在考虑重叠活动的情况下计算用户 "busy duration"

Question

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

我有一组数据如下所示：

| 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       |

以上输出可以通过以下查询生成：

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
;

我需要计算每个用户的总“忙碌”时间，同时考虑到一些活动相互重叠。使用现有查询，我将得到每个用户的总持续时间，jsmith 为 125，bjones 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持续时间：

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

如有任何帮助，我们将不胜感激。

Answer 1

您可以先对分钟进行逆轴旋转，然后通过使用 NOT EXISTS 免除 non-overlapping 间隔（由于这种情况，我没有考虑天间隔，您如果需要其他计算案例，可以添加 EXTRACT( hour FROM max_end_time - min_start_time )*3600 )

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

Answer 2

我会用 gaps-and-islands 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

我们的想法是使用 window 总和来构建具有相同用户和重叠时间段的记录组。然后，您可以计算每个“岛”的终点和起点之间的差异，最后对每个用户进行汇总。

Answer 3

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）。然后按用户求和分组：

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

Answer 4

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：

在考虑重叠活动的情况下计算用户 "busy duration"

Calculating users' "busy duration" taking into account overlapping activities

sql

oracle

datetime

window-functions

gaps-and-islands