SQL从标志中获取3个相邻的动作而不重复

如何解决SQL从标志中获取3个相邻的动作而不重复

我有一个与 question#66044663 有点相似但更复杂的问题。

这是我的虚拟数据。

我想从每个用户的标志中获取 3 个相邻的操作（没有重复）。

这是描述我想法的图表。

这是我想要的：

如何实现 sql（我使用 Google Bigquery）？ 我知道函数 LAG 可能是一个解决方案，但我不知道如何避免重复操作。

希望有人能点亮我。感谢一百万！

这是生成数据集的代码。

WITH
src_table AS (
SELECT 'Jack' AS User,1 AS Sequence,'Eat' AS Action,'' AS Flag UNION ALL
SELECT 'Jack' AS User,2 AS Sequence,'Work' AS Action,3 AS Sequence,'Sleep' AS Action,'Flag A' AS Flag UNION ALL
SELECT 'Jack' AS User,4 AS Sequence,'Exercise' AS Action,'Flag B' AS Flag UNION ALL
SELECT 'Kenny' AS User,'Run' AS Action,'' AS Flag UNION ALL
SELECT 'Kenny' AS User,'Flag C' AS Flag UNION ALL
SELECT 'Kenny' AS User,5 AS Sequence,'Flag D' AS Flag UNION ALL
SELECT 'May' AS User,'Flag A' AS Flag
)

解决方法

考虑以下

select user,actions.action_sequence,flag  from (
  select *,(
    select as struct count(1) actions_count,string_agg(action,' >> ' order by grp) action_sequence
    from (
      select action,grp from t.arr group by action,grp
    )) actions
  from (
    select *,array_agg(struct(action,grp)) 
      over(partition by user order by grp desc range between current row and 2 following) arr
    from (
      select *,countif(change) over(partition by user order by sequence) grp
      from (
        select *,action != lag(action) over(partition by user order by sequence) change
        from src_table
      )
    )
  ) t
)
where flag != '' 
and actions.actions_count = 3
# order by user,sequence

如果应用于您问题中的样本数据 - 输出为

注意：上述解决方案适用于任意数量的相邻操作（无重复） - 您只需要在两个不同的位置更改它（2 和 3）

over(partition by user order by grp desc range between current row and 2 following) arr    

和

and actions.actions_count = 3   

,

您可以使用 RANK() 对重复项进行排序，然后过滤 RANK() = 1 以获取每个重复项的第一个（或最后一个）。那么问题就归结为你提到的另一个问题。

,

这与您之前的查询类似。如果我假设具有相同动作的相邻行最多有一个标志，那么我们可以使用间隙和岛屿方法。 . .然后滞后。

第一步是：

select user,min(sequence) as seqnuence,action,max(flag) as flag
from (select t.*,row_number() over (partition by user order by sequence) as seqnum
      from t
     ) t
group by user,sequence - seqnum;

然后，以此作为“基础”数据，我们可以使用滞后：

with cte as (
      select user,max(flag) as flag
      from (select t.*,row_number() over (partition by user order by sequence) as seqnum
            from t
           ) t
      group by user,sequence - seqnum
     )
select user,prev_action,prev_action_2,flag
from (select t.*,lag(action) over (partition by user order by sequence) as prev_action,lag(action,2) over (partition by user order by sequence) as prev_action2
      from t
     ) t
where prev_action is not null;

如果具有相同活动的用户可以有不同的标志，如果您能提出一个新问题，我将不胜感激。在新问题中，如果您包含 SELECT 语句来生成正在使用的示例数据，将会很有帮助。