如何选择 COUNT DISTINCT 为 MAX 的行？

Posted 2023-02-16

【中文标题】如何选择 COUNT DISTINCT 为 MAX 的行？

【英文标题】：How to SELECT lines where COUNT DISTINCT is MAX?

【发布时间】：2021-03-06 07:56:40

【问题描述】：

这可能是一个简单的问题，但基本上我想在 SQL SERVER 中只选择出现次数最多的行。 我有一个产生以下数据的查询：

SELECT [JOB ROLE], [CITY], COUNT(DISTINCT([EMPLOYEE_ID])) as [COUNT]
FROM MyTable
GROUP BY [JOB ROLE], [CITY]

我想达到以下结果，仅选择 EMPLOYEE_ID 的 COUNT DISTINCT 为 MAX 的行：

非常感谢您就如何以最简单的方式到达这里的友好帮助和建议！

【问题讨论】：

DISTINCT 不是一个函数，它是一个集合量词。跳过那些额外的括号，只需写 COUNT(DISTINCT [EMPLOYEE_ID]) 以使代码更清晰。

多组共享最大应该怎么办？是否有平局规则，或者所有组都返回，或者，...

【参考方案1】：

您可以使用TOP WITH TIES 和RANK()：

SELECT TOP (1) WITH TIES [JOB ROLE], [CITY], COUNT(DISTINCT [EMPLOYEE_ID]) as [COUNT]
FROM MyTable
GROUP BY [JOB ROLE], [CITY]
ORDER BY RANK() OVER (ORDER BY COUNT(DISTINCT [EMPLOYEE_ID]) DESC;

或者，使用子查询和RANK()：

SELECT [JOB ROLE], [CITY], [COUNT]
FROM (SELECT [JOB ROLE], [CITY],
             COUNT(DISTINCT [EMPLOYEE_ID]) as [COUNT],
             RANK() OVER (ORDER BY COUNT(DISTINCT [EMPLOYEE_ID]) DESC) as seqnum
      FROM MyTable
      GROUP BY [JOB ROLE], [CITY]
     ) jc
WHERE seqnum = 1;

编辑：

这个问题似乎变成了每个城市最常见的工作角色。上面的查询很容易修改：

SELECT TOP (1) WITH TIES [JOB ROLE], [CITY], COUNT(DISTINCT [EMPLOYEE_ID]) as [COUNT]
FROM MyTable
GROUP BY [JOB ROLE], [CITY]
ORDER BY RANK() OVER (PARTITION BY [JOB ROLE] ORDER BY COUNT(DISTINCT [EMPLOYEE_ID]) DESC;

或者，使用子查询和RANK()：

SELECT [JOB ROLE], [CITY], [COUNT]
FROM (SELECT [JOB ROLE], [CITY],
             COUNT(DISTINCT [EMPLOYEE_ID]) as [COUNT],
             RANK() OVER (PARTITION BY [JOB ROLE], ORDER BY COUNT(DISTINCT [EMPLOYEE_ID]) DESC) as seqnum
      FROM MyTable
      GROUP BY [JOB ROLE], [CITY]
     ) jc
WHERE seqnum = 1;

【讨论】：

感谢您的建议！不幸的是，这仅返回给我一行，选择最常见的 [JOB ROLE] 和 [CITY]，而不是最常见的 [CITY] 的所有 [JOB ROLE] 值。

@gonzika 这不是你要求的吗？在查看所需的输出时，我认为这是您的问题，所以我认为这个答案是有效的

@GordonLinoff 问题与首次发布时相同：***.com/posts/64971175/revisions

【参考方案2】：

如果你不关心平局，你可以使用 FIRST_VALUE() 窗口函数：

SELECT DISTINCT [JOB ROLE], 
       FIRST_VALUE([CITY]) OVER (PARTITION BY [JOB ROLE] ORDER BY COUNT(DISTINCT([EMPLOYEE_ID])) DESC)
FROM MyTable
GROUP BY [JOB ROLE], [CITY]