LAG函数和GROUP BY [英] LAG function and GROUP BY

问题描述

我有一个这样的表

 event_id |          date          
----------+------------------------
  1703702 | 2013-06-25 07:50:57-04
  3197588 | 2013-06-25 07:51:57-04
 60894420 | 2013-06-25 07:52:57-04
 60894420 | 2013-06-25 07:53:57-04
   183503 | 2013-06-25 07:54:57-04
 63116743 | 2013-06-25 07:55:57-04
 63110451 | 2013-06-25 07:56:57-04
 63116743 | 2013-06-25 07:57:57-04
 63116743 | 2013-06-25 07:58:57-04

我想应用滞后功能

我想要这样的东西：

SELECT event_id, difference
FROM ( 
  SELECT event_id, date - lag(date) over (order by date) as
  difference FROM table GROUP BY event_id
) t;

但是我不能将GROUP BY与LAG功能一起使用。我想要类似以下结果：

I cannot however use GROUP BY with the LAG function. I'd like a result similar to the following:

63116743, {120, 60}
60894420, {60}
...
...

所以有120第一个ID的事件之间有60s的窗口，第二个id的事件之间有60s的窗口。

So there was a 120s and 60s window between the events for the first id, and a 60s window for the second id.

有没有办法做到这一点？只要我可以将其最终放入数组，输出格式就不太重要。我正在使用Postgres 9.1

Is there a way to do this? The output format is not too important as long as I can get it into an array in the end. I'm using Postgres 9.1

推荐答案

WITH diffs as (
    SELECT
        event_id,
        date - lag(date) over (partition BY event_id ORDER BY date) as difference
    FROM
        TABLE
)
SELECT
    event_id,
    array_agg( difference ) as all_diffs
FROM
    diffs
GROUP BY event_id;

应该工作。

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