CODEWITHSUNDEEP
pythondjangoperformanceobjectdjango-models
Leopd
Leopd

Reputation: 42779

Fastest way to get the first object from a queryset in django?

Often I find myself wanting to get the first object from a queryset in Django, or return None if there aren't any. There are lots of ways to do this which all work. But I'm wondering which is the most performant.

qs = MyModel.objects.filter(blah = blah)
if qs.count() > 0:
    return qs[0]
else:
    return None

Does this result in two database calls? That seems wasteful. Is this any faster?

qs = MyModel.objects.filter(blah = blah)
if len(qs) > 0:
    return qs[0]
else:
    return None

Another option would be:

qs = MyModel.objects.filter(blah = blah)
try:
    return qs[0]
except IndexError:
    return None

This generates a single database call, which is good. But requires creating an exception object a lot of the time, which is a very memory-intensive thing to do when all you really need is a trivial if-test.

How can I do this with just a single database call and without churning memory with exception objects?

Upvotes: 307

Views: 377264

Answers (9)

Super Kai - Kazuya Ito
Super Kai - Kazuya Ito

Reputation: 1

You can get the first object with these ways as shown below:

MyModel.objects.first()
MyModel.objects.all().first()
MyModel.objects.all()[0]
MyModel.objects.filter().first()
MyModel.objects.filter()[0]

In addition, you can get the last object with these ways as shown below:

MyModel.objects.last()
MyModel.objects.all().last()
MyModel.objects.filter().last()

Upvotes: 0

cod3monk3y
cod3monk3y

Reputation: 9853

Django 1.6 (released Nov 2013) introduced the convenience methods first() and last() which swallow the resulting exception and return None if the queryset returns no objects.

Upvotes: 495

stormlifter
stormlifter

Reputation: 3851

You can use array slicing:

Entry.objects.all()[:1].get()

Which can be used with .filter():

Entry.objects.filter()[:1].get()

You wouldn't want to first turn it into a list because that would force a full database call of all the records. Just do the above and it will only pull the first. You could even use .order_by() to ensure you get the first you want.

Be sure to add the .get() or else you will get a QuerySet back and not an object.

Upvotes: 174

Nick Cuevas
Nick Cuevas

Reputation: 1748

This could work as well:

def get_first_element(MyModel):
    my_query = MyModel.objects.all()
    return my_query[:1]

if it's empty, then returns an empty list, otherwise it returns the first element inside a list.

Upvotes: 8

levi
levi

Reputation: 22707

Now, in Django 1.9 you have first() method for querysets.

YourModel.objects.all().first()

This is a better way than .get() or [0] because it does not throw an exception if queryset is empty, Therafore, you don't need to check using exists()

Upvotes: 58

Nikolay Fominyh
Nikolay Fominyh

Reputation: 9256

If you plan to get first element often - you can extend QuerySet in this direction: 

class FirstQuerySet(models.query.QuerySet):
    def first(self):
        return self[0]


class ManagerWithFirstQuery(models.Manager):
    def get_query_set(self):
        return FirstQuerySet(self.model)

Define model like this: 

class MyModel(models.Model):
    objects = ManagerWithFirstQuery()

And use it like this:

 first_object = MyModel.objects.filter(x=100).first()

Upvotes: 7

Nauman Tariq
Nauman Tariq

Reputation: 1410

It can be like this

obj = model.objects.filter(id=emp_id)[0]

or

obj = model.objects.latest('id')

Upvotes: 4

Ari
Ari

Reputation: 3127

You should use django methods, like exists. Its there for you to use it.

if qs.exists():
    return qs[0]
return None

Upvotes: 3

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799560

r = list(qs[:1])
if r:
  return r[0]
return None

Upvotes: 48

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