SQL result dense Rank in revolving group pattern

Say I have a table like

store	date	is_open
Bay	1/1/2022	true
Bay	1/2/2022	true
Bay	1/3/2022	true
Bay	1/4/2022	false
Bay	1/5/2022	false
Bay	1/6/2022	false
Bay	1/7/2022	true
Bay	1/8/2022	true
Bay	1/9/2022	true
Walmart	1/7/2022	true
Walmart	1/8/2022	false
Walmart	1/9/2022	true

I want them to use partition by and get the rank of the group such as

store	date	is_open	group
Bay	1/1/2022	true	1
Bay	1/2/2022	true	1
Bay	1/3/2022	true	1
Bay	1/4/2022	false	2
Bay	1/5/2022	false	2
Bay	1/6/2022	false	2
Bay	1/7/2022	true	3
Bay	1/8/2022	true	3
Bay	1/9/2022	true	3
Walmart	1/7/2022	true	1
Walmart	1/8/2022	false	2
Walmart	1/9/2022	true	3

I started trying partition by store and is_open but really confuse what to use in order by clause any help would be appreciated.

Answer

You can use LAG() to detect start of a group.

with cte AS (
    SELECT t.*, case when lag(is_open) OVER (PARTITION BY store ORDER BY date) = is_open then 0 else 1 end sflag
    FROM yourTable t
)
SELECT store, date, is_open, sum(sflag) over(PARTITION BY store ORDER BY date) grp
FROM cte
ORDER BY store, date;

db<>fiddle