Product of distinct values before a certain date - sql

Question

I have a table with schema:
date | item_id | factor
----------------------
20180710 | 1 | 0.1
20180711 | 1 | 0.1
20180712 | 1 | 2
20180713 | 1 | 2
20180714 | 1 | 2
20180710 | 2 | 0.1
20180711 | 2 | 0.1
20180712 | 2 | 5
20180713 | 2 | 5
20180714 | 2 | 10
The factor for each item_id can change on any date. On each date, I need to calculate the product of all the distinct factors for each item_id up to that date (inclusive), so the final output for the above table should be:
date | id | cumulative_factor
20180710 | 1 | 0.1
20180711 | 1 | 0.1
20180712 | 1 | 0.2
20180713 | 1 | 0.2
20180714 | 1 | 0.2
20180710 | 2 | 0.1
20180711 | 2 | 0.1
20180712 | 2 | 0.5
20180713 | 2 | 0.5
20180714 | 2 | 5
Logic:
On 20180711, for id=1, the distinct factors is 0.1 only, so the cumulative factor is 0.1.
On 20180714, for id=1, the distinct factors are 0.1 and 2, so the cumulative factor is 0.1*2 = 0.2.
On 20180714, for id=2, the distinct factors are 0.1, 5 and 10, so the cumulative factor is 0.1*5*10 = 5.
I've tried
select a.id, a.date, b.cum_factor
from factor_table a
left join (
select id, date, ISNULL(EXP(SUM(distinct log_factor)),1) as cum_factor
from factor_table
where date < a.date
) b
on a.id=b.id and a.date = b.date
but I get the error
a.date not found

Answer 1

there isn't a product aggregate function in SQL Server.
However, you can emulate it with EXP ( SUM ( LAG ( value ) ) )
please refer to in-line query for the comments
; with
cte as
(
-- this cte set the factor to 1 if it is same as previous row
-- as you wanted `product of distinct`
select *,
factor2 = CASE WHEN LAG(factor) OVER (PARTITION BY id
ORDER BY [date]) IS NULL
OR LAG(factor) OVER (PARTITION BY id
ORDER BY [date]) <> factor
THEN factor
ELSE 1
END
from factor_table
),
cte2 as
(
-- this cte peform SUM( LOG ( factor ) ) only. except EXP()
select *, factor3 = SUM(LOG(factor2)) OVER (PARTITION BY id
ORDER BY [date])
from cte
)
-- EXP() is not a window function, so it has to do it in separately in another level
select *, EXP(factor3) as cumulative_factor
from cte2
Note : LAG() required SQL Server 2012 or later

Answer 2

Something seems wrong with multiplying distinct factors. You can pretty easily express this using window functions:
select f.id, f.date, f.cum_factor
exp(sum(distinct log(log_factor) over (partition by f.id order by f.date)))
from factor_table f;
To get around the limitation on distinct:
select f.id, f.date, f.cum_factor
exp(sum(log(case when seqnum = 1 then log_factor end) over (partition by f.id order by f.date)))
from (select t.*,
row_number() over (partition by id, log_factor order by date) as seqnum
from factor_table f
) f;