为每对 Col1 和 Col2 找到最常出现的 Col3

Question

给定一个有 4 列的 table myTable，假设 Col1、Col2、Col3 和 Col4:

A X 5 B
A Y 5 C
A X 7 D
A Y 3 E 
A X 7 F

我需要为每对 (col1, col2).

找到出现次数最多的 col3

所以这个例子的结果将是：

A X 7   D/F  -- D or F
A Y 5/3 C/E  -- It can be 5 and C or 3 and E

所以我写了一个类似这样的查询：

select Col1,Col2,Col3 
from myTable M 
group by Col1,Col2,Col3 
having Col3 =
     (select Col3 
      from myTable N 
      where M.Col1=N.col1 
      group by Col3 
      order by Col3 desc limit 1);

但是查询没有给出想要的结果。
此外，我不知道如何获得 Col4 作为 group by 子句，我不想根据 Col4.

进行分组

对于每个 (Col1, Col2) 对，我想要单个 Col4 与出现的最大值 Col3。

Answer 1

一种方法是在聚合查询之上使用 row_number() window 函数：

SELECT col1, col2, col3
FROM   (SELECT col1, col2, col3, 
        ROW_NUMBER () OVER (PARTITION BY col1, col2 ORDER BY cnt DESC) AS rn
        FROM (SELECT   col1, col2, col3, COUNT(*) AS cnt
              FROM     mytable
              GROUP BY col1, col2, col3) t
       ) q
WHERE  rn = 1

Answer 2

你只需要一个带有 DISTINCT ON:

的子查询

SELECT DISTINCT ON (col1, col2)
       col1, col2, col3, min(col4) As col4
FROM   tbl
GROUP  BY col1, col2, col3
ORDER  BY col1, col2, count(*) DESC, col3;

通过这种方式，每个 (col1, col2) 得到一个单行行 最常见 col3（“最常见”的多个并列的最小值）和最小 col4 与 col3 一致。

类似地，要使 all 符合条件 col3，您可以在子查询中使用 window function rank()，该子查询也会在 after聚合：

SELECT col1, col2, col3, col4_list
FROM  (
   SELECT col1, col2, col3, count(*) AS ct, string_agg(col4, '/') AS col4_list
        , rank() OVER (PARTITION BY col1, col2 ORDER BY count(*) DESC) AS rnk
   FROM   tbl
   GROUP  BY col1, col2, col3
   ) sub
WHERE  rnk = 1
ORDER  BY col1, col2, col3;

这行得通，因为您可以运行 window 函数 over 聚合函数。
如果数据类型不是 character type.

，则转换为 text

或者，列表中每个 (col1, col2) 的所有符合条件 col3，加上第二个列表中所有匹配的 col4：

SELECT col1, col2
     , string_agg(col3::text, '/') AS col3_list  -- cast if necessary
     , string_agg(col4_list,  '/') AS col4_list
FROM  (
   SELECT col1, col2, col3, count(*) AS ct, string_agg(col4, '/') AS col4_list
        , rank() OVER (PARTITION BY col1, col2 ORDER BY count(*) DESC) AS rnk
   FROM   tbl
   GROUP  BY col1, col2, col3
   ) sub
WHERE  rnk = 1
GROUP  BY col1, col2
ORDER  BY col1, col2, col3_list;

更多解释的相关答案：

Select first row in each GROUP BY group?
Best way to get result count before LIMIT was applied
Get the distinct sum of a joined table column

Amazon Redshift 解决方案

row_number() 可用，所以这应该有效：

SELECT col1, col2, col3, col4
FROM  (
   SELECT col1, col2, col3, min(col4) AS col4
        , row_number() OVER (PARTITION BY col1, col2
                             ORDER BY count(*) DESC, col3) AS rn
   FROM   tbl
   GROUP  BY col1, col2, col3
   ) sub
WHERE  rn = 1
ORDER  BY col1, col2;

或者如果 window 不允许聚合函数上的函数，使用另一个子查询

SELECT col1, col2, col3, col4
FROM  (
   SELECT *, row_number() OVER (PARTITION BY col1, col2
                                ORDER BY ct DESC, col3) AS rn
   FROM (
      SELECT col1, col2, col3, min(col4) AS col4, COUNT(*) AS ct
      FROM   tbl
      GROUP  BY col1, col2, col3
      ) sub1
   ) sub2
WHERE  rn = 1;

这会选择最小的 col3 如果超过一个并列为最大计数。而最小的col4为各自的col3.

SQL Fiddle 在 Postgres 9.3 中演示所有内容。

为每对 Col1 和 Col2 找到最常出现的 Col3

Find most occurring Col3 for every pair of Col1 and Col2

sql

database

postgresql

greatest-n-per-group

amazon-redshift

Amazon Redshift 解决方案