How can I use the LAG function and return the same value if the subsequent value in a row is duplicated? - sql-server

Question

I am using the LAG function to move my values one row down.
However, I need to use the same value as previous if the items in source column is duplicated:
ID | SOURCE | LAG | DESIRED OUTCOME
1 | 4 | - | -
2 | 2 | 4 | 4
3 | 3 | 2 | 2
4 | 3 | 3 | 2
5 | 3 | 3 | 2
6 | 1 | 3 | 3
7 | 4 | 1 | 1
8 | 4 | 4 | 1
As you can see, for instance in ID range 3-5 the source data doesn't change and the desired outcome should be fed from the last row with different value (so in this case ID 2).

Answer 1

Sql server's version of lag supports an expression in the second argument to determine how many rows back to look. You can replace this with some sort of check to not look back e.g.
select lagged = lag(data,iif(decider < 0,0,1)) over (order by id)
from (values(0,1,'dog')
,(1,2,'horse')
,(2,-1,'donkey')
,(3,2,'chicken')
,(4,23,'cow'))f(id,decider,data)
This returns the following list
null
dog
donkey
donkey
chicken
Because the decider value on the row with id of 2 was negative.

Answer 2

Well, first lag may not be the tool for the job. This might be easier to solve with a recursive CTE. Sql and window functions work over set. That said, our goal here is to come up with a way of describing what we want. We'd like a way to partition our data so that sequential islands of the same value are part of the same set.
One way we can do that is by using lag to help us discover if the previous row was different or not.
From there, we can now having a running sum over these change events to create partitions. Once we have partitions, we can assign a row number to each element in the partition. Finally, once we have that, we can now use the row number to look
back that many elements.
;with d as (
select * from (values
(1,4)
,(2,2)
,(3,3)
,(4,3)
,(5,3)
,(6,1)
,(7,4)
,(8,4)
)f(id,source))
select *,lag(source,rn) over (order by Id)
from (
select *,rn=row_number() over (partition by partition_id order by id)
from (
select *, partition_id = sum(change) over (order by id)
from (
select *,change = iif(lag(source) over (order by id) != source,1,0)
from d
) source_with_change
) partitioned
) row_counted
As an aside, this an absolutely cruel interview question I was asked to do once.