Question

我的表中有以下数据：

SELECT category, value FROM test

 
| category | value |
+----------+-------+
|    1     |   1   | 
|    1     |   3   |
|    1     |   4   |
|    1     |   8   |

现在我正在使用两个单独的查询。

获得平均值：

  SELECT category, avg(value) as Average
    FROM test
GROUP BY category

 
 | category | value |
 +----------+-------+
 |    1     |   4   |

获得中位数：

SELECT DISTINCT category, 
                PERCENTILE_CONT(0.5) 
                   WITHIN GROUP (ORDER BY value) 
                   OVER (partition BY category) AS Median
           FROM test

 
 | category | value |
 +----------+-------+
 |    1     |  3.5  |

有没有办法在一个查询中合并它们？

注意：我知道我也可以使用两个子查询获得中位数，但我更喜欢使用PERCENTILE_CONT函数来获取它。

Answer 1

AVG也是一个窗口函数：

select 
distinct
category,
avg(value) over (partition by category) as average,
PERCENTILE_CONT(0.5) 
                   WITHIN GROUP (ORDER BY value) 
                   OVER (partition BY category) AS Median
                   from test

Answer 2

我会以稍微不同的方式处理这个问题：

select category, avg(value) as avg,
       avg(case when 2 * seqnum in (cnt, cnt + 1, cnt + 2) then value end) as median
from (select t.*, row_number() over (partition by category order by value) as seqnum,
             count(*) over (partition by category) as cnt
      from test t
     ) t
group by category;

Answer 3

我希望对这个问题有一个更彻底的回答，经过一些挖掘后，在对Dwain Camps的多种方法的详尽分析中发现它，以防其他人发现它有用：

Calculating the Median Value within a Partitioned Set Using T-SQL

我去了＆＃34;他的＆＃34;第四个解决方案（他正在组合/测试其他人的方法），这很容易理解并且确实表现良好：

WITH Counts AS
(
   SELECT ID, c=COUNT(*)
   FROM #MedianValues
   GROUP BY ID
)
SELECT a.ID, Median=AVG(0.+N)
FROM Counts a
CROSS APPLY
(
   SELECT TOP(((a.c - 1) / 2) + (1 + (1 - a.c % 2)))
      N, r=ROW_NUMBER() OVER (ORDER BY N)
   FROM #MedianValues b
   WHERE a.ID = b.ID
   ORDER BY N
) p
WHERE r BETWEEN ((a.c - 1) / 2) + 1 AND (((a.c - 1) / 2) + (1 + (1 - a.c % 2)))
GROUP BY a.ID;

Sql Server 2012中一个查询的平均值和中位数

3 个答案: