The Cartesian Product Trap in Django ORM

#django #python #postgres #sql

I hit a performance wall at work recently after adding a single extra Count() on a Django queryset that looked harmless. Luckily, the problematic code didn't make it into the production environment, but I had to think hard to figure out what went wrong and find a solution.

The Demo Models

All the following models and entities are made up by me to illustrate the problem.

Imagine we have a PostgreSQL database storing information about separate stores, their departments, employees and products of the stores:

The DB tables can contain much more fields, but in our case we only care about relations between the models.

So in Django, models.py would look like this:

from django.db.models import Model, CharField, ForeignKey, CASCADE

class Store(Model):
    name = CharField(max_length=200)
    # ...

class Department(Model):
    store = ForeignKey(
        Store, related_name="departments", on_delete=CASCADE
    )
    # ...

class Employee(Model):
    department = ForeignKey(
        Department, related_name="employees", on_delete=CASCADE
    )
    # ...

class Product(Model):
    store = ForeignKey(
        Store, related_name="products", on_delete=CASCADE
    )
    # ...

Seed data (roughly what bit me):

2 stores,
10 departments per store,
500 employees per department - 10 000 employees overall,
2500 products per store.

In one of the places of our system, we show a list of the stores, including the amount of employees in each store. Pretty easy with Django ORM, right?

stores = Store.objects.annotate(
    total_employees=Count("departments__employees")
).values(
    "id",
    # ...
    "total_employees"
)

This query is relatively fast and works like a charm.

The Problem

Now let's imagine we were asked to add another counter: a number of products per store. We already have total_employees, why not just add total_products?

Well, most likely we already have some unit test for our piece of code, which checks the logic on a small amount of data, and before releasing the code, we can figure out that another JOIN was added, some data is duplicated, and instead of just COUNT(...), we switch to COUNT(DISTINCT ...), eventually coming up with something like this:

stores = Store.objects.annotate(
    total_employees=Count("departments__employees",
                          distinct=True),
    total_products=Count("products", distinct=True),
).values(
    "id",
    # ...
    "total_employees",
    "total_products",
)

Looks safe to commit, push, wait for the green tests on CI, merge and deploy.

However, after the deployment you'll almost immediately see that everything hangs. As I said, there's not that many stores, only 2 for now, not that many employees, and not that many departments.

And it takes, like, 10 seconds to fetch the numbers! What's wrong with it?

Let's take a closer look at the generated SQL for this seemingly innocent queryset:

SELECT 
    "shop_store"."id",
    COUNT(DISTINCT "shop_product"."id") AS "total_products", 
    COUNT(DISTINCT "shop_employee"."id") AS "total_employees"
FROM "shop_store"
  LEFT OUTER JOIN "shop_product"
    ON "shop_store"."id" = "shop_product"."store_id"
  LEFT OUTER JOIN "shop_department"
    ON "shop_store"."id" = "shop_department"."store_id"
  LEFT OUTER JOIN "shop_employee"
    ON "shop_department"."id" = "shop_employee"."department_id" 
GROUP BY "shop_store"."id"

And let's check the actual query plan:

GroupAggregate ... rows=2 ...
  ->  Nested Loop Left Join ... rows=25000000 ...
  ...
...
Execution Time: 11376.072 ms

Translation: these three JOINs turn into a 25-million-row cartesian mess before GROUP BY and COUNT(DISTINCT):

Products × Departments × Employees
= 2500 × 10 × 500
= 12 500 000  (per store)
× 2 stores
= 25 000 000 joined rows

The Fix

There are multiple ways of handling this case, but the easiest fix is to use subqueries:

subquery_products = Subquery(
    Product.objects.filter(store_id=OuterRef("pk"))
        .values("store_id")
        .annotate(count=Count("pk"))
        .values("count"),
    output_field=IntegerField()
)
subquery_employees = Subquery(
    Employee.objects.filter(department__store_id=OuterRef("pk"))
        .values("department__store_id")
        .annotate(count=Count("pk"))
        .values("count"),
    output_field=IntegerField()
)
stores = Store.objects.annotate(
    total_products=Coalesce(subquery_products, 0),
    total_employees=Coalesce(subquery_employees, 0),
).values("id", "total_products", "total_employees")

SQL query:

SELECT "shop_store"."id",
       COALESCE((
         SELECT COUNT(U0."id") AS "count"  
         FROM "shop_product" U0  
         WHERE U0."store_id" = "shop_store"."id"  
         GROUP BY U0."store_id"), 0
       ) AS "total_products",
       COALESCE((
         SELECT COUNT(U0."id") AS "count"
         FROM "shop_employee" U0
           INNER JOIN "shop_department" U1
             ON U0."department_id" = U1."id"
         WHERE U1."store_id" = "shop_store"."id"
         GROUP BY U1."store_id"), 0
       ) AS "total_employees"
FROM "shop_store";

Now this one takes a couple of milliseconds with pretty modest and predictable plan:

Seq Scan on shop_store ... rows=2 ...
  SubPlan 1
     -> Seq Scan on shop_product u0 ... rows=2500 ...
  SubPlan 2
     -> Hash Join ... rows=5000 ...
        -> Seq Scan on shop_employee u0_1 ... rows=10000 ...
        -> Hash ... rows=10 ...
...
Execution Time: 5.600 ms

No giant intermediate data sets, just two tiny scans:

before: 11 376 ms (~11 seconds)
after: 5.6 ms (2000x faster)

Takeaways

COUNT(DISTINCT) with multi-branch LEFT JOINs makes the database loop through the entire cartesian product.
Correlated subqueries aggregate each branch separately and scale linearly with data size.
Always test aggregate queries against production-sized data before you ship.

Top comments (2)

david duymelinck • Jul 10

I don't know how to do it django ORM, but another way to write the query is with UNION.

SELECT 
    store_id,
    SUM(total_products) AS total_products,
    SUM(total_employees) AS total_employees
FROM (
    SELECT 
        store_id,
        COUNT(DISTINCT id) AS total_products,
        0 AS total_employees
    FROM shop_product
    GROUP BY store_id

    UNION ALL

    SELECT 
        d.store_id,
        0 AS total_products,
        COUNT(DISTINCT e.id) AS total_employees
    FROM shop_department d
    LEFT JOIN shop_employee e ON d.id = e.department_id
    GROUP BY d.store_id
) counts
GROUP BY store_id;

I find aggregate queries so awkward.

Dmitrii Doroshev • Jul 13 • Edited

@xwero I don't think it's possible to compose a Django queryset with UNION in FROM, but I see that your query is way more efficient, especially when there's more than just 2 stores.

I created 32 stores, 160000 employees and 80000 products, and checked the plans. Here's the plan of your query with FROM ( ... UNION ALL ... ):