SQL, rank for each instance of a partition - sql

Question

I am trying to to create a rank for each instance of a status occurring, for example
ID
Status
From_date
To_date
rank
1
Available
2022-01-01
2022-01-02
1
1
Available
2022-01-02
2022-01-03
1
1
Unavailable
2022-01-03
2022-01-10
2
1
Available
2022-01-10
2022-01-20
3
For each ID, for each instance of a status occurring, by from_date ascending.
I want to do this as i see this as the best way of getting to the final result i want which is
ID
Status
From_date
To_date
rank
1
Available
2022-01-01
2022-01-03
1
1
Unavailable
2022-01-03
2022-01-10
2
1
Available
2022-01-10
2022-01-20
3
I tried dense_rank(partition by id order by status, from_date but can see now why that wouldnt work. Not sure how to get to this result.

Answer 1

So with this CTE for the data:
with data(ID, Status, From_date, To_date) as (
select * from values
(1, 'Available', '2022-01-01', '2022-01-02'),
(1, 'Available', '2022-01-02', '2022-01-03'),
(1, 'Unavailable', '2022-01-03', '2022-01-10'),
(1, 'Available', '2022-01-10', '2022-01-20')
)
the first result, being rank can be done with CONDITIONAL_CHANGE_EVENT:
select *
,CONDITIONAL_CHANGE_EVENT( Status ) OVER ( PARTITION BY ID ORDER BY From_date ) as rank
from data;
ID
STATUS
FROM_DATE
TO_DATE
RANK
1
Available
2022-01-01
2022-01-02
0
1
Available
2022-01-02
2022-01-03
0
1
Unavailable
2022-01-03
2022-01-10
1
1
Available
2022-01-10
2022-01-20
2
and thus the keeps the first of each rank can be achieved with a QUALIFY/ROW_NUMBER, because the CONDITIONAL_CHANGE is a complex operation, needs wrapping in a sub-select, so the answer is not as short as I would like:
select * from (
select *
,CONDITIONAL_CHANGE_EVENT( Status ) OVER ( PARTITION BY ID ORDER BY From_date ) as rank
from data
)
qualify row_number() over(partition by id, rank ORDER BY From_date ) = 1
gives:
ID
STATUS
FROM_DATE
TO_DATE
RANK
1
Available
2022-01-01
2022-01-02
0
1
Unavailable
2022-01-03
2022-01-10
1
1
Available
2022-01-10
2022-01-20
2
Also, the final result minus the ranking can be done with:
select *
from data
qualify nvl(Status <> lag(status) over ( PARTITION BY ID ORDER BY From_date ), true)
ID
STATUS
FROM_DATE
TO_DATE
1
Available
2022-01-01
2022-01-02
1
Unavailable
2022-01-03
2022-01-10
1
Available
2022-01-10
2022-01-20
and thus a rank can be added at the end
select *
,rank() over ( PARTITION BY ID ORDER BY From_date ) as rank
from (
select *
from data
qualify nvl(Status <> lag(status) over ( PARTITION BY ID ORDER BY From_date ), true)
)
ID
STATUS
FROM_DATE
TO_DATE
RANK
1
Available
2022-01-01
2022-01-02
1
1
Unavailable
2022-01-03
2022-01-10
2
1
Available
2022-01-10
2022-01-20
3

Answer 2

This is a typical gaps-and-island problem, where islands are groups of consecutive rows that have the same status.
Here is one way to solve it with window functions:
select id, status,
min(from_date) from_date, max(to_date) to_date,
row_number() over (partition by id order by min(from_date)) rn
from (
select t.*,
row_number() over (partition by id order by from_date) rn1,
row_number() over (partition by id, status order by from_date) rn2
from mytable t
) t
group by id, status, rn1 - rn2
order by min(from_date)
This worked by ranking rows within two different partitions (with a without the status) ; the difference between the row numbers define the islands.

Answer 3

You can group consecutive status using conditional_change_event, then collapse the dates using min and max, and finally use row_number() to rank the events
with cte as
(select *,conditional_change_event(status) over (partition by id order by from_date) as rn
from t)
select id,
status,
min(from_date) as from_date,
max(to_date) as to_date,
row_number() over (partition by id, order by min(from_date), max(to_date)) as rank
from cte
group by id, status, rn
order by rank