Question

I’ve got the following problen: I know SQL and I don’t know how to work with SQLAlchemy but I need to change it in 1 place in the project that I’ve inherited.
So, I’ve got this:

ModelCategories = request.sa.query(
    Model.category_id
    , Category.name
    , Category.alias).distinct().join(Category).order_by(Category.alias
    , Category.name )

And it generates a rather slow request:

SELECT DISTINCT 
  model.category_id AS model_category_id
  , category.name AS category_name
  , category.alias AS category_alias 
FROM model 
JOIN category ON category.id = model.category_id 
ORDER BY category.alias, category.name

And I need to change it with this:

SELECT 
  model.category_id AS model_category_id
  , category.name AS category_name
  , category.alias AS category_alias 
FROM ( SELECT DISTINCT model_category_id ) as model 
JOIN category ON category.id = model.category_id 
ORDER BY category.alias, category.name

But in terms of SQLAlchemy as is the first request.

Answer 1

First of all check the SQL execution plan. If you have an index on the model.category_id column, the query should not really be slow.
Otherwise, following options are available:

Option-1: almost your current solution

# python
ModelCategories = session.query(Category).distinct().join(Model).order_by(Category.alias, Category.name)

# SQL
SELECT DISTINCT category.id AS category_id, category.name AS category_name, category.alias AS category_alias 
FROM category 
JOIN model ON category.id = model.category_id 
ORDER BY category.alias, category.name

This is like your current solution but somewhat cleaner in my view. I assume the performance issue might come from the fact that all table Model is used in the query, and this is also why you need to use distinct.

Option-2: use any() on relationship

# python (assumption: model mapping has relationship defined between Category and Model
mapper(Category, category_table, properties={
    'models': relationship(Model, backref="category") })

# python
ModelCategories = session.query(Category).filter(Category.models.any()).order_by(Category.alias, Category.name)

# SQL
SELECT category.id AS category_id, category.name AS category_name, category.alias AS category_alias 
FROM category 
WHERE EXISTS (SELECT 1 
    FROM model 
    WHERE category.id = model.category_id)
ORDER BY category.alias, category.name

This should boost your performance already. I prefer this to following option-3 as it is again more clean code

Option-3: use subquery

# python
q = select([Model.category_id]).distinct().alias("subq")
ModelCategories = session.query(Category).join(q, Category.id==q.c.category_id)

# SQL
SELECT category.id AS category_id, category.name AS category_name, category.alias AS category_alias 
FROM category
JOIN (SELECT DISTINCT model.category_id AS category_id FROM model) AS subq 
  ON category.id = subq.category_id
ORDER BY category.alias, category.name

This should give you exactly the SQL you asked for. As mentioned, I personally prefer version-2.