在SnowFlake中按窗口分区使用条件

Question

一些相关内容：根据 Snowflake 中的日期和窗口函数过滤查询

我需要创建一个查询来计算

id

在 -+ 90 天窗口中出现的次数，与此类似，但作为窗口函数，这可能吗？

WITH fake_data(id, DATE) as (
    SELECT * FROM VALUES
    -- this id has visted once
    (1, '2022-04-14'::date),
    -- this id has visited 3 times
    (3, '2022-01-13'::date),
    (3, '2022-03-13'::date),
    (3, '2022-05-13'::date),
    -- this id is a huge vistor
    (5, '2022-01-01'::date),
    (5, '2022-02-01'::date),
    (5, '2022-05-01'::date),
    (5, '2022-06-01'::date),
    (5, '2022-08-01'::date)
)
select * from (
    select 
    count_if("change" between -90 and 90) over (partition by ID, t1.DATE) "c",
    *
    from fake_data as t1
    left outer join lateral (
        select t1.DATE - t.DATE "change", t.DATE "t_DATE" 
        from fake_data AS t
        where t1.id = t.id and t1.DATE - t.DATE between -90 and 9
     ) as t2
    order by ID, t1.DATE, "change"
)
where "change" = 0;

结果（

change

和

t_DATE

仅供参考）：

c	身份证	日期	改变	t_DATE
1	1	2022-04-14	0	2022-04-14
2	3	2022-01-13	0	2022-01-13
3	3	2022-03-13	0	2022-03-13
2	3	2022-05-13	0	2022-04-13
2	5	2022-01-01	0	2022-01-01
3	5	2022-02-01	0	2022-02-01
3	5	2022-05-01	0	2022-05-01
3	5	2022-06-01	0	2022-06-01
2	5	2022-08-01	0	2022-08-01

这是我喜欢做的，但似乎当前行的日期不可用（或者我可以使用别名）：

select
  count_if(DATE - d between -90 and 90) over (partition by id, DATE  as d) as "c",
  id,
  date
from fake_data;

Answer 1

嗯，

即使不是你想要的，你的SQL也可以这样写：

select 
    t1.DATE - t.DATE as change
    ,count_if(abs(t1.DATE - t.DATE) <= 90) over (partition by t1.ID, t1.DATE) as c
    ,t1.*
    ,t.date as t_date
from fake_data as t1
left join fake_data as t
    on t1.id = t.id and abs(t1.DATE - t.DATE) <= 90
qualify change = 0
order by t1.ID, t1.DATE, change

但是考虑到 join 与你的 count_if 相同，也可以写成：

select 
    t1.DATE - t.DATE as change
    ,count(*) over (partition by t1.ID, t1.DATE) as c
    ,t1.*
    ,t.date as t_date
from fake_data as t1
left join fake_data as t
    on t1.id = t.id and abs(t1.DATE - t.DATE) <= 90
qualify change = 0
order by t1.ID, t1.DATE, change

但是给定的窗口函数没有您希望的时间范围的“这一行”，您可以通过使用 Javascript UDTF 来解决这个问题，并为每一行构建一个内存集，并通过它进行计数，然后在决赛中发出它，然后加入它。

此时，您不妨分解数据并在原始 SQL 中进行等连接，与自连接相比，在 +90、-90 天的时间里，对于海量数据来说，这可能仍然相当快

因此对于海量数据，这应该表现得更好：

WITH fake_data(id, DATE) as (
    SELECT * FROM VALUES
    -- this id has visted once
    (1, '2022-04-14'::date),
    -- this id has visited 3 times
    (3, '2022-01-13'::date),
    (3, '2022-03-13'::date),
    (3, '2022-05-13'::date),
    -- this id is a huge vistor
    (5, '2022-01-01'::date),
    (5, '2022-02-01'::date),
    (5, '2022-05-01'::date),
    (5, '2022-06-01'::date),
    (5, '2022-08-01'::date)
), range as (
    select row_number() over (order by null)-91 as rn
    from table(generator(ROWCOUNT => 181))
), exploded as (
    select
        id, 
        dateadd('day', e.rn, d.date) as t_date
    from fake_data as d
    cross join range as e
)
select
    f.*
    ,count(t_date) as c
from fake_data as f
join exploded as e
    on f.id = e.id and f.date = t_date
group by f.id, f.date
order by f.id, f.DATE
;

身份证	日期	C
1	2022-04-14	1
3	2022-01-13	2
3	2022-03-13	3
3	2022-05-13	2
5	2022-01-01	2
5	2022-02-01	3
5	2022-05-01	3
5	2022-06-01	3
5	2022-08-01	2

Answer 2

Snowflake 现在支持窗口函数中的 RANGE 子句。示例数据：

输入表：访客数据(

visitor_data

)

VISITOR_ID	DATE_VISITED
1	2022-04-14
3	2022-01-13
3	2022-03-13
3	2022-05-13
5	2022-01-01
5	2022-02-01
5	2022-05-01
5	2022-06-01
5	2022-08-01

SQL 查询

select 
  *
  , count(*) over (partition by visitor_id
      order by date_visited
      range between interval '90 day' preceding and current row
    ) as count_90_days
from visitor_data;

查询输出：

VISITOR_ID	DATE_VISITED	COUNT_90_DAYS
3	2022-01-13	1
3	2022-03-13	2
3	2022-05-13	2
1	2022-04-14	1
5	2022-01-01	1
5	2022-02-01	2
5	2022-05-01	2
5	2022-06-01	2
5	2022-08-01	2