[Django]-GROUP_CONCAT equivalent in Django

2๐Ÿ‘

โœ…

The Django ORM does not support this; if you donโ€™t want to use raw SQL then youโ€™ll need to group and join.

57๐Ÿ‘

You can create your own Aggregate Function (doc)

from django.db.models import Aggregate

class Concat(Aggregate):
    function = 'GROUP_CONCAT'
    template = '%(function)s(%(distinct)s%(expressions)s)'

    def __init__(self, expression, distinct=False, **extra):
        super(Concat, self).__init__(
            expression,
            distinct='DISTINCT ' if distinct else '',
            output_field=CharField(),
            **extra)

and use it simply as:

query_set = Fruits.objects.values('type').annotate(count=Count('type'),
                       name = Concat('name')).order_by('-count')

I am using django 1.8 and mysql 4.0.3

๐Ÿ‘คShashank Singla

20๐Ÿ‘

NOTICE that Django (>=1.8) provides Database functions support.
https://docs.djangoproject.com/en/dev/ref/models/database-functions/#concat

Here is an enhanced version of Shashank Singla

from django.db.models import Aggregate, CharField

class GroupConcat(Aggregate):
    function = 'GROUP_CONCAT'
    template = '%(function)s(%(distinct)s%(expressions)s%(ordering)s%(separator)s)'

    def __init__(self, expression, distinct=False, ordering=None, separator=',', **extra):
        super(GroupConcat, self).__init__(
            expression,
            distinct='DISTINCT ' if distinct else '',
            ordering=' ORDER BY %s' % ordering if ordering is not None else '',
            separator=' SEPARATOR "%s"' % separator,
            output_field=CharField(),
            **extra
        )

Usage:

LogModel.objects.values('level', 'info').annotate(
    count=Count(1), time=GroupConcat('time', ordering='time DESC', separator=' | ')
).order_by('-time', '-count')
๐Ÿ‘คWeizhongTu

9๐Ÿ‘

Use GroupConcat from the Django-MySQL package (
https://django-mysql.readthedocs.org/en/latest/aggregates.html#django_mysql.models.GroupConcat ) which I maintain. With it you can do it simply like:

>>> from django_mysql.models import GroupConcat
>>> Fruits.objects.annotate(
...     count=Count('type'),
...     name_list=GroupConcat('name'),
... ).order_by('-count').values('type', 'count', 'name_list')
[{'type': 'apple', 'count': 2, 'name_list': 'fuji,mac'},
 {'type': 'orange', 'count': 1, 'name_list': 'navel'}]
๐Ÿ‘คAdam Johnson

4๐Ÿ‘

If you are using PostgreSQL, you can use ArrayAgg to aggregate all of the values into an array.

https://www.postgresql.org/docs/9.5/static/functions-aggregate.html

๐Ÿ‘คJin

3๐Ÿ‘

If you donโ€™t mind doing this in your template the Django template tag regroup accomplishes this

๐Ÿ‘คdaseme

3๐Ÿ‘

As of Django 1.8 you can use Func() expressions.

query_set = Fruits.objects.values('type').annotate(
    count=Count('type'),
    name=Func(F('name'), 'GROUP_BY')
).order_by('-count')
๐Ÿ‘คJoel Davis

3๐Ÿ‘

Similar Aggregate Function for PostgreSQL in case someone needs:

from django.db.models import Aggregate, CharField

class GroupConcat(Aggregate):
    function = "STRING_AGG"
    template = "%(function)s(%(expressions)s::text, %(separator)s%(ordering)s)"

    def __init__(self, expression, ordering=None, separator=',', **extra):
        super().__init__(
            expression,
            ordering=" ORDER BY %s" % ordering if ordering is not None else "",
            separator="'%s'" % separator,
            output_field=CharField(),
            **extra
        )

queryset = Fruits.objects.values('type').annotate(
    count=Count('type'),
    name=GroupConcat('name')
).order_by('-count')
๐Ÿ‘คRezolventa

1๐Ÿ‘

Not supported by Django ORM, but you can build your own aggregator.

Itโ€™s actually pretty straightforward, here is a link to a how-to that does just that, with GROUP_CONCAT for SQLite: http://harkablog.com/inside-the-django-orm-aggregates.html

Note however, that it might be necessary to handle different SQL dialects separately. For example, the SQLite docs say about group_concat:

The order of the concatenated elements is arbitrary

While MySQL allows you to specify the order.

I guess that may be a reason why GROUP_CONCAT itโ€™s not implemented in Django at the moment.

๐Ÿ‘คfrnhr

1๐Ÿ‘

To complete the answer of @WeizhongTu, Notice that you can not use the keyword SEPARATOR with SQLITE. In cases where you are using MySQL and SQLite for your tests, you can write :

class GroupConcat(Aggregate):
    function = 'GROUP_CONCAT'
    separator = ','

    def __init__(self, expression, distinct=False, ordering=None, **extra):
        super(GroupConcat, self).__init__(expression,
                                          distinct='DISTINCT ' if distinct else '',
                                          ordering=' ORDER BY %s' % ordering if ordering is not None else '',
                                          output_field=CharField(),
                                          **extra)

    def as_mysql(self, compiler, connection, separator=separator):
        return super().as_sql(compiler,
                              connection,
                              template='%(function)s(%(distinct)s%(expressions)s%(ordering)s%(separator)s)',
                              separator=' SEPARATOR \'%s\'' % separator)

    def as_sql(self, compiler, connection, **extra):
        return super().as_sql(compiler,
                              connection,
                              template='%(function)s(%(distinct)s%(expressions)s%(ordering)s)',
                              **extra)
๐Ÿ‘คJboulery

1๐Ÿ‘

I just wanted to say a word of caution if you go with any of the proposed solutions for MySQL: by default, MySQL will truncate GROUP_CONCAT result to 1024 characters. Iโ€™ve spent some time figuring out why I was getting non-existent ids (they were truncated existent ids).

You can avoid the limitation by setting group_concat_max_len in Django settings. An example is here: Include multiple statements in Django's raw queries

๐Ÿ‘คSoid

0๐Ÿ‘

this is the best way of working in Django ORM

f1 = Fruits.objects.values('type').annotate(count = Count('type'),namelist= GroupConcat('namelist')).distinct()

Leave a comment