如何为最多包含 5 个时间相关成员的每个分区创建平均值？

Question

我的目标是 select 平均恰好 5 条记录，前提是它们满足左连接标准到另一个 table。假设我们有 table 个（左）记录：

RECNUM   ID    DATE         JOB
1      | cat | 2019.01.01 | meow
2      | dog | 2019.01.01 | bark

我们有 table 两个 （右）记录：

RECNUM   ID    Action_ID    DATE         REWARD
1      | cat | 1          | 2019.01.02 | 20
2      | cat | 99         | 2018.12.30 | 1
3      | cat | 23         | 2019.12.28 | 20       
4      | cat | 54         | 2018.01.01 | 20
5      | cat | 32         | 2018.01.02 | 20
6      | cat | 21         | 2018.01.03 | 20
7      | cat | 43         | 2018.12.28 | 1
8      | cat | 65         | 2018.12.29 | 1
9      | cat | 87         | 2018.09.12 | 1
10     | cat | 98         | 2018.10.11 | 1 
11     | dog | 56         | 2018.09.01 | 99 
12     | dog | 42         | 2019.09.02 | 99

结果应该return:

ID  | AVG(Reward_from_latest_5_jobs)
cat | 1

满足的条件应该是：对于 left table 中的每个 JOB，尝试在 right table 并计算它们的平均值。所以换句话说，狗叫了，我们不知道要给他什么奖励，我们试着统计他最近得到的五次奖励的平均值。如果找到少于 5 个，则不要 return anything/put null，如果更多，则丢弃最旧的。

我想要的方式是这样的：

         SELECT a."ID", COUNT(b."Action_ID"), AVG(b."REWARD")  
         FROM 
             ( 
                SELECT "ID", "DATE"
                 FROM :left_table
             ) a  

              LEFT JOIN

             ( 
                SELECT "ID", "Action_ID", "DATE", "REWARD"
                 FROM :right_table
             ) b 

             ON(
                    a."ID" = b."ID" 
               )    
         WHERE a."DATE" > b."DATE" 
         GROUP BY a."ID"
         HAVING COUNT(b."Action_ID") >= 5;

但随后它会计算所有符合条件的 Action_ID(s)，而不仅仅是最近的五个。你能告诉我如何达到预期的结果吗？我可以使用 sub-tables 而不必在一个 SQL 语句中完成。此用例不允许使用过程。非常感谢任何意见。

Answer 1

您可以使用 window 函数，然后聚合：

select 
    id,
    avg(reward) avg_reward
from (
    select 
        t1.id, 
        t2.reward, 
        count(*) over(partition by t1.id) cnt,
        rank() over(partition by t1.id order by t2.date desc) rn
    from leftable t1
    inner join righttable t2 on t1.id = t2.id and t2.date >= t1.date
) t
where cnt >= 5 and rn <= 5
group by id

内部查询根据您的要求加入 table，对每个 id 的可用记录总数进行 window 计数，并对每个 [=11= 的记录进行排名] 递减 date.

然后外部查询筛选至少有 5 条记录的 ids，并为每个 id.

计算前 5 条记录的平均值

Answer 2

使用window函数获得前5名：

select id, avg(reward)
from (select r.*,
             row_number() over (partition by l.id order by r.date desc) as seqnum
      from table1 l join
           table2 r
           on l.id = r.id and l.date > r.date
     ) r
where seqnum <= 5
group by id
having count(*) >= 5;

然后是一个 having 子句来过滤掉那些没有五行的 ID。

Answer 3

这里是一个连接的方法（如果你想做更多的连接，只需对每个连接重复这个方法

  SELECT ONE.ID, 
         CASE WHEN MAX(J1.RN) < 5 THEN NULL ELSE AVG(J1.REWARD) END AS REWARD_AVG
         -- we could also use count
       --CASE WHEN COUNT(*) = 5 THEN AVG(J1.REWARD) ELSE NULL END AS REWARD_AVG
  FROM TABLE_ONE ONE
  JOIN (
    SELECT
      ID,
      REWARD,
      ROW_NUMBER() OVER (PARTITION BY ID ORDER BY DATE DESC) AS RN
    FROM TABLE_TWO
    WHERE TABLE_TWO.DATE < ONE.DATE
  ) AS J1 ON J1.ID = ONE.ID and RN <= 5 -- take first five only
  GROUP BY ONE.ID

如何为最多包含 5 个时间相关成员的每个分区创建平均值？

How to create an average per partitions containing a maximum of 5 time dependent members?

sql

select

sql-scripts

data-partitioning

hana-sql-script