Django: Efficient model for ordered list of objects

Question

I have two classes: A and B. Each A has an ordered ranking of Bs, and each B has an ordered ranking of As.

My models.py:

class A(models.Model):
    name = models.CharField(max_length=200)
    ...

class B(models.Model):
    name = models.CharField(max_length=200)
    ...

class Ranking(models.Model):
    a = models.ForeignKey(A)
    b = models.ForeignKey(B)
    rankofa = models.IntegerField()
    rankofb = models.IntegerField()

Examples of what I need to do:

Find the number of As that a B has ranked.
Iterate in order through all Bs ranked by a particular A and evaluate some property of B.
Go to an A's lowest-ranked B, and find the ranking given by that B to that A.

I'd like to set up the rankings as an array/list of objects (or some way that does not keep making expensive database calls), but not sure how to do this as a Django models/database solution.

Any and all thoughts appreciated.

Marat · Accepted Answer · 2012-08-31 00:53:05Z

What you need is intermediary M2M relation

class Seller(models.Model):
    name = models.CharField(max_length=200)

class Customer(models.Model):
    name = models.CharField(max_length=200)
    feedbacks = models.ManyToManyField(Seller, through='Feedback', related_name='feedbacks')

class Feedback(models.Model):
    seller = models.ForeignKey(Seller)
    customer = models.ForeignKey(Customer)
    seller_feedback = models.IntegerField()
    customer_feedback = models.IntegerField()

Find the number of sellers customer has ranked:

customer.feedbacks.all().count()

Iterate in order through all customers ranked by a particular seller and evaluate some customer property.

for customer in seller.feedbacks.all():
    do_something(customer)

Go to an sellers's lowest-ranked customer, and find the ranking given by that customer to that a seller.

def lowes_feedback_response(seller):
    try:
        return models.Feedback.filter(seller=seller).order_by('-seller_feedback')[0].customer_feedback
    except models.Feedback.DoesNotExist:
        return None

I didn't evaluate the code but you get the basic idea

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