获得顶部和底部 25% 的平均值

Question

我有一张表格，上面有员工名单和他们售出的单位数量。

我想获得前 25 个百分位的平均销售量和后 25 个百分位的平均销售量。

我已经创建了我的数据的表示SLQ Fiddle

我真的不知道如何开始？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 个百分位并且不能限制为 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 中高效地完成这一切？

Answer 1

这是一个使用了我从from this question学到的诡计的解决方案。

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

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

见SQL Fiddle。

这个想法是使用MySQL: LIMIT by a percentage of the amount of records? 解决方案来获取百分位数。基于此（以及在pdw answer），我们创建了一个输出，其中仅显示前 25% 和后 75%。

最后，我们计算和求和以获得您请求的值。

---

注意这在命令之上运行：

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

Answer 3

用这个逻辑怎么样：

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