first value in group by

Question

I want to set a column in the group by table based on the first (actually, only) value in the group.

Specifically, given a table

id   good
1    t
1    t
2    f
3    t

I want to produce the table

id    multiplicity  goodN
1     2             0
2     1             0
3     1             1

where goodN is 1 if and only if multiplicity is 1 and good is t:

select id, count(*) as multiplicity,
       if (count(*) > 1, 0, if(good = 't', 1, 0)) as goodN
from ...

The question is: how do I extract the first (in my case, only) value of good from the group?

PS. is there a cheaper way to test that the group has size 1 than count(*)=1?

Answer 1

If the count is 1, then both the MAX(good) and MIN(good) will be the "first" row in the group.

select id, count(*) as multiplicity,
       if (count(*) > 1, 0, if(max(good) = 't', 1, 0)) as goodN
from ...

Answer 2

If it were MySQL then simply:

SELECT id,
       good,
       count(*) AS multiplicity,
       if((count(*) > 1 AND good = 't'), @flag := 1, @flag := 0) AS goodN
  FROM goods
GROUP BY good, id;

PS: Let me know if you want me to delete my answer.

Answer 3

I don't have experience with hiveql, but this does what you want in mySql. I think should work the same in hiveql.

select t.id, count(*) as multiplicity, 
      (select case count(*) when 1 then 1 else 0 end
         from table1
        where id = t.id
          and good = 't') as goodN
  from table1 t
 group by t.id

sqlFiddle

Answer 4

Select good from... Where id=(select min(id) from... Where good >0)

And

Group by id having count(id)=1

You can test count(goods) instead of count(*) and at the end add group by id. To take only record with moltiplicity 1 add having count(goods)=1