Snowflake SQL count distinct with window function

Question

I have a dataset with weekly customer and store information,

question- I have to calculate features such as total unique customers in last 1,2,3,4,5,6..so on weeks as of current week.

I am getting error while using count distinct of customers column over window function -

I tried concat function to create array which also didn't work- thanks for help!

SELECT STORE,WEEK,

count(distinct Customers) over (partition by STORE order by WEEK rows between 1 preceding and 1 preceding) as last_1_week_customers,
count(distinct Customers) over (partition by STORE order by WEEK rows between 2 preceding and 2 preceding) as last_2_week_customers
    from TEST_TABLE
    group by STORE,WEEK

error- SQL compilation error: distinct cannot be used with a window frame or an order.

how can I fix this error?

Input

CREATE TABLE TEST_TABLE (STORE STRING,WEEK STRING,Customers STRING);


INSERT INTO TEST_TABLE VALUES

('A','1','AA'),
('A','1','DD'),
('A','2','AA'),
('A','2','BB'),
('A','2','CC'),
('A','3','AA'); 

Output

Answer 1

Hmm... I think you don't really need window functions here at all...

First of all, we can start off with a simple grouping:

select
    store,
    week,
    count(distinct customers) as cnt
from
    test_table
where
    week >= [this week's number minus 5]
group by
   store, week

This will result in a simple table:

store	week	cnt
A	1	2
A	2	3
A	3	1

At this point I'd ask you to consider if maybe this is already enough. It could be that you can already use the data in this format for whatever purpose you need. But if not, then we can further modify this to get a "pivoted" output.

In this query, replace ${w} with this week's number:

select
    store,
    count(distinct case when week=${w} then customers else null end) as cnt_now,
    count(distinct case when week=${w-1} then customers else null end) as cnt_minus_1,
    count(distinct case when week=${w-2} then customers else null end) as cnt_minus_2,
    count(distinct case when week=${w-3} then customers else null end) as cnt_minus_3,
    count(distinct case when week=${w-4} then customers else null end) as cnt_minus_4,
    count(distinct case when week=${w-5} then customers else null end) as cnt_minus_5
from
    test_table
where
    week >= {$w-5}
group by
   store

Remember - COUNT() and COUNT(DISTINCT) only count NON-NULL values.

store	cnt_now	cnt_minus_1	cnt_minus_2	cnt_minus_3	cnt_minus_4	cnt_minus_5
A	1	2	3	4	5	6
B	9	8	7	6	5	4