SQL count distinct over partition by cumulatively - sql

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