如何根据时间戳列为一组行生成唯一 ID？

Question

我有下表：

Prospect_ID	Channel	Timestamp
23455	Digital	01/01/2020 10:00:00
23455	Digital	01/01/2020 10:14:10
23455	Call	01/01/2020 10:30:10
55869	Digital	02/01/2020 09:00:01
55869	Digital	02/01/2020 14:00:00
55869	Call	02/01/2020 14:10:00

我想生成一个填充有唯一 ID 的列，该 ID 将每个潜在客户 ID 的时间戳在 30 分钟内分组在一起。

输出应如下所示：

Prospect_ID	Channel	Timestamp	Timestamp_id
23455	Digital	01/01/2020 10:00:00	abc1
23455	Digital	01/01/2020 10:14:10	abc1
23455	Call	01/01/2020 10:30:10	abc1
55869	Digital	01/01/2020 10:31:01	abc2
55869	Digital	02/01/2020 14:00:00	abc3
55869	Call	02/01/2020 14:10:00	abc3

我为此使用 Snowflake DB。

提前致谢！

Answer 1

您可以使用lag() 和累积和：

select t.*,
       sum(case when prev_timestamp >= timestamp - interval '30 minute' then 0 else 1 end) over (partition by prospectid order by timestamp) as timestamp_id
from (select t.*,
             lag(timestamp) over (partition by prospectid order by timestamp) as prev_timestamp
      from t
     ) t;

如果您希望所有用户都有一个唯一编号，您可以调整逻辑：

select t.*,
       sum(case when prev_timestamp >= timestamp - interval '30 minute' then 0 else 1 end) over (order by prospectid, timestamp) as timestamp_id

Answer 2

您可以将分析函数 lag() 和 sum() 与 UDF 一起使用。

CREATE OR REPLACE FUNCTION timestamp_UDF() 
RETURNS table(PROSPECT_ID INTEGER, CHANNEL VARCHAR2(10),
MY_TIMESTAMP timestamp, timestamp_id VARCHAR2(10))
AS  
$$
  select t.prospect_id,t.channel, t.my_timestamp,
  'abc '||sum(case when prev_timestamp >= my_timestamp - interval '30 minute' then 0 else 1 end) over (order by my_timestamp) as timestamp_id
   from (select t.*,
   lag(my_timestamp) over (order by my_timestamp) as prev_timestamp
    from table_name as t
   ) t
$$
;

select * from table(timestamp_UDF());