SQL get rank of ordered data - sql

Question

I have a data set that looks like this (ordered by date):
date
value
first_id
second_id
2020-01-01
10
1
1
2020-01-02
15
1
1
2020-01-03
5
1
2
2020-01-04
75
2
2
2020-01-05
101
2
2
2020-01-06
12
1
1
2020-01-07
5
1
1
2020-01-08
14
1
2
I need to get an aggregation when values are the same for the same first_id and second_id in a sequence, lets say max(value), so I can get:
max_value
first_id
second_id
15
1
1
5
1
2
101
2
2
12
1
1
14
1
2
If you do max(value) and group by, same first_id and second_id combinations will give just one row (regardless of date ordering).
I was thinking to add RANK when one of ids changes, e.g:
date
value
first_id
second_id
rank
2020-01-01
10
1
1
1
2020-01-02
15
1
1
1
2020-01-03
5
1
2
2
2020-01-04
75
2
2
3
2020-01-05
101
2
2
3
2020-01-06
12
1
1
4
2020-01-07
5
1
1
4
2020-01-08
14
1
2
5
But I don't know how to get that rank as well since same id combinations are considered together.

Answer 1

You can use lag() and a cumulative sum to define the groups and then aggregate. You can see the groups if you run this query:
select t.*,
sum(case when prev_date = prev_date2 then 0 else 1 end) over (order by date) as grp
from (select t.*,
lag(date) over (order by date) as prev_date,
lag(date) over (partition by first_id, second_id order by date) as prev_date2
from t
) t;
The logic is saying that a new group starts when the previous date does not have the same values of the two id columns.
Then the aggregation is:
with grps as (
select t.*,
sum(case when prev_date = prev_date2 then 0 else 1 end) over (order by date) as grp
from (select t.*,
lag(date) over (order by date) as prev_date,
lag(date) over (partition by first_id, second_id order by date) as prev_date2
from t
) t
)
select first_id, second_id, max(value), min(date), max(date)
from grps
group by grp