Get the difference between top and last value in result set - sql-server

Question

My table looks something like
deal_id test_value run_date
820117648 1.2 2014-03-31
820117648 1.33 2014-04-30
820117648 1.33 2014-05-30
820117648 1.26 2014-06-30
820117648 1.11 2014-07-31
820117648 0.58 2014-09-30
820117648 1.64 2014-10-31
820117648 0.64 2014-11-28
820117648 3.65 2014-12-31
820117648 3.8 2015-03-11
820117649 0.64 2014-09-31
820117649 0.23 2014-10-31
820117649 0.64 2014-11-28
820117649 3.65 2014-12-31
820117649 3.8 2015-03-11
SELECT deal_id,test_value,run_date FROM ems.cdotests
WHERE run_date >= Dateadd(month, -4 Getdate())
I am trying to grab the last 4 run_date record from the whole set and then
I need to find the difference between the testValue on the first rundate of my select criteria and the last rundate i.e for deal_id 820117649 3.8 - 0.23 = 3.57 and for deal_id 820117648 it should be 3.8 -1.64 = 2.16
This is the large table with multiple deal id and several run_date associated with each dealid and may have data for the last 15 years or so
Any suggestion would be really helpfull

Answer 1

Try this solution. It will not be bound to particular deal_id.
DECLARE #t TABLE ( ID INT, V MONEY, D DATE )
INSERT INTO #t
VALUES ( 1, 1, '20140101' ),
( 1, 4, '20140102' ),
( 1, 2, '20140103' ),
( 1, 7, '20140104' ),
( 2, 5, '20140104' ),
( 2, 8, '20140110' ),
( 2, 11, '20140105' );
WITH cte
AS ( SELECT ID ,
D ,
V ,
MIN(D) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) AS Min ,
MAX(D) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) AS Max
FROM #t
)
SELECT c1.ID ,
c2.V - c1.V AS V
FROM cte c1
JOIN cte c2 ON c1.ID = c2.ID AND c1.D < c2.D
WHERE ( c1.D = c1.MIN OR c1.D = c1.MAX ) AND ( c2.D = c2.MIN OR c2.D = c2.MAX )
Output:
ID V
1 6.00
2 3.00
You have changed you question. For latest 6 month, add filter CTE above main CTE:
;WITH filter
AS ( SELECT ID ,
D ,
V ,
ROW_NUMBER() OVER ( PARTITION BY ID ORDER BY D DESC ) AS RN
FROM #t
),
cte
AS ( SELECT ID ,
D ,
V ,
MIN(D) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) AS Min ,
MAX(D) OVER ( PARTITION BY ID ORDER BY D ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) AS Max
FROM filter
WHERE RN <= 6
)
SELECT c1.ID ,
c2.V - c1.V AS V
FROM cte c1
JOIN cte c2 ON c1.ID = c2.ID AND c1.D < c2.D
WHERE ( c1.D = c1.MIN OR c1.D = c1.MAX )
AND ( c2.D = c2.MIN OR c2.D = c2.MAX )
AND (c2.V - c1.V > 5)

Answer 2

You can use window functions
WITH C AS(
SELECT ROW_NUMBER() OVER (PARTITION BY deal_Id ORDER BY deal_Id) Rn
,deal_Id
,test_value
,run_date
FROM ems.cdotests
)
SELECT id1 deal_id, t2 - t1 diff
FROM (SELECT c1.deal_id id1
,c1.test_value t1
,c2.deal_id id2
, c2.test_value t2
FROM C c1
INNER JOIN C c2 ON c1.deal_id = c2.deal_id
WHERE c1.Rn = 1
AND c2.Rn = (SELECT TOP 1 MAX(Rn) FROM C GROUP BY deal_id)
) t

Answer 3

So late to answer, but a way is this:
;with t as (
select *,
row_number() over (partition by deal_id order by run_date desc) seq
from ems.cdotests
)
select
t1.deal_id, t1.run_date run_date1, t1.test_value test_value1,
t2.run_date run_date4, t2.test_value test_value4,
t1.test_value - t2.test_value diff
from t t1
join t t2
on t1.deal_id = t2.deal_id
where t1.seq = 1
and t2.seq = 4
-- as an extra condition of those should have at least 5 records:
and exists (select 1 from t ti where ti.deal_id = t1.deal_id and ti.seq > 5)
[SQL Fiddle]