如何连续统计SQL中所有大于或等于自身的？

Question

假设我有一个 table 包含每个设备类型和设备年龄的设备的设备 ID，我如何做一个 Count Distinct 至少具有该设备年龄的设备 ID .

例如，假设这是我们拥有的所有数据：

equipment_type	equipment_id	equipment_age
Screwdriver	A123	1
Screwdriver	A234	2
Screwdriver	A345	2
Screwdriver	A456	2
Screwdriver	A567	3

我希望输出为：

equipment_type	equipment_age	count_of_equipment_at_least_this_age
Screwdriver	1	5
Screwdriver	2	4
Screwdriver	3	1

原因是有 5 把螺丝刀至少使用了 1 天，4 把螺丝刀至少使用了 2 天，只有 1 把螺丝刀至少使用了 3 天。

到目前为止，我只能计算属于每个 equipment_age 的设备数量（如下所示的查询），但不是“至少 equipment_age”。

SELECT
equipment_type,
equipment_age,
COUNT(DISTINCT equipment_id) as count_of_equipments
FROM equipment_table
GROUP BY 1, 2

Answer 1

使用自连接方法：

SELECT
    e1.equipment_type,
    e1.equipment_age,
    COUNT(*) AS count_of_equipments
FROM equipment_table e1
INNER JOIN equipment_table e2
    ON e2.equipment_type = e1.equipment_type AND
       e2.equipment_age >= e1.equipment_age
GROUP BY 1, 2
ORDER BY 1, 2;

Answer 2

GROUP BY 将 COUNT 的范围限制在组中的行，即它不会让您到达其他行（equipment_age 大于当前组的行）。所以你需要一个子查询或窗口函数来获取它们。一种方式：

SELECT
   equipment_type,
   equipment_age,
   (Select COUNT(*) 
    from equipment_table cnt 
     where cnt.equipment_type = a.equipment_type
       AND cnt.equipment_age >= a.equipment_age
     )  as count_of_equipments
FROM equipment_table a
GROUP BY 1, 2, 3

不过，我不确定您的环境是否支持这种语法。如果没有，请告诉我们我们会找到其他方法。

Answer 3

考虑以下join-less解决方案

select distinct
  equipment_type,
  equipment_age,
  count(*) over equipment_at_least_this_age as count_of_equipment_at_least_this_age
from equipment_table 
window equipment_at_least_this_age as (
  partition by equipment_type 
  order by equipment_age 
  range between current row and unbounded following
)

如果应用于您问题中的示例数据 - 输出为

如何连续统计SQL中所有大于或等于自身的？

How to consecutively count everything greater than or equal to itself in SQL?

sql

google-bigquery