窗函数和整体平均

Question

我正在尝试在每一行中获取像这样的

overall average

数字。您能帮助我如何使用 Snowflake SQL 最好地实现这一目标吗？

Overall Average

应该是

(300+150+100)/3

身份证	描述	SUB_ID	价值	按 ID 平均	整体平均
1	ABC	10	300	300	183.33
1	ABC	20	150	300	183.33
1	ABC	30	450	300	183.33
2	防御	10	150	150	183.33
2	防御	20	150	150	183.33
3	EFG	30	100	100	183.33
3	EFG	10	180	100	183.33
3	EFG	20	20	100	183.33

我能够使用窗口函数得到

Average by ID

AVG (Value) Over (partition by ID)

Answer 1

所有值的平均值：

将

PARTITION BY

更改为常数，例如

null

或

true

select 
    $1 AS id,
    $2 AS value,
    avg(value) over (partition by id) as avg_by_id,
    avg(value) over (partition by true) as overall_avg
from values 
    (1,300),
    (1,150),
    (1,450),
    (2,150),
    (2,150),
    (3,100),
    (3,180),
    (3,20);

平均 平均：

你要求的是平均值，这个值静态地被认为是垃圾值”，所以它是可以计算的，但它通常也是没有意义的。所以应该避免。

执行此操作的其中之一是：

with example_data(id, value) as (
    select * from values
        (1,300),
        (1,150),
        (1,450),
        (2,150),
        (2,150),
        (3,100),
        (3,180),
        (3,20)
)
select a.* 
    ,b.*
from (
    select 
        id,
        value,
        avg(value) over (partition by id) as avg_by_id,
        avg(value) over (partition by true) as overall_avg_correct
    from example_data
) as a
cross join (        
    select avg(id_avg) as avg_avg
    from (
        select avg(value) as id_avg from example_data group by id
    )
) as b;

或者它可以作为基于 SELECT 的子选择来完成：

select 
    id,
    value,
    avg(value) over (partition by id) as avg_by_id,
    avg(value) over (partition by true) as overall_avg_correct,
    (        
        select avg(id_avg)
        from (
            select avg(value) as id_avg from example_data group by id
        )
    ) as avg_avg
from example_data

Answer 2

在 SQL 中，这可以通过窗口函数和 OVER PARTITION 来完成

SELECT *
  , AVG (Value) OVER (PARTITION BY [ID]) AS [AverageByID]
  , AVG (Value) OVER() AS [OverallAverage]
FROM [data]