SQL count distinct over partition by cumulatively

Question

I am using AWS Athena (Presto based) and I have this table named base :

id	category	year	month
1	a	2021	6
1	b	2022	8
1	a	2022	11
2	a	2022	1
2	a	2022	4
2	b	2022	6

I would like to craft a query that counts the distinct values of the categories per id, cumulatively per month and year, but retaining the original columns:

id	category	year	month	sumC
1	a	2021	6	1
1	b	2022	8	2
1	a	2022	11	2
2	a	2022	1	1
2	a	2022	4	1
2	b	2022	6	2

I've tried doing the following query with no success:

SELECT id, 
       category, 
       year, 
       month, 
       COUNT(category) OVER (PARTITION BY id, ORDER BY year, month) AS sumC FROM base;

This results in 1, 2, 3, 1, 2, 3 which is not what I'm looking for. I'd rather need something like a COUNT(DISTINCT) inside a window function, though it's not supported as a construct.

I also tried the DENSE_RANK trick:

  DENSE_RANK() OVER (PARTITION BY id ORDER BY category) 
+ DENSE_RANK() OVER (PARTITION BY id ORDER BY category) 
- 1 as sumC

Though, because there is no ordering between year and month , it just results in 2, 2, 2, 2, 2, 2 .

Any help is appreciated!

Answer 1

One option is

• creating a new column that will contain when each " category " is seen for the first time (partitioning on " id ", " category " and ordering on " year ", " month ")
• computing a running sum over this column, with the same partition

WITH cte AS (
    SELECT *, 
           CASE WHEN ROW_NUMBER() OVER(
                         PARTITION BY id, category
                         ORDER     BY year, month) = 1
                THEN 1 
                ELSE 0 
           END AS rn1
    FROM base
    ORDER BY id, 
             year_, 
             month_
)
SELECT id,
       category,
       year_,
       month_,
       SUM(rn1) OVER(
            PARTITION BY id
            ORDER     BY year, month 
       ) AS sumC
FROM cte

Does it work for you?