SQL选择最大值（计数（A列、B列））

我一直在为这件事伤脑筋。我想我很接近了。（Oracle，SQL）

我有一张像下面这样的桌子

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

每个公司可以解析为多个代码。 我想做的是创建一个SQL语句，对于每个公司，它将为我提供出现频率最高的代码。所以在我的例子中，我得到

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

---

mj

您可以通过以下方式完成：

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

分解如下：

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

这让我们了解每个公司的代码计数：

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

因此，我们希望最受欢迎。我们通过一种分析方法来实现这一点：

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)

给予：

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

给予

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

因此，我们需要说明其中RN=1和c=1（即只有一行具有该数量的REC）。因此，我们最终得出：

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，c>1检查在顶部（因为我们不想筛选出行，只需将它们标记为不明确即可。

您可以使用以下方法：

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

分解如下：

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

这让我们了解每个公司的代码计数：

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

因此，我们希望最受欢迎。我们通过分析：

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)

给予：

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

给予

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

因此，我们需要说明其中RN=1和c=1（即只有一行具有该数量的REC）。因此，我们最终得出：

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，c>1检查在顶部（因为我们不想筛选出行，只需将它们标记为不明确的行。

试试看

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) 

是小提琴

试试看

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) 

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
)

---

是fiddle

Oracle数据库11g企业版11.2.0.1.0-64位产品Oracle数据库11g企业版11.2.0.1.0-64位产品

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
)