Address.objects.filter(user__group__id=1).order_by('-id')
unique_users_ids = []
unique_addresses = []

for address in addresses:
    if address.user.id not in unique_users_ids:
        unique_addresses.append(address)
        unique_users_ids.append(address.user.id)

print unique_addresses # TA-DA!

But it seems too much for a simple thing like a group by (damn you Django).

Is there a easy way to achieve this?

Answer 1

By using .distinct() with a field name

Django has also a .distinct(..) function that takes as input column the column names that should be unique. Alas most database systems do not support this (only PostgreSQL to the best of my knowledge). But in PostgreSQL we can thus perform:

# Limited number of database systems support this
addresses = (Address.objects
                    .filter(user__group__id=1)
                    .order_by('-id')
                    )

By using two queries

Another way to handle this is by first having a query that works over the users, and for each user obtains the largest address_id :

address_ids = (User.objects
                   
                   .filter()
                   .values_list())

So now for every user, we have calculated the largest corresponding address_id , and we eliminate User s that have no address. We then obtain the list of id s.

In a second step, we then fetch the addresses:

addresses = Address.objects.filter()