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?

Use the convenience methods .first() and .last():

MyModel.objects.filter(blah=blah).first()

They both swallow the resulting exception and return None if the queryset returns no objects.

These were added in Django 1.6, which was released in Nov 2013.

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.

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

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()