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  1. PO
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    <p>Seems like you could overload the <code>save</code> of the <code>Vote</code> class and have it update the corresponding <code>Link</code> object. Something like this should work well:</p> <pre><code>from datetime import datetime, timedelta class Link(models.Model): category = models.ForeignKey(Category) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) fame = models.PositiveIntegerField(default = 1) title = models.CharField(max_length = 256) url = models.URLField(max_length = 2048) #a field to keep the most recently calculated popularity popularity = models.FloatField(default = None) def CalculatePopularity(self): """ Add a shorcut to make life easier ... this is used by the overloaded save() method and can be used in a management function to do a mass-update periodically """ ts = datetime.now()-self.created th = ts.seconds/60/60 self.popularity = (self.user_set.count()-1)/((th+2)**1.5) def save(self, *args, **kwargs): """ Modify the save function to calculate the popularity """ self.CalculatePopularity() super(Link, self).save(*args, **kwargs) def __unicode__(self): return self.title class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) karma_delta = models.SmallIntegerField() def save(self, *args, **kwargs): """ Modify the save function to calculate the popularity of the Link object """ self.link.CalculatePopularity() super(Vote, self).save(*args, **kwargs) def __unicode__(self): return str(self.karma_delta) </code></pre> <p>This way every time you call a link_o.save() or vote_o.save() it will re-calculate the popularity. You have to be a little careful because when you call <code>Link.objects.all().update('updating something')</code> then it won't call our overloaded <code>save()</code> function. So when I use this sort of thing I create a management command which updates all of the objects so they're not too out of date. Something like this will work wonderfully:</p> <pre><code>from itertools import imap imap(lambda x:x.CalculatePopularity(), Link.objects.all().select_related().iterator()) </code></pre> <p>This way it will only load a single Link object into memory at once ... so if you have a giant database it won't cause a memory error.</p> <p>Now to do your ranking all you have to do is:</p> <pre><code>Link.objects.all().order_by('-popularity') </code></pre> <p>It will be super-fast since all of you Link items have already calculated the popularity.</p>
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    1. POImplementing a popularity algorithm in Django
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    1. USJudoWill
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    1. CODoes this mean that everytime a link is voted on, every single link row is updated?
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      1. USTheLizardKing
    2. COno ... it would only update the link that was voted on ... so it wouldn't be that processor intensive
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