Delete rows in table that are sum of other rows per group - sql

Question

Group rows by T, and in each group find the row that is the largest or smallest (if values are negative) sum of other rows from that group, and delete that row (one for each group), if group does not have enough elements to find sum or enough but none of the rows indicates sum of others nothing happens
CREATE TABLE Test (
T varchar(10),
V int
);
INSERT INTO Test
VALUES ('A', 4),
('B', -5),
('C', 5),
('A', 2),
('B', -1),
('C', 10),
('A', 2),
('B', -4),
('C', 5),
('D', 0);
expected result:
A 2
A 2
B -1
B -4
C 5
C 5
D 0

Answer 1

Like the comments, the requirements seem strange. The below code assumes that the summing is already pre-populated and merely removes the largest/smallest as long as the highest value is not 0.
if object_id('tempdb..#test') is not null
drop table #test
CREATE TABLE #Test (
T varchar(10),
V int
);
INSERT INTO #Test
VALUES ('A', 4), ('B', -5), ('C', 5), ('A', 2), ('B', -1), ('C', 10), ('A', 2), ('B', -4), ('C', 5), ('D', 0);
if object_id('tempdb..#test2') is not null
drop table #test2
SELECT
T,
V,
ABS(V) as absV
INTO #TEST2
FROM #TEST
SELECT * FROM #TEST2
if object_id('tempdb..#max') is not null
drop table #max
SELECT
T,
MAX(absV) AS MaxAbsV
INTO #Max
FROM #TEST2
GROUP BY T
HAVING MAX(AbsV) != 0
DELETE #TEST2
FROM #TEST2
INNER JOIN #MAX ON #TEST2.T = #MAX.T AND #TEST2.absV = #Max.MaxAbsV
SELECT * FROM #TEST2
ORDER BY T ASC

Answer 2

; with cte as
(
select T, V,
R = row_number() over (partition by T order by ABS(V) desc),
C = count(*) over (partition by T)
from Test
)
delete c
from cte c
inner join
(
select T, S = sum(V)
from cte
where R <> 1
group by T
) s on c.T = s.T
where c.C >= 3
and c.R = 1
and c.V = s.S

Answer 3

Using ABS and NOT Exists
DECLARE #Test TABLE (
T varchar(10),
V int
);
INSERT INTO #Test
VALUES ('A', 4), ('B', -5), ('C', 5), ('A', 2), ('B', -1), ('C', 10), ('A', 2), ('B', -4), ('C', 5), ('D', 0);
;WITH CTE as (
select T,max(ABS(v ))v from #Test
WHERE V <> 0
GROUP BY T )
SELECT T,V FROM #Test T where NOT exists (Select 1 FROM cte WHERE T = T.T AND v = ABS(T.V) )
ORDER BY T.T

Answer 4

Determine first if the rows are positive or negative by checking if SUM(V) is positive. And then determine if the smallest or largest value is equal to the SUM of the other rows, by subtracting from SUM(V) the MIN(V) if negative or MAX(V) if positive:
DELETE t
FROM Test t
INNER JOIN (
SELECT
T,
SUM(V) - CASE WHEN SUM(V) >= 0 THEN MAX(V) ELSE MIN(V) END AS ToDelete
FROM Test
GROUP BY T
HAVING COUNT(*) >= 3
) a
ON a.T = t.T
AND a.ToDelete = t.V
ONLINE DEMO

Answer 5

You can use the below query to get the required output :-
select * into #t1 from test
select * from
(
select TT.T as T,TT.V as V
from test TT
JOIN
(select T,max(abs(V)) as V from #t1
group by T) P
on TT.T=P.T
where abs(TT.V) <> P.V
UNION ALL
select A.T as T,A.V as V from test A
JOIN(
select T,count(T) as Tcount from test
group by T
having count(T)=1) B on A.T=B.T
) X order by T
drop table #t1

Answer 6

You are looking for a value per group that is the sum of all the group's other values. E.g. 4 of (2,2,4) or -5 of (-5,-4,-1).
This is usually only one record per group. But it can be multiple times the same number. Here are examples for ties: (0,0) or (-2,2,4,4), or (-2,-2,4,4,4) or (-10,3,3,3,3,4).
As you see, you are looking in any way for values that equal half of the group's total sum. (Of course. We are looking for n+n, where one n is in one record and the other n is the sum of all the other records.)
The only special case is when there is only one value in the group which is zero. That we don't want to delete of course.
Here is an update statement that cannot deal with ties, but would delete all maximum values instead of just one:
delete from test
where 2 * v =
(
select case when count(*) = 1 then null else sum(v) end
from test fullgroup
where fullgroup.t = test.t
);
In order to deal with ties you would need artificial row numbers, so as to delete only one record of all candidates.
with candidates as
(
select t, v, row_number() over (partition by t order by t) as rn
from
(
select
t, v,
sum(v) over (partition by t) as sumv,
count(*) over (partition by t) as cnt
from test
) comparables
where sumv = 2 * v and cnt > 1
)
delete
from candidates
where rn = 1;
SQL fiddle: http://sqlfiddle.com/#!6/6d97e/1

Answer 7

See if below query helps:
DELETE [Audit].[dbo].[Test] FROM [Audit].[dbo].[Test] as AA
INNER JOIN (select T,
CASE
WHEN MAX(V) < 0 THEN MIN(V)
WHEN MIN(V) > 0 THEN MAX(V) ELSE MAX(V)
END as MAX_V,
CASE
WHEN SUM(V) > 0 THEN SUM(V) - MAX(V)
WHEN SUM(V) < 0 THEN SUM(V) - MIN(V) ELSE SUM(V)
END as SUM_V_REST
from [Audit].[dbo].[Test]
Group by T
Having Count(V) > 1) as BB ON AA.T = BB.T and AA.V = BB.MAX_V