How to identify positive minimum or negative maximum in a column for a key? - sql

Question

I have the following columns - Person_ID Days. For one person id, multiple days are possible. Something like this:
Person_Id Days
1000 100
1000 200
1000 -50
1000 -10
1001 100
1001 200
1001 50
1001 10
1002 -50
1002 -10
I need to address the following scenarios:
If all values for days column are positive, I need minimum of the days for a person_id. If the days column has both positive and negative, I need minimum of positive. If all negatives, I need maximum of negative.
The output like:
Person_id Days
1000 100
1001 10
1002 -10
I tried using case statement, but I am unable to use a same column in the condition as well as grouping.

Answer 1

Try this (Postgres 9.4+):
select person_id, coalesce(min(days) filter (where days > 0), max(days))
from a_table
group by 1
order by 1;

Answer 2

Oracle Setup:
CREATE TABLE table_name ( Person_Id, Days ) AS
SELECT 1000, 100 FROM DUAL UNION ALL
SELECT 1000, 200 FROM DUAL UNION ALL
SELECT 1000, -50 FROM DUAL UNION ALL
SELECT 1000, -10 FROM DUAL UNION ALL
SELECT 1001, 100 FROM DUAL UNION ALL
SELECT 1001, 200 FROM DUAL UNION ALL
SELECT 1001, 50 FROM DUAL UNION ALL
SELECT 1001, 10 FROM DUAL UNION ALL
SELECT 1002, -50 FROM DUAL UNION ALL
SELECT 1002, -10 FROM DUAL;
Query:
SELECT person_id, days
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY person_id
ORDER BY SIGN( ABS( days ) ),
SIGN( DAYS ) DESC,
ABS( DAYS )
) AS rn
FROM table_name t
)
WHERE rn = 1;
Output:
PERSON_ID DAYS
---------- ----------
1000 100
1001 10
1002 -10

Answer 3

Oracle solution:
with
input_data ( person_id, days) as (
select 1000, 100 from dual union all
select 1000, 200 from dual union all
select 1000, -50 from dual union all
select 1000, -10 from dual union all
select 1001, 100 from dual union all
select 1001, 200 from dual union all
select 1001, 50 from dual union all
select 1001, 10 from dual union all
select 1002, -50 from dual union all
select 1002, -10 from dual
)
select person_id,
NVL(min(case when days > 0 then days end), max(days)) as days
from input_data
group by person_id;
PERSON_ID DAYS
---------- ----------
1000 100
1001 10
1002 -10
For each person_id, if there is at least one days value that is strictly positive, then the min will be taken over positive days only and will be returned by NVL(). Otherwise the min() will return null, and NVL() will return max() over all days (all of which are, in this case, negative or 0).

Answer 4

select Person_id, min(abs(days)) * days/abs(days) from table_name
group by Person_id
-- + handle zero_divide .. SORRY.. the above works only in MySQL .
Something like this will work anywhere which is equivalent of above query:
select t.Person_id , min(t.days) from table_name t,
(select Person_id, min(abs(days)) as days from table_name group by Person_id) v
where t.Person_id = v.Person_id
and abs(t days) = v.days
group by Person_id;
OR
select id, min(Days) from (
select Person_id, min(abs(Days)) as Days from temp group by Person_id
union
select Person_id, max(Days) as Days from temp group by Person_id
) temp
group by Person_id;

Answer 5

You can do this by using GroupBy clause in sql server. Take a look into below query:-
CREATE TABLE #test(Person_Id INT, [Days] INT)
DECLARE #LargestNumberFromTable INT;
INSERT INTO #test
SELECT 1000 , 100 UNION
SELECT 1000 , 200 UNION
SELECT 1000 , -50 UNION
SELECT 1000 , -10 UNION
SELECT 1001 , 100 UNION
SELECT 1001 , 200 UNION
SELECT 1001 , 50 UNION
SELECT 1001 , 10 UNION
SELECT 1002 , -50 UNION
SELECT 1002 , -10
SELECT #LargestNumberFromTable = ISNULL(MAX([Days]), 0)
FROM #test
SELECT Person_Id
,CASE WHEN SUM(IIF([Days] > 0,[Days] , 0)) = 0 THEN MAX([Days]) -- All Negative
WHEN SUM([Days]) = SUM(IIF([Days] > 0, [Days], 0)) THEN MIN ([Days]) -- ALL Positive
WHEN SUM([Days]) <> SUM(IIF([Days] > 0, [Days], 0)) THEN MIN(IIF([Days] > 0, [Days], #LargestNumberFromTable)) --Mix (Negative And positive)
END AS [Days]
FROM #test
GROUP BY Person_Id
DROP TABLE #test