On Thu, Feb 10, 2011 at 2:29 PM, Alex Robbins <[email protected]> wrote: > Yeah, you'll definitely want to find some aggregate to do the sorting.
Ok, didn't know it. I'll take a look at it... > The only way to sort by a custom method is doing an in-python sort, > which is going to be much slower than having the db sort for you. Yeah, I tought so but it seems (to me) the only way of doing this... > If you post the score models, we could probably help more. The score models are from django-voting: http://django-voting.googlecode.com/svn/trunk/voting/models.py http://django-voting.googlecode.com/svn/trunk/voting/managers.py Thanks for the help! Andres > > Alex > > On Feb 9, 8:49 am, "Casey S. Greene" <[email protected]> wrote: >> I haven't used django-voting but it sounds to me like you want something >> like: >> Link.objects.aggregate(Avg(score = 'vote__score')).order_by('score') >> >> If I recall correctly you can chain aggregate and order_by. >> >> Anyway, that example and this link should get you started at >> least:http://docs.djangoproject.com/en/dev/topics/db/aggregation/ >> >> Hope this helps! >> Casey >> >> >> >> >> >> >> >> On Wed, 2011-02-09 at 10:08 +0100, Andres Lucena wrote: >> > On Tue, Feb 8, 2011 at 6:00 PM, Andres Lucena <[email protected]> >> > wrote: >> > > Dear Gurus, >> >> > > I've made a custom method for getting the score (from django-voting) >> > > for a giving Model: >> >> > > class Link(models.Model): >> > > episode = models.ForeignKey("Episode", related_name="links") >> > > url = models.CharField(max_length=255, unique=True, db_index=True) >> >> > > def __unicode__(self): >> > > return self.url >> >> > > def get_score(self): >> > > return Vote.objects.get_score(self)['score'] >> >> > > Now I want to make a custom manager to getting the top-scored links >> > > for the given episode. AFAIK, you can't sort by a custom method, so >> > > I'm trying to apply the ordering through sorted(), like this links >> > > says: >> >> > >http://stackoverflow.com/questions/981375/using-a-django-custom-model... >> > >http://stackoverflow.com/questions/883575/custom-ordering-in-django >> >> > > So, what I have now is this: >> >> > > class LinkGetTopScores(models.Manager): >> > > def get_top_score(self): >> > > return sorted(self.filter(episode=self.episode), key=lambda n: >> > > n.get_score) >> >> > > class Link(models.Model): >> > > episode = models.ForeignKey("Episode", related_name="links") >> > > url = models.CharField(max_length=255, unique=True, db_index=True) >> > > get_top_score = LinkGetTopScores() >> > > .... >> >> > > So of course this isn't working because of the self.episode stuff... >> > > But I've to filter somehow by episode (the ForeignKey), and I don't >> > > know how. Is there anyway of doing this?? What I'm doing is right or >> > > there would be an easier way of doing this? >> >> > I noticed that the .filter isn't necesary, so now I have this: >> >> > class LinkGetTopScores(models.Manager): >> > def get_top_score(self): >> > return sorted(self.all(), key=lambda n: n.get_score) >> >> > But it don't sort by score, and I don't know what I'm doing wrong :S >> >> > Any idea? >> >> > Thanks, >> > Andres >> >> > > Thank you, >> > > Andres > > -- > You received this message because you are subscribed to the Google Groups > "Django users" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/django-users?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
