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

Question

I've been trying to work out a solution to this problem but so far haven't been able to work it out. I'm using Oracle.
I have a set of data that looks like this:
| 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 |
The above output can be generated from the following query:
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
;
I need to calculate the total "busy" time per user, taking into account that some of the activities overlap each other. Using the existing query I'll get a total duration per user of 125 for jsmith and 50 for bjones, However since some of the activities overlapped this doesn't reflect the total amount of time the users were busy.
The output I'm looking for is the total busy duration per day by user:
| USER | DATE | DURATION |
|--------|------------|----------|
| jsmith | 2020-08-24 | 95 |
| bjones | 2020-08-24 | 45 |
Any help with this would be greatly appreciated.

Answer 1

You can unpivot the minutes first, and then exempt the non-overlapping interval through use of NOT EXISTS (I didn't consider the day interval due to this case, you can add EXTRACT( hour FROM max_end_time - min_start_time )*3600 if needed for other calculation cases )
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

I would address this with gaps-and-islands techniques rather than recursion:
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
The idea is to build groups of records having the same user and overlapping periods, using a window sum. You can then take the difference between the end and start of each "island", and finally aggregate per user.

Answer 3

The below code requires at least 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";
This should solve your problem if you are interested in match_recognize
output:

Answer 4

Many possible solutions. Here is another one: using CTE, first calculate the clean end time (if following start time is earlier than end time take following start time instead) using the LEAD function. Then sum and group by user:
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