考虑重叠活动计算用户的“忙碌时长”答案

【问题标题】：Calculating users' "busy duration" taking into account overlapping activities考虑重叠活动计算用户的“忙碌时长”
【发布时间】：2020-12-13 19:57:57
【问题描述】：

我一直在尝试解决这个问题，但到目前为止还没有解决。我正在使用 Oracle。

我有一组如下所示的数据：

| USER   | ACTIVITY   | START_TIME      | END_TIME        | DURATION |
|--------|------------|-----------------|-----------------|----------|
| jsmith | Front Desk | 2020-08-24 8:00 | 2020-08-24 9:30 | 90       |
| jsmith | Phones     | 2020-08-24 8:15 | 2020-08-24 8:45 | 30       |
| jsmith | Phones     | 2020-08-24 9:45 | 2020-08-24 9:50 | 5        |
| bjones | Phones     | 2020-08-24 9:00 | 2020-08-24 9:10 | 10       |
| bjones | Front Desk | 2020-08-24 9:05 | 2020-08-24 9:15 | 10       |
| bjones | Phones     | 2020-08-24 9:15 | 2020-08-24 9:45 | 30       |

上面的输出可以从以下查询生成：

SELECT
    USER,
    ACTIVITY,
    START_TIME,
    END_TIME,
    DURATION
FROM USER_ACTIVITIES
    WHERE USER IN ('jsmith', 'bjones')
    AND START_TIME BETWEEN '2020-08-24 00:00:00' AND '2020-08-25 00:00:00'
    ORDER BY USER, START_TIME, END_TIME
;

考虑到一些活动相互重叠，我需要计算每个用户的总“忙碌”时间。使用现有查询，我将得到 jsmith 的每个用户的总持续时间为 125，bjones 的总持续时间为 50，但是由于某些活动重叠，这并不能反映用户忙碌的总时间。

我正在寻找的输出是用户每天的总忙碌时长：

| USER   | DATE       | DURATION |
|--------|------------|----------|
| jsmith | 2020-08-24 | 95       |
| bjones | 2020-08-24 | 45       |

对此的任何帮助将不胜感激。

【问题讨论】：

标签： sql oracle datetime window-functions gaps-and-islands

【解决方案1】：

您可以先取消旋转分钟，然后通过使用 NOT EXISTS 来免除非重叠间隔（由于这种情况，我没有考虑天间隔，如果需要，您可以添加 EXTRACT( hour FROM max_end_time - min_start_time )*3600其他计算案例)

WITH  t AS
(  
SELECT "user" , MIN(start_time) AS min_start_time, MAX(end_time) AS max_end_time
  FROM user_activities
 GROUP BY "user" 
), t2 AS
(
 SELECT "user", min_start_time + NUMTODSINTERVAL(level, 'minute') AS minutes
   FROM t  
CONNECT BY level <= EXTRACT( hour FROM max_end_time - min_start_time )*60 + 
                    EXTRACT( minute FROM max_end_time - min_start_time ) 
    AND PRIOR SYS_GUID() IS NOT NULL
    AND PRIOR "user" = "user"
)
SELECT "user", COUNT(*) AS "Duration"
  FROM t2  
 WHERE EXISTS ( SELECT * 
                  FROM user_activities 
                 WHERE minutes BETWEEN start_time and end_time
                   AND "user" = t2."user" )
 GROUP BY "user"

Demo

【讨论】：

【解决方案2】：

我会用间隙和孤岛技术而不是递归来解决这个问题：

select usr, sum(duration) * 24 * 60 duration 
from (
    select usr, max(end_time) - min(start_time) duration
    from (
        select 
            ua.*,
            sum(case when start_time <= lag_end_time then 0 else 1 end) over(partition by usr order by start_time) grp
        from (
            select 
                ua.*,
                lag(end_time) over(partition by usr order by start_time) lag_end_time
            from user_activities ua
        ) ua
    ) ua
    group by usr, grp
) ua
group by usr

这个想法是使用窗口总和来构建具有相同用户和重叠时期的记录组。然后，您可以获取每个“岛”的结束和开始之间的差异，最后按用户聚合。

【讨论】：

此解决方案是否适用于您必须回顾不止一行的情况？例如。事件一：1:00 到 2:00，事件二：1:10 到 1:20，事件三：1:30 到 1:45 在这种情况下，事件 3 会被排除在总数之外吗？

【解决方案3】：

以下代码至少需要 12c：

WITH user_activities( "user", activity, start_time, end_time ) AS
(
  SELECT 'jsmith', 'Front Desk', timestamp'2020-08-24 08:00:00' , timestamp'2020-08-24 09:30:00' FROM dual UNION ALL
  SELECT 'jsmith', 'Phones'    , timestamp'2020-08-24 08:15:00' , timestamp'2020-08-24 08:45:00' FROM dual UNION ALL
  SELECT 'jsmith', 'Phones'    , timestamp'2020-08-24 09:45:00' , timestamp'2020-08-24 09:50:00' FROM dual UNION ALL
  SELECT 'bjones', 'Phones'    , timestamp'2020-08-24 09:00:00' , timestamp'2020-08-24 09:10:00' FROM dual UNION ALL
  SELECT 'bjones', 'Front Desk', timestamp'2020-08-24 09:05:00' , timestamp'2020-08-24 09:15:00' FROM dual UNION ALL
  SELECT 'bjones', 'Phones'    , timestamp'2020-08-24 09:15:00' , timestamp'2020-08-24 09:45:00' FROM dual 
)
select  "user", sum(durations) as durations
from     
(
    select "user",  extract(hour from (end_time - start_time)) * 60 + extract(minute from (end_time - start_time)) as durations
    from    user_activities 
    match_recognize
    (
        partition by "user"
        order by     start_time, end_time
        measures     first(start_time) start_time, max(end_time) as end_time
        pattern      (a* b)
        define       a as max(end_time) >= next(start_time)
    ) 
)
group by "user";

如果您对 match_recognize 感兴趣，这应该可以解决您的问题

输出：

【讨论】：

当我使用数据运行此查询多天时，结果仍然返回示例数据中的第一个日期。有没有办法修改它以返回每天的结果？

【解决方案4】：

许多可能的解决方案。这是另一个：使用 CTE，首先使用 LEAD 函数计算干净的结束时间（如果后续开始时间早于结束时间，则取后续开始时间）。然后按用户汇总和分组：

WITH sampledata (username,activity,start_time,end_time)
AS
(
  SELECT 'jsmith', 'Front Desk' ,'2020-08-24 8:00','2020-08-24 9:30' FROM DUAL UNION ALL
  SELECT 'jsmith', 'Phones'     ,'2020-08-24 8:15','2020-08-24 8:45' FROM DUAL UNION ALL
  SELECT 'jsmith', 'Phones'     ,'2020-08-24 9:45','2020-08-24 9:50' FROM DUAL UNION ALL
  SELECT 'bjones', 'Phones'     ,'2020-08-24 9:00','2020-08-24 9:10' FROM DUAL UNION ALL
  SELECT 'bjones', 'Front Desk' ,'2020-08-24 9:05','2020-08-24 9:15' FROM DUAL UNION ALL
  SELECT 'bjones', 'Phones'     ,'2020-08-24 9:15','2020-08-24 9:45' FROM DUAL  
), clean_sampledata (username,activity,start_time,end_time)
AS
(
SELECT 
  username,
  activity,
  TO_DATE(start_time,'YYYY-MM-DD HH24:MI'),
  TO_DATE(end_time,'YYYY-MM-DD HH24:MI')
  FROM sampledata
), clear_overlapped (username,activity,start_time,clean_end_time)
AS
(
SELECT 
  username,
  activity,
  start_time,
  NVL(LEAST(LEAD(start_time)  OVER (PARTITION BY username ORDER BY start_time),end_time),end_time)
  FROM clean_sampledata
), cleaned_minutes_per_username (username,mins)
AS
(
SELECT 
  username,
  ROUND((clean_end_time - start_time) * 1440)
  FROM clear_overlapped
)
SELECT
  username,
  SUM(mins)
  FROM cleaned_minutes_per_username
  GROUP BY username ;

bjones  45
jsmith  50

【讨论】：