[Django]-Get distinct values of Queryset by field

With Alex Answer I also have the n:1 for each item. Even with a distinct() clause.

It’s weird because this is returning the good numbers of items :

Visit.objects.filter(stuff).values("ip_address").distinct().count()

But when I iterate over “Visit.objects.filter(stuff).values(“ip_address”).distinct()” I got much more items and some duplicates…

EDIT :

The filter clause was causing me troubles. I was filtering with another table field and a SQL JOIN was made that was breaking the distinct stuff.
I used this hint to see the query that was really used :

q=Visit.objects.filter(myothertable__field=x).values("ip_address").distinct().count()
print q.query

I then reverted the class on witch I was making the query and the filter to have a join that doesn’t rely on any “Visit” id.

hope this helps

The question is different from what the title suggests. If you want set-like behavior from the database, you need something like this.

x = Visit.objects.all().values_list('ip_address', flat=True).distinct()

It should give you something like this for x.

[1.2.3.4, 2.3.4.5, ...]

Where

len(x) == len(set(x))

Returns True