如何使用 SQL 按给定时间 window 对事件时间进行排名？

Question

我正在努力寻找一种使用 SQL 对事件进行排名的方法。目标是每当事件发生超过 delta 秒（例如 1 秒）之前的观察时增加排名。到目前为止，我的尝试如下所示：

select a.event_time, a.user_name, a.object_name, a.rnk, case when a.ddif <= 1000 then 0 else 1 end as new_query,
            case when a.ddif <= 1000 then 0 else rnk end as new_rnk
from (
    select *, rank() OVER (PARTITION BY user_name ORDER BY event_time) AS rnk,
              date_diff('second',lag(event_time) OVER (PARTITION BY user_name ORDER BY event_time),event_time) as ddif
    from tmp
    ) a

但它只给了我以下结果，我仍然不知道如何实现 yellow 中的结果（它们中的任何一个都对我来说很完美）。

如果有任何帮助，我将不胜感激。

请注意：我正在使用 Presto DB，因此我仅限于此查询引擎。

Answer 1

使用lag() 和累积总和来定义组。然后分配行号：

select t.*,
       row_number() over (partition by user_name, grp order by event_time) as seqnum
from (select t.*,
             sum(case when prev_et > event_time - interval '1' second
                      then 0 else 1
                 end) over (partition by user_name order by event_time) as grp
      from (select t.*,
                   lag(event_time) over (partition by user_name order by event_time) as prev_et
            from tmp t
           ) t
     ) t;

Answer 2

您可以使用 lag() 和 window sum():

select
    t.*,
    sum(case when event_time <= lag_event_time + interval '1' second then 0 else 1 end) rnk 
from (
    select 
        t.*, 
        lag(event_time) over(order by event_time partition by user_name) lag_event_time
    from mytable t
) t

Answer 3

感谢所有好的提示，它们为我指明了最终解决方案的方向，即：

select a.*, sum (case when a.ddif <= 1 then 0 else 1 end) over (partition by user_name order by event_time) as acc_rnk
    from (
        select *, date_diff('second',lag(event_time) OVER (PARTITION BY user_name ORDER BY event_time),event_time) as ddif
        from tmp
        ) a

如何使用 SQL 按给定时间 window 对事件时间进行排名？

How to rank event times by a given time window using SQL?

sql

window-functions

presto

gaps-and-islands