窗口函数在逻辑标准上增加列

Question

我正在寻找一种方法来根据以下标准添加与用户无关的“会话”列。

1) 每个用户定义自己的会话

2) +10 分钟的时间间隔为给定用户启动一个新会话。

因此对于输入：

| user_id | datetime_col     |
------------------------------
|1        |  01/01/2020 13:00|
|1        |  01/01/2020 13:01|
|1        |  01/01/2020 13:02|
|1        |  01/01/2020 13:20|
|1        |  01/01/2020 13:21|
|1        |  01/01/2020 13:22|
|1        |  01/01/2020 13:23|
|2        |  01/01/2020 13:00|
|2        |  01/01/2020 13:01|
|2        |  01/01/2020 13:02|
|2        |  01/01/2020 13:03|
|2        |  01/01/2020 13:04|
|3        |  01/01/2020 13:00|
|3        |  01/01/2020 13:01|
|3        |  01/01/2020 13:02|
|3        |  01/01/2020 13:03|
|3        |  01/01/2020 13:04|

我想要以下输出：

| user_id | datetime_col     | seesion_id|
------------------------------------------
|1        |  01/01/2020 13:00|     0     |
|1        |  01/01/2020 13:01|     0     |
|1        |  01/01/2020 13:02|     0     |
|1        |  01/01/2020 13:20|     1     |
|1        |  01/01/2020 13:21|     1     |
|1        |  01/01/2020 13:22|     1     |
|1        |  01/01/2020 13:23|     1     |
|2        |  01/01/2020 13:00|     2     |
|2        |  01/01/2020 13:01|     2     |
|2        |  01/01/2020 13:02|     2     |
|2        |  01/01/2020 13:03|     2     |
|2        |  01/01/2020 13:04|     2     |
|3        |  01/01/2020 13:00|     3     |
|3        |  01/01/2020 13:01|     3     |
|3        |  01/01/2020 13:02|     3     |
|3        |  01/01/2020 13:03|     3     |
|3        |  01/01/2020 13:04|     3     |

从概念上讲，我想计算每个用户每行之间的时间差，然后在每次 user_id 更改或时间滞后大于 10 分钟时递增。我可以做类似的事情

SELECT *,
DATE_PART('minutes', datetime_col - LAG(datetime_col, 1) OVER (PARTITION BY user_id ORDER BY datetime_col)) AS grp
FROM Table1

然后得到

| user_id | datetime_col     | grp       |
------------------------------------------
|1        |  01/01/2020 13:00|   (null)  |
|1        |  01/01/2020 13:01|     1     |
|1        |  01/01/2020 13:02|     1     |
|1        |  01/01/2020 13:20|    18     |
|1        |  01/01/2020 13:21|     1     |
|1        |  01/01/2020 13:22|     1     |
|1        |  01/01/2020 13:23|     1     |
|2        |  01/01/2020 13:00|   (null)  |
|2        |  01/01/2020 13:01|     1     |
|2        |  01/01/2020 13:02|     1     |
|2        |  01/01/2020 13:03|     1     |
|2        |  01/01/2020 13:04|     1     |
|3        |  01/01/2020 13:00|   (null)  |
|3        |  01/01/2020 13:01|     1     |
|3        |  01/01/2020 13:02|     1     |
|3        |  01/01/2020 13:03|     1     |
|3        |  01/01/2020 13:04|     1     |

但是从这里我被难住了，我该如何解决这个问题？

Answer 1

您可以使用lag() 和累积总和计算每个用户的单独会话：

select t.*,
       count(*) filter (where prev_dt < datetime_col - interval '10 minute') over (partition by user_id order by datetime_col) as session_id
from (select t.*,
             lag(datetime_col) over (partition by user_id order by datetime_col) as prev_dt
      from table1 t
     ) t;

滞后获取上一个日期。然后，仅计算相隔超过 10 分钟的转换以获得session_id。

这对我来说最有意义。

但您似乎想先按USER_ID 累积这些，然后按日期/时间（总体而言，按日期/时间对我来说更有意义）。这看起来很奇怪，但你可以使用dense_rank()：

with s as (
      select t.*,
             count(*) filter (where prev_dt < datetime_col - interval '10 minute') over (partition by user_id order by datetime_col) as session_id
      from (select t.*,
                   lag(datetime_col) over (partition by user_id order by datetime_col) as prev_dt
            from table1 t
           ) t
     )
select s.*,
        dense_rank() over (order by user_id, session_id) - 1 as new_session_id
from  s
order by user_id, datetime_col;

Here 是一个 dbfiddle。