Get running time from table with start / stop event datetime only - sql

Question

I need your help to get the total running duration per day from a table when I record only start and stop events:
id
ts
event
1
2020-12-26 09:00:00.589016
0
2
2020-12-26 10:25:00.589016
1
3
2020-12-26 19:30:45.644092
0
4
2020-12-26 22:30:00.554092
1
0 = stop event
1 = start event
The difficulty here is to compute the duration between start and stop events but also:
if a start event is the day before, include the duration between midnight and the first start event (in this example 9h)
Any idea to achieve it ?

Answer 1

Assuming your Times are already in a datetime_64 format as shown below:
ts event
id
1 2020-12-25 23:55:09.589016 1
2 2020-12-26 00:05:18.589016 0
3 2020-12-26 09:00:00.589016 1
4 2020-12-26 10:25:00.589016 0
5 2020-12-26 19:30:45.644092 1
6 2020-12-26 22:30:00.554092 0
You can do the following:
dfs = df.loc[df.event == 1]
dfs = dfs.rename(columns={"ts": "Start"})
dfs.reset_index(drop= True, inplace=True)
dfnd = df.loc[df.event==0]
dfnd = dfnd.rename(columns={"ts": "Stop"})
dfnd.reset_index(drop= True, inplace=True)
dfdur = dfnd.Stop - dfs.Start
Which Yields the following:
0 0 days 00:10:09
1 0 days 01:25:00
2 0 days 02:59:14.910000

Answer 2

For each row with event = 0 and non existing previous row of the same day with event = 1 create another row with ts at midnight of the same day.
Similarly for each row with event = 1 and non existing next row of the same day with event = 0 create another row with ts at 23:59:59.99999 of the same day.
This can be done in a CTE.
Then use window function LAG() for each row with event = 0 to get the starting time and with strftime() calculate the difference and finally aggregate on all the differences of each day:
WITH cte AS (
SELECT ts, event FROM tablename
UNION ALL
SELECT datetime(date(t.ts)), 1
FROM tablename t
WHERE event = 0 AND NOT EXISTS (SELECT 1 FROM tablename WHERE event = 1 AND date(ts) = date(t.ts) AND ts < t.ts)
UNION ALL
SELECT date(t.ts) || ' 23:59:59.999999', 0
FROM tablename t
WHERE event = 1 AND NOT EXISTS (SELECT 1 FROM tablename WHERE event = 0 AND date(ts) = date(t.ts) AND ts > t.ts)
)
SELECT date(ts) date,
SUM(strftime('%s', ts) - strftime('%s', prev_ts)) total
FROM (
SELECT *, LAG(ts) OVER (ORDER BY ts) prev_ts
FROM cte
)
WHERE event = 0
GROUP BY date
You will get the total per day in seconds.
If you want better accuracy you can use the function julianday() instead of strftime():
..............................
SELECT date(ts) date,
SUM(julianday(ts) - julianday(prev_ts)) * 24 * 3600 total
..............................
Or, a more efficient way:
WITH cte AS (
SELECT *,
LAG(ts, 1, date(ts)) OVER (PARTITION BY date(ts) ORDER BY ts) start_ts,
event = 1 AND LEAD(ts) OVER (PARTITION BY date(ts) ORDER BY ts) IS NULL flag
FROM tablename
)
SELECT date(ts) date,
SUM(
CASE flag
WHEN 0 THEN strftime('%s', ts) - strftime('%s', start_ts)
WHEN 1 THEN strftime('%s', date(ts) || ' 23:59:59.999999') - strftime('%s', ts)
END
) total
FROM cte
WHERE event = 0 OR flag = 1
GROUP BY date
Note that this code works only if all datetimes are in the format YYYY-MM-DD hh:mm:ss.ssssss (I noticed that in your sample data there is a value that is not of that format: '2020-12-26 9:00:00.589016').
See the demo.
Results:
> date | total
> :--------- | ----:
> 2020-12-26 | 70544

Answer 3

You can find the difference between the start and stop times for each interval, and then sum the latter result, grouped by the day:
with _events as (select row_number() over (order by t1.id) r, substr(t1.ts, 0, instr(t1.ts, " ")) day, t1.* from test t1),
events as (select (select sum(e2.event = 1 and e2.r < e1.r) from _events e2) c, e1.* from _events e1)
select day_r.day, sum(diff) from (
select e3.day, (julianday(max(e3.ts)) - julianday(min(e3.ts)))*24*60*60 diff
from events e3
group by e3.c
)
day_r group by day_r.day;