SQL选择最大值（计数（列A，列B））

Question

I've been banging my head on this one. I think I'm close. (Oracle, SQL)

I've got a table that looks like the following.

Company    Code
Apple      A
Google     A
Microsoft  B
Apple      C
Google     B
Microsoft  B
Apple      C
Google     C
Microsoft  B

Each company can resolve to multiple codes. What I want to do is create a SQL statement that for each company will give me the company with code that appears the most frequently. So in my example I'd get

Apple      C
Google     <nothing since there's no clear max>
Microsoft  B

What I've put together so far is the following. This query returns to me the company with the code that appears the most, however if I have a tie between two codes for a company, I get both back. For Google in my example I'd get (Google, A), (Google, B), (Google, C). I want nothing returned.

I believe that I can join the whole thing again with itself and some additional where clauses to filter out the duplicate companies, however I was wondering if there is a better way to do this. What's killing me is the aggregation functions along with the group by since sometimes I get an Oracle single-group group error. Any suggestions are appreciated.

SELECT m1, c.company sp1, b.code ul1 FROM
  (SELECT MAX(c1) m1, company FROM
    (SELECT COUNT(company||code) c1, company, code FROM table GROUP BY company, code ) a
  GROUP BY company) c
left OUTER JOIN
  (SELECT COUNT(company||code) c1, company, code FROM table GROUP BY company, code ) b
ON c.company=b.company and
m1=b.c1;

mj

Answer 1

you can do it with this:

select company, case when c > 1 then null else code1 end code1
  from (select company, code1, recs, count(*) over (partition by company, recs ) c, 
           row_number() over (partition by company order by recs desc) rn
  from (select company, code1, count(*) recs
          from table
         group by company, code1))
 where rn = 1
 order by 1

breaking this down:

select company, code1, count(*) recs
 from table
 group by company, code1

this gets us each company with their code count:

COMPANY   C       RECS
--------- - ----------
Google    A          1
Google    B          1
Microsoft B          3
Apple     A          1
Apple     C          2
Google    C          1

from this we want the most popular. we do this with an analyic:

select company, code1, recs,
       row_number() over (partition by company order by recs desc) rn
  from (select company, code1, count(*) recs
          from t1
         group by company, code1)

giving:

COMPANY   C       RECS         RN
--------- - ---------- ----------
Apple     C          2          1 <- we want all rn= "1" rows
Apple     A          1          2
Google    A          1          1<- we want all rn= "1" rows
Google    B          1          2
Google    C          1          3
Microsoft B          3          1<- we want all rn= "1" rows

but now if theres duplicates (as google has)..we count(*) the rows that have RN=1.

select company, code1, recs,
       row_number() over (partition by company order by recs desc) rn,
       count(*) over (partition by company, recs ) c
  from (select company, code1, count(*) recs
          from t1
         group by company, code1)

giving

COMPANY   C       RECS         RN          C
--------- - ---------- ---------- ----------
Apple     C          2          1          1
Apple     A          1          2          1
Google    A          1          1          3
Google    B          1          2          3
Google    C          1          3          3
Microsoft B          3          1          1

so we need to say where RN=1 and also c = 1 (ie only ONE row had that number of recs. so we end up with:

select company, case when c > 1 then null else Code1 end Code1
  from (select company, code1, recs, count(*) over (partition by company, recs ) c, 
           row_number() over (partition by company order by recs desc) rn
  from (select company, code1, count(*) recs
          from t1
         group by company, code1))
 where rn = 1

ie rn = 1 and the c > 1 check is in the case at the top (as we don't want to filter rows out, just mark them as ambiguous.

Answer 2

Try

with
tcount as
(
select t.company, t.code, count(*) c
from table1 t 
group by t.company, t.code) 

select distinct tt.company,
decode(count(tt.company) over(partition by tt.company),
1,
tt.code,
null)
from tcount tt
where tt.c =
(select max(c) from tcount tti where tt.company = tti.company) 

Here is a fiddle

Answer 3

SELECT company,CASE WHEN count_ > 1 THEN NULL ELSE code END
(
SELECT company,MAX(code) code,count(1) count_
(SELECT company,code,rank() OVER(PARTITION BY company ORDER BY count_ DESC) rn FROM
(
select
       company,code,count(1) count_
from   table
group by
       company,code
) 
)
where rn = 1
GROUP BY 
   company
)