SQLAlchemy ORM 如何计算组内查询PERCENTILE_DISC的中位数（按{列}排序）？

Question

I am using SQLAlchemy 1.4 ORM with a Postgres database and can't find a way to use the ORM to calculate the median of a column.

This is the SQL statement I want to replicate:

select PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY ${column}) AS ${column}_median from  ${table};

Here is an example of the output when querying the database directly with raw SQL:

select PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY used) AS used_median from speed_limit;
 used_median 
--------------
           18
(1 row)

In the SA docs this is suggested:

func.unnest(
    func.percentile_disc([0.25,0.5,0.75,1]).within_group(user_table.c.name)

This does not meet my requirement as it not using the SA ORM.

I have tried doing something like this in Python utilizing the ORM:

session.query(func.percentile_disc(0.5).within_group(SpeedLimit.used)).scalar()

This does not give the expected result 18 , it returns 8 .

When I switch scalar for all like this:

session.query(func.percentile_disc(0.5).within_group(SpeedLimit.used)).all()

This doesn't return the correct response either, it returns [(8,)]

I have not been able to find anything in the documentation about creating this query through the ORM.

If anyone can point me in the right direction it would be appreciated: :)

Answer 1

I just realized I was using the wrong func call!

The ORM query should read

session.query(func.percentile_cont(0.5).within_group(SpeedLimit.used)).scalar()

This gives the correct response.