Good ways to sort a queryset? - Django
Solution 1:
What about
import operator
auths = Author.objects.order_by('-score')[:30]
ordered = sorted(auths, key=operator.attrgetter('last_name'))
In Django 1.4 and newer you can order by providing multiple fields.
Reference: https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by
order_by(*fields)
By default, results returned by a QuerySet
are ordered by the ordering tuple given by the ordering
option in the model’s Meta. You can override this on a per-QuerySet basis by using the order_by
method.
Example:
ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]
The result above will be ordered by score
descending, then by last_name
ascending. The negative sign in front of "-score"
indicates descending order. Ascending order is implied.
Solution 2:
I just wanted to illustrate that the built-in solutions (SQL-only) are not always the best ones. At first I thought that because Django's QuerySet.objects.order_by
method accepts multiple arguments, you could easily chain them:
ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]
But, it does not work as you would expect. Case in point, first is a list of presidents sorted by score (selecting top 5 for easier reading):
>>> auths = Author.objects.order_by('-score')[:5]
>>> for x in auths: print x
...
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)
Using Alex Martelli's solution which accurately provides the top 5 people sorted by last_name
:
>>> for x in sorted(auths, key=operator.attrgetter('last_name')): print x
...
Benjamin Harrison (467)
James Monroe (487)
Gerald Rudolph (464)
Ulysses Simpson (474)
Harry Truman (471)
And now the combined order_by
call:
>>> myauths = Author.objects.order_by('-score', 'last_name')[:5]
>>> for x in myauths: print x
...
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)
As you can see it is the same result as the first one, meaning it doesn't work as you would expect.
Solution 3:
Here's a way that allows for ties for the cut-off score.
author_count = Author.objects.count()
cut_off_score = Author.objects.order_by('-score').values_list('score')[min(30, author_count)]
top_authors = Author.objects.filter(score__gte=cut_off_score).order_by('last_name')
You may get more than 30 authors in top_authors this way and the min(30,author_count)
is there incase you have fewer than 30 authors.