Django filter queryset __in for *every* item in list


Let’s say I have the following models

class Photo(models.Model):
    tags = models.ManyToManyField(Tag)

class Tag(models.Model):
    name = models.CharField(max_length=50)

In a view I have a list with active filters called categories.
I want to filter Photo objects which have all tags present in categories.

I tried:


But this matches any item in categories, not all items.

So if categories would be [‘holiday’, ‘summer’] I want Photo’s with both a holiday and summer tag.

Can this be achieved?

Asked By: Sander van Leeuwen



This also can be done by dynamic query generation using Django ORM and some Python magic 🙂

from operator import and_
from django.db.models import Q

categories = ['holiday', 'summer']
res = Photo.filter(reduce(and_, [Q(tags__name=c) for c in categories]))

The idea is to generate appropriate Q objects for each category and then combine them using AND operator into one QuerySet. E.g. for your example it’d be equal to

res = Photo.filter(Q(tags__name='holiday') & Q(tags__name='summer'))
Answered By: demalexx


One option is, as suggested by jpic and sgallen in the comments, to add .filter() for each category. Each additional filter adds more joins, which should not be a problem for small set of categories.

There is the aggregation approach. This query would be shorter and perhaps quicker for a large set of categories.

You also have the option of using custom queries.

Some examples

Test setup:

class Photo(models.Model):
    tags = models.ManyToManyField('Tag')

class Tag(models.Model):
    name = models.CharField(max_length=50)

    def __unicode__(self):

In [2]: t1 = Tag.objects.create(name='holiday')
In [3]: t2 = Tag.objects.create(name='summer')
In [4]: p = Photo.objects.create()
In [5]: p.tags.add(t1)
In [6]: p.tags.add(t2)
In [7]: p.tags.all()
Out[7]: [<Tag: holiday>, <Tag: summer>]

Using chained filters approach:

In [8]: Photo.objects.filter(tags=t1).filter(tags=t2)
Out[8]: [<Photo: Photo object>]

Resulting query:

In [17]: print Photo.objects.filter(tags=t1).filter(tags=t2).query
SELECT "test_photo"."id"
FROM "test_photo"
INNER JOIN "test_photo_tags" ON ("test_photo"."id" = "test_photo_tags"."photo_id")
INNER JOIN "test_photo_tags" T4 ON ("test_photo"."id" = T4."photo_id")
WHERE ("test_photo_tags"."tag_id" = 3  AND T4."tag_id" = 4 )

Note that each filter adds more JOINS to the query.

Using annotation approach:

In [29]: from django.db.models import Count
In [30]: Photo.objects.filter(tags__in=[t1, t2]).annotate(num_tags=Count('tags')).filter(num_tags=2)
Out[30]: [<Photo: Photo object>]

Resulting query:

In [32]: print Photo.objects.filter(tags__in=[t1, t2]).annotate(num_tags=Count('tags')).filter(num_tags=2).query
SELECT "test_photo"."id", COUNT("test_photo_tags"."tag_id") AS "num_tags"
FROM "test_photo"
LEFT OUTER JOIN "test_photo_tags" ON ("test_photo"."id" = "test_photo_tags"."photo_id")
WHERE ("test_photo_tags"."tag_id" IN (3, 4))
GROUP BY "test_photo"."id", "test_photo"."id"
HAVING COUNT("test_photo_tags"."tag_id") = 2

ANDed Q objects would not work:

In [9]: from django.db.models import Q
In [10]: Photo.objects.filter(Q(tags__name='holiday') & Q(tags__name='summer'))
Out[10]: []
In [11]: from operator import and_
In [12]: Photo.objects.filter(reduce(and_, [Q(tags__name='holiday'), Q(tags__name='summer')]))
Out[12]: []

Resulting query:

In [25]: print Photo.objects.filter(Q(tags__name='holiday') & Q(tags__name='summer')).query
SELECT "test_photo"."id"
FROM "test_photo"
INNER JOIN "test_photo_tags" ON ("test_photo"."id" = "test_photo_tags"."photo_id")
INNER JOIN "test_tag" ON ("test_photo_tags"."tag_id" = "test_tag"."id")
WHERE ("test_tag"."name" = holiday  AND "test_tag"."name" = summer )
Answered By: Davor Lucic

Another approach that works, although PostgreSQL only, is using django.contrib.postgres.fields.ArrayField:

Example copied from docs:

>>> Post.objects.create(name='First post', tags=['thoughts', 'django'])
>>> Post.objects.create(name='Second post', tags=['thoughts'])
>>> Post.objects.create(name='Third post', tags=['tutorial', 'django'])

>>> Post.objects.filter(tags__contains=['thoughts'])
<QuerySet [<Post: First post>, <Post: Second post>]>

>>> Post.objects.filter(tags__contains=['django'])
<QuerySet [<Post: First post>, <Post: Third post>]>

>>> Post.objects.filter(tags__contains=['django', 'thoughts'])
<QuerySet [<Post: First post>]>

ArrayField has some more powerful features such as overlap and index transforms.

Answered By: Sander van Leeuwen

If we want to do it dynamically, followed the example:

tag_ids = [,]
qs = Photo.objects.all()

for tag_id in tag_ids:
    qs = qs.filter(tag__id=tag_id)    

print qs
Answered By: tarasinf

I use a little function that iterates filters over a list for a given operator an a column name :

def exclusive_in (cls,column,operator,value_list):         
    myfilter = column + '__' + operator
    query = cls.objects
    for value in value_list:
    return query  

and this function can be called like that:


it also work with any class and more tags in the list; operators can be anyone like ‘iexact’,’in’,’contains’,’ne’,…

Answered By: David
queryset = Photo.objects.filter(tags__name="vacaciones") | Photo.objects.filter(tags__name="verano")

If you struggled with this problem as i did and nothing mentioned helped you, maybe this one will solve your issue

Instead of chaining filter, in some cases it would be better just to store ids of previous filter

tags = [1, 2]
for tag in tags:
    ids = list(queryset.filter(tags__id=tag).values_list("id", flat=True))
    queryset = queryset.filter(id__in=ids)

Using this approach will help you to avoid stacking JOIN in SQL query:

Answered By: Alexander Lekontsev

My solution:
let say
author is list of elements that need to match all item in list, so:

        for a in author:
            queryset = queryset.filter(authors__author_first_name=a)
                if not queryset:
Answered By: simon
for category in categories:
    query = Photo.objects.filter(tags_name=category)

this piece of code , filters your photos which have all the tags name coming from categories.

Answered By: fateme akrami