2

在 MSSQL2008 中，我尝试使用经典的中值查询从公用表表达式中计算一列数字的中值，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，当我希望中值列一直向下为“x1”时，“中值”列与“数字”列相同。我使用一个类似的表达式来计算模式，它在同一个公用表表达式上工作得很好。

4

3 回答 3

3

Here's a slightly different way to do it:

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
于 2010-07-09T22:36:14.390 回答
1

您的查询的问题是您正在做

SELECT TOP 1 cte.number FROM...

but it isn't correlated with the sub query it is correlated with the Outer query so the subquery is irrelevant. Which explains why you simply end up with the same value all the way down. Removing the cte. (as below) gives the median of the CTE. Which is a constant value. What are you trying to do?

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

Returns

NUMBER      median
----------- -----------
0           1023
1           1023
2           1023
3           1023
4           1023
5           1023
6           1023
7           1023
于 2010-07-09T22:32:13.337 回答
1

This is not an entirely new answer as it mostly expands on Mark Byer's answer, but there are a couple of options for simplifying the query even further.

The first thing is to really make use of CTE's. Not only can you have multiple CTE's, but they can refer to each other. With this in mind, we can create an additional CTE to compute the median based on the results of the first. This encapsulates the median computation and leaves the actual SELECT to do only what it needs to do. Note that the ROW_NUMBER() had to be moved into the first CTE.

;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

And to further reduce complexity, you "could" use a custom CLR Aggregate to handle the Median (such as the one provided in the free SQL# library at http://www.SQLsharp.com/ [which I am the author of]).

;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
于 2011-01-21T21:27:51.343 回答