I want to take the last 10 instances of a model and have this code:
Model.objects.all().order_by('-id')[:10] Is it true that firstly pick up all instances, and then take only 10 last ones? Is there any more effective method?
- 12docs.djangoproject.com/en/dev/topics/db/queries/…Ignacio Vazquez-Abrams– Ignacio Vazquez-Abrams2011-07-04 16:38:10 +00:00Commented Jul 4, 2011 at 16:38
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7 Answers
Django querysets are lazy. That means a query will hit the database only when you specifically ask for the result.
So until you print or actually use the result of a query you can filter further with no database access.
As you can see below your code only executes one sql query to fetch only the last 10 items.
In [19]: import logging In [20]: l = logging.getLogger('django.db.backends') In [21]: l.setLevel(logging.DEBUG) In [22]: l.addHandler(logging.StreamHandler()) In [23]: User.objects.all().order_by('-id')[:10] (0.000) SELECT "auth_user"."id", "auth_user"."username", "auth_user"."first_name", "auth_user"."last_name", "auth_user"."email", "auth_user"."password", "auth_user"."is_staff", "auth_user"."is_active", "auth_user"."is_superuser", "auth_user"."last_login", "auth_user"."date_joined" FROM "auth_user" ORDER BY "auth_user"."id" DESC LIMIT 10; args=() Out[23]: [<User: hamdi>] 
Comments
Actually I think the LIMIT 10 would be issued to the database so slicing would not occur in Python but in the database.
See limiting-querysets for more information.
2 Comments
Mike 'Pomax' Kamermans
Note that this won't work for querysets that also need filtering, as you cannot filter after slicing.
Vyachez
So filter first than slice it. Thanks Davor for link!
Looks like the solution in the question doesn't work with Django 1.7 anymore and raises an error: "Cannot reorder a query once a slice has been taken"
According to the documentation https://docs.djangoproject.com/en/dev/topics/db/queries/#limiting-querysets forcing the “step” parameter of Python slice syntax evaluates the Query. It works this way:
Model.objects.all().order_by('-id')[:10:1] Still I wonder if the limit is executed in SQL or Python slices the whole result array returned. There is no good to retrieve huge lists to application memory.
1 Comment
sonus21
Even this solution doesn't work with django>=1.8 tested.
Yes. If you want to fetch a limited subset of objects, you can with the below code:
Example:
obj=emp.objects.all()[0:10] The beginning 0 is optional, so
obj=emp.objects.all()[:10] The above code returns the first 10 instances.
Comments
The simple answer for filter issue
Notification.objects.filter(user=request.user).order_by("-id")[:limit] Just put order_by and then [:limit]
Comments
Django 1.9.1
print qs.query and see LIMIT
pos_random = Position.objects.filter( gps_time__gte = start_time, gps_time__lte = end_time, vehicle = v)[:**LIMIT_NUMBER**] 
Comments
As an addition and observation to the other useful answers, it's worth noticing that actually doing [:10] as slicing will return the first 10 elements of the list, not the last 10...
To get the last 10 you should do [-10:] instead (see here). This will help you avoid using order_by('-id') with the - to reverse the elements.
5 Comments
bparker
I tried this and got "Negative indexing is not supported."
bersam
@DarkCygnus
Product.objects.filter(~Q(price=0))[-5:] cause me same error: "Negative indexing is not supported."
valem
This doesn't work in django on a queryset: code.djangoproject.com/ticket/13089 If you convert the queryset to a list it will work.
FalseDev
@valem Converting the queryset to a list will override django's lazy evaluation and cause huge datasets to be fetched from the database
valem
@Navaneethan yes that's true. but not really relevant to this answer