计算 SQL 公用表表达式中列的中位数

Question

在 MSSQL2008 中，我尝试使用经典的中位数查询从公用表表达式中计算一列数字的中位数，如下所示：

WITH cte AS
(
   SELECT number
   FROM table
) 

SELECT cte.*,
(SELECT 
  (SELECT (   
    (SELECT TOP 1 cte.number  
     FROM     
     (SELECT TOP 50 PERCENT cte.number     
      FROM cte
      ORDER BY cte.number) AS medianSubquery1   
    ORDER BY cte.number DESC)  
    +   
  (SELECT TOP 1 cte.number
   FROM     
    (SELECT TOP 50 PERCENT cte.number    
     FROM cte   
     ORDER BY cte.number DESC) AS medianSubquery2   
   ORDER BY cte.number ASC) ) / 2)) AS median

FROM cte
ORDER BY cte.number

我得到的结果集如下：

NUMBER    MEDIAN
x1        x1
x1        x1
x1        x1
x2        x2
x3        x3

换句话说，当我希望中位数列一直向下为“x1”时，“中位数”列与“数字”列相同。我使用类似的表达式来计算模式，它在同一个公用表表达式上运行良好。

Answer 1

这不是一个全新的答案，因为它主要扩展了 Mark Byer 的答案，但有几个选项可以进一步简化查询。

首先要真正利用 CTE。您不仅可以拥有多个 CTE，而且它们可以相互引用。考虑到这一点，我们可以创建一个额外的 CTE 来根据第一个结果计算中位数。这封装了中值计算，让实际的 SELECT 只做它需要做的事情。请注意，ROW_NUMBER() 必须移到第一个 CTE 中。

;WITH cte AS
(
   SELECT number, ROW_NUMBER() OVER(ORDER BY number) AS rn
   FROM table1
),
med AS
(
    SELECT AVG(number) AS median
    FROM cte
    WHERE cte.rn = ((SELECT COUNT(*) FROM cte) + 1) / 2
    OR cte.rn = ((SELECT COUNT(*) FROM cte) + 2) / 2
)
SELECT cte.number, med.median
FROM cte
CROSS JOIN med

为了进一步降低复杂性，您“可以”使用自定义 CLR 聚合来处理中位数（例如免费 SQL# 库中提供的 http://www.SQLsharp.com/ [我是作者]）。

;WITH cte AS
(
   SELECT number
   FROM table1
),
med AS
(
    SELECT  SQL#.Agg_Median(cte.number) AS median
    FROM    cte
)
SELECT cte.number, med.median
FROM cte
CROSS JOIN med

Answer 2

你的查询的问题是你在做

SELECT TOP 1 cte.number FROM...

但它与子查询不相关，它与外部查询相关，因此子查询无关紧要。这就解释了为什么你最终只会得到相同的值。删除 cte.（如下）给出 CTE 的中值。这是一个常数值。你想做什么？

WITH cte AS
    ( SELECT NUMBER
    FROM master.dbo.spt_values
    WHERE TYPE='p'
    )

SELECT cte.*,
(SELECT 
  (SELECT (   
    (SELECT TOP 1 number  
     FROM     
     (SELECT TOP 50 PERCENT cte.number     
      FROM cte
      ORDER BY cte.number) AS medianSubquery1   
    ORDER BY number DESC)  
    +   
  (SELECT TOP 1 number
   FROM     
    (SELECT TOP 50 PERCENT cte.number    
     FROM cte   
     ORDER BY cte.number DESC) AS medianSubquery2   
   ORDER BY number ASC) ) / 2)) AS median
FROM cte
ORDER BY cte.number

返回

NUMBER      median
----------- -----------
0           1023
1           1023
2           1023
3           1023
4           1023
5           1023
6           1023
7           1023

Answer 3

这里有一个稍微不同的方法：

WITH cte AS
(
   SELECT number
   FROM table1
)
SELECT T1.number, T3.median
FROM cte T1, 
(
    SELECT AVG(number) AS median
    FROM
    (
        SELECT number, ROW_NUMBER() OVER(ORDER BY number) AS rn
        FROM cte
    ) T2
    WHERE T2.rn = ((SELECT COUNT(*) FROM table1) + 1) / 2
    OR T2.rn = ((SELECT COUNT(*) FROM table1) + 2) / 2
) T3