Hi folks,
what I'm trying to do is this:
get the 30 Authors with highest score (
Author.objects.order_by('-score')[:30]
)order the authors by
last_name
Any suggestions?
Hi folks,
what I'm trying to do is this:
get the 30 Authors with highest score ( Author.objects.order_by('-score')[:30]
)
order the authors by last_name
Any suggestions?
What about
import operator
auths = Author.objects.order_by('-score')[:30]
ordered = sorted(auths, key=operator.attrgetter('last_name'))
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.
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.