Question

我有以下MySQL查询，它完美运行：

select 
    count(*) as `# of Data points`, 
    name, 
    max((QNTY_Sell/QNTYDelivered)*1000) as `MAX Thousand Price`,
    min((QNTY_Sell/QNTYDelivered)*1000) as `MIN Thousand Price`,
    avg((QNTY_Sell/QNTYDelivered)*1000) as `MEAN Thousand Price` 
from 
    table_name 
where 
    year(date) >= 2012 and 
    name like "%the_name%" and 
    QNTYDelivered > 0 and 
    QNTY_Sell > 0 
group by name 
order by name;

现在我还希望添加一个结果列，它为每行提供MEDIAN数据。在SELECT下，在完美的世界中，这将是这样的：

median((QNTY_Sell/QNTYDelivered)*1000) as `MEDIAN Thousand Price`

在Google上搜索MySQL中值函数让我得到了这个答案，如果您对整个表的数据集的中位数感兴趣，这似乎没问题：Simple way to calculate median with MySQL

这里的不同之处在于我正在通过name列对表格中的数据进行分组，并希望获得按此列分组的每行数据的中位数。

有谁知道允许我这样做的好玩法？

谢谢！

Answer 1

即使没有内置函数，也可以使用MySQL中的GROUP BY计算中位数。

考虑一下表：

Acrington   200.00
Acrington   200.00
Acrington   300.00
Acrington   400.00
Bulingdon   200.00
Bulingdon   300.00
Bulingdon   400.00
Bulingdon   500.00
Cardington  100.00
Cardington  149.00
Cardington  151.00
Cardington  300.00
Cardington  300.00

对于每一行，您可以计算较少的类似项目的数量。您还可以计算有多少值小于或等于：

name        v       <   <=
Acrington   200.00  0   2
Acrington   200.00  0   2
Acrington   300.00  2   3
Acrington   400.00  3   4
Bulingdon   200.00  0   1
Bulingdon   300.00  1   2
Bulingdon   400.00  2   3
Bulingdon   500.00  3   4
Cardington  100.00  0   1
Cardington  149.00  1   2
Cardington  151.00  2   3
Cardington  300.00  3   5
Cardington  300.00  3   5

使用查询

SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
             , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
  FROM sale o

当小于或等于的数量是项目数量的一半时，将出现中值

Acrington 有4件商品。其中一半为2，范围为0..2（对应200.00），范围为2..3（对应300.00）
Bullingdon 也有4项。 2的范围是1..2（值300.00）和2..3（值400.00）
Cardington 有5件商品。值2.5介于2和3之间，对应于Cardington 151。

中值是返回的最小值和最大值的平均值：

SELECT cs.name,v
   FROM
   (SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
                 , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
      FROM sale o) cs JOIN
   (SELECT name,COUNT(1)*.5 as cn
      FROM sale
      GROUP BY name) cc ON cs.name=cc.name
 WHERE cn between ls and lse

给出了：

Acrington   200.00
Acrington   200.00
Acrington   300.00
Bulingdon   300.00
Bulingdon   400.00
Cardington  151.00

最后我们可以得到中位数：

SELECT name,(MAX(v)+MIN(v))/2 FROM
(SELECT cs.name,v
   FROM
   (SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v<o.v AND name=o.name) as ls
                 , (SELECT COUNT(1) FROM sale WHERE v<=o.v AND name=o.name) as lse
      FROM sale o) cs JOIN
   (SELECT name,COUNT(1)*.5 as cn
      FROM sale
     GROUP BY name) cc ON cs.name=cc.name
 WHERE cn between ls and lse
 ) AS medians
GROUP BY name

给予

Acrington   250.000000
Bulingdon   350.000000
Cardington  151.000000

Answer 2

我发现这样做的唯一方法是通过字符串操作：
使用GROUP_CONCAT创建所有值的列表，然后使用缩进SUBSTRING_INDEX获取中值

SELECT
    count(*) AS `# of Data points`,
    name,
    max((QNTY_Sell/QNTYDelivered)*1000) AS `MAX Thousand Price`,
    min((QNTY_Sell/QNTYDelivered)*1000) AS `MIN Thousand Price`,
    avg((QNTY_Sell/QNTYDelivered)*1000) AS `MEAN Thousand Price`
  , CASE (count(*) % 2)
    WHEN 1 THEN SUBSTRING_INDEX(
      SUBSTRING_INDEX(
        group_concat((QNTY_Sell/QNTYDelivered)*1000 
                      ORDER BY (QNTY_Sell/QNTYDelivered)*1000 SEPARATOR ',')
      , ',', (count(*) + 1) / 2)
    , ',', -1)
    ELSE (SUBSTRING_INDEX(
      SUBSTRING_INDEX(
        group_concat((QNTY_Sell/QNTYDelivered)*1000 
                      ORDER BY (QNTY_Sell/QNTYDelivered)*1000 SEPARATOR ',')
      , ',', count(*) / 2)
    , ',', -1)
  + SUBSTRING_INDEX(
      SUBSTRING_INDEX(
        group_concat((QNTY_Sell/QNTYDelivered)*1000 
                      ORDER BY (QNTY_Sell/QNTYDelivered)*1000 SEPARATOR ',')
      , ',', (count(*) + 1) / 2)
    , ',', -1)) / 2
    END median
FROM
    sales
WHERE
    year(date) >= 2012 AND
    name LIKE "%art.%" AND
    QNTYDelivered > 0 AND
    QNTY_Sell > 0
GROUP BY name
ORDER BY name;

CASE需要检查我们是否有一个中值，奇数个值，或两个中值，偶数个值，在第二种情况下，中位数是两个值的平均值

SQLFiddle

在MySQL中使用MEDIAN和MAX，MIN和AVG功能

2 个答案: