无法计算中位数 - SQL Server 2017

Question

我正在尝试计算每个类别中交易的中位数。一些注意事项（因为下面的数据集是更大数据集的一小段）：

一名员工可以属于多个类别
每笔交易的中位数应 > 0
并非每个人都出现在每个类别中

数据是这样设置的：

|  Person | Category | Transaction |
|:-------:|:--------:|:-----------:|
| PersonA |   Sales  |      27     |
| PersonB |   Sales  |      75     |
| PersonC |   Sales  |      87     |
| PersonD |   Sales  |      36     |
| PersonE |   Sales  |      70     |
| PersonB |   Buys   |      60     |
| PersonC |   Buys   |      92     |
| PersonD |   Buys   |      39     |
| PersonA |    HR    |      59     |
| PersonB |    HR    |      53     |
| PersonC |    HR    |      98     |
| PersonD |    HR    |      54     |
| PersonE |    HR    |      70     |
| PersonA |   Other  |      46     |
| PersonC |   Other  |      66     |
| PersonD |   Other  |      76     |
| PersonB |   Other  |      2      |

理想的输出应该是这样的：

| Category | Median | Average |
|:--------:|:------:|:-------:|
|   Sales  |   70   |    59   |
|   Buys   |   60   |    64   |
|    HR    |   59   |    67   |
|   Other  |   56   |    48   |

我可以通过以下方式获得平均值：

SELECT
    Category,
    AVG(Transaction) AS Average_Transactions
FROM
    table
GROUP BY
    Category

效果很好！

This post 试图帮我找到中位数。我写的是：

SELECT
    Category,
    PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY Transaction) OVER (PARTITION BY Category) AS Median_Transactions
FROM
    table
GROUP BY
    Category

但是我得到一个错误：

Msg 8120: Column 'Transactions' is invalid in the select list because it is not contained in either an aggregate function or the **GROUP BY** clause

我该如何解决这个问题？

Answer 1

您可以使用 SELECT DISTINCT 做您想做的事：

SELECT DISTINCT Category,
       PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY Transaction) OVER (PARTITION BY Category) AS Median_Transactions
FROM table;

不幸的是，SQL 服务器不提供 PERCENTILE_ 函数作为 window 函数，并且没有 MEDIAN() 聚合函数。您也可以使用子查询和计数来执行此操作。

Answer 2

这不是最佳解决方案，但这是您的解决方案

SELECT DISTINCT
       category,
       PERCENTILE_DISC(0.5)WITHIN GROUP(ORDER BY val) OVER (PARTITION BY category) AS Median_Transactions,
       AVG(val) OVER (PARTITION BY d.category)  [AVG]
FROM #data d;

Answer 3

我不认为这很漂亮，但它确实有效。没花时间打磨

with 
avg_t as 
( select category, avg(sales) as avg_sales 
  from sample 
   group by 1), 
mn as 
(  select category, avg(sales) as median_sales 
   from (
     select category, sales , 
            row_number() over  (partition by category order by sales asc) as r , 
            count(person) over (partition by category) as total_count 
     from  sample 
         ) mn_sub 
    where (total_count % 2 = 0 and r in ( (total_count/2), ((total_count/2)+1)) ) or 
           (total_count % 2 <> 0 and r = ((total_count+1)/2))
         
    group by 1 
)
select avg_t.category, avg_t.avg_sales, mn.median_sales  
 from avg_t 
 inner join mn 
 on avg_t.category=mn.category

无法计算中位数 - SQL Server 2017

Unable to calculate median - SQL Server 2017

sql

sql-server

aggregate-functions

median