即使没有内置函数,你也可以用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

随着查询

SELECT name,v, (SELECT COUNT(1) FROM sale WHERE v

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

FROM sale o

当小于或等于计数是项目数量的一半时,将出现中值

> Acrington有4件商品.其中一半是2,其范围为0..2(对应于200.00),也在2..3范围内(对应于300.00)

> Bullingdon也有4件物品. 2在1..2(值300.00)和2..3(值400.00)的范围内

> Cardington有5件衣服.值2.5在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

, (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

这使：

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

, (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