Question

我有一张表，列出了员工名单和他们销售的单位数量。

我希望销售前25％的平均销售量和25％的平均销售量。

我创建了数据SLQ Fiddle

的表示

我真的不知道如何开始这个？我看到的所有示例都是针对SQL Server而不是MySQL。这就是我的想法。

我想要25个百分点并且不能限制25个项目。基本上它会涉及：

1) #_of_employees = The number of total employees. 

2) #_of_employees_in_25_percentile =  #_of_employees*0.25 

3) Calculate the sum of the units sold by the top/bottom 25 percentile (limit #_of_employees_in_25_percentile)

4) Divide the sum by #_of_employees_in_25_percentile to get the average.

如何在MySQL中有效地完成所有这些工作？

Answer 1

这是一个使用我学到的狡猾技巧的解决方案from this question。

SELECT id, unit_sold, n * 100 / @total AS percentile
FROM (
  SELECT id, unit_sold, @total := @total + unit_sold AS n
  FROM mydata, (SELECT @total := 0) AS total
  ORDER BY unit_sold ASC
) AS t

SQL Fiddle.

Answer 2

如何使用这种逻辑：

全选，按百分位排序（DESC），限制为25
全选，按百分位（ASC）排序，限制为25

这是您正在寻找的逻辑类型吗？

示例查询：

$q1 = mysql_query(SELECT * FROM table_name ORDER BY percentile DESC LIMIT 25)
$q2 = mysql_query(SELECT * FROM table_name ORDER BY percentile ASC LIMIT 25)

Answer 3

这个怎么样？

SELECT
  SUM(unit_sold) AS sum_tot, SUM(unit_sold)/count(id) AS average,
  SUM(CASE WHEN percentile<25 THEN unit_sold ELSE 0 END) AS sum_top25,
  SUM(CASE WHEN percentile<25 THEN 1 ELSE 0 END) AS count_top25,
  SUM(CASE WHEN percentile<25 THEN unit_sold ELSE 0 END)/SUM(CASE WHEN percentile<25 THEN 1 ELSE 0 END) AS average_top25,
  SUM(CASE WHEN percentile>75 THEN unit_sold ELSE 0 END) AS sum_bottom25,
  SUM(CASE WHEN percentile>75 THEN 1 ELSE 0 END) AS count_bottom25,
  SUM(CASE WHEN percentile>75 THEN unit_sold ELSE 0 END)/SUM(CASE WHEN percentile>75 THEN 1 ELSE 0 END) AS average_bottom25
FROM
(SELECT
  id, unit_sold, c * 100 / @counter AS percentile
FROM
   (SELECT
      m.*, @counter:=@counter+1 AS c 
   FROM
     (SELECT @counter:=0) AS initvar, mydata AS m
   ORDER BY unit_sold desc
   ) AS t
WHERE
  c <= (25/100 * @counter)
  OR c >= (75/100 * @counter)
) AS t2

输出：

SUM_TOT AVERAGE     SUM_TOP25   COUNT_TOP25 AVERAGE_TOP25 SUM_BOTTOM25  COUNT_BOTTOM25  AVERAGE_BOTTOM25
850     283.3333    500         1            500          350           2               175

请参阅SQL Fiddle。

我们的想法是使用MySQL: LIMIT by a percentage of the amount of records?解决方案来获得百分位数。基于此（以及pdw answer），我们创建了一个输出，其中我们只显示前25％和后75％。

最后，我们计算并求和以获得您请求的值。

请注意，这会在命令之上运行：

SELECT
  id, unit_sold, c * 100 / @counter AS percentile
FROM
   (SELECT
      m.*, @counter:=@counter+1 AS c 
   FROM
     (SELECT @counter:=0) AS initvar, mydata AS m
   ORDER BY unit_sold desc
   ) AS t
WHERE
  c <= (25/100 * @counter)
  OR c >= (75/100 * @counter)

谁的输出是：

ID  UNIT_SOLD   PERCENTILE
d   500         20
a   250         80
e   100         100

获得最高和最低25％的平均值

3 个答案: