id_person transaction internation_in internation_out
1 456465 2015-01-01 2015-02-01
2 564564 2015-02-03 2015-04-02
3 4564654 2015-01-01 2015-01-05
4 4564646 2015-01-01 2015-02-04
4 4564656 2015-03-01 2015-04-15
4 87899465 2015-05-16 2015-05-25
5 56456456 2015-01-01 2105-01-08
5 45456546 2015-02-04 2015-03-04
I want to know how to group by id_person the difference (Interval in hours) between the internation_out from the first transaction with the internation_in of the next transaction.
I probe with lag and lead but I can't group by id_person
I Want this Result using id_person 4 for example
id_person transaction Gap
4 4564646 Null
4 4564656 The result of (2015-02-04- 2015-03-01)
4 87899465 The result of (2015-04-15- 2015-05-16)
If your time periods are not overlapping (and yours are not), then there is a simple calculation for the gaps: it is the total number of days from the beginning to the end minus the total on each row. So, you don't need lead() or lag():
select id_person,
(case when count(*) > 1
then (max(internation_out) - min(internation_in) -
sum(internation_out - internation_in)
)
end) as gap_duration
from table t
group by id_person;
Note that this returns NULL if there is only one row for the person. If you want 0, then you don't need the case.
Related
I have a bunch of timestamps grouped by ID and type in the sample data shown below.
I would like to find overlapped time between start_time and end_time columns in seconds for each group of ID and between each lead and follower combinations. I would like to show the overlap time only for the first record of each group which will always be the "lead" type.
For example, for the ID 1, the follower's start and end times in row 3 overlap with the lead's in row 1 for 193 seconds (from 09:00:00 to 09:03:13). the follower's times in row 3 also overlap with the lead's in row 2 for 133 seconds (09:01:00 to 2020-05-07 09:03:13). That's a total of 326 seconds (193+133)
I used the partition clause to rank rows by ID and type and order them by start_time as a start.
How do I get the overlap column?
row# ID type start_time end_time rank. overlap
1 1 lead 2020-05-07 09:00:00 2020-05-07 09:03:34 1 326
2 1 lead 2020-05-07 09:01:00 2020-05-07 09:03:13 2
3 1 follower 2020-05-07 08:59:00 2020-05-07 09:03:13 1
4 2 lead 2020-05-07 11:23:00 2020-05-07 11:33:00 1 540
4 2 follower 2020-05-07 11:27:00 2020-05-07 11:32:00 1
5 3 lead 2020-05-07 14:45:00 2020-05-07 15:00:00 1 305
6 3 follower 2020-05-07 14:44:00 2020-05-07 14:44:45 1
7 3 follower 2020-05-07 14:50:00 2020-05-07 14:55:05 2
In your example, the times completely cover the total duration. If this is always true, you can use the following logic:
select id,
(sum(datediff(second, start_time, end_time) -
datediff(second, min(start_time), max(end_time)
) as overlap
from t
group by id;
To add this as an additional column, then either use window functions or join in the result from the above query.
If the overall time has gaps, then the problem is quite a bit more complicated. I would suggest that you ask a new question and set up a db fiddle for the problem.
Tried this a couple of way and got it to work.
I first joined 2 tables with individual records for each type, 'lead' and 'follower' and created a case statement to calculate max start time for each lead and follower start time combination and min end time for each lead and follower end time combination. Stored this in a temp table.
CASE
WHEN lead_table.start_time > follower_table.start_time THEN lead_table.start_time
WHEN lead_table.start_time < follower_table.start_time THEN patient_table.start_time_local
ELSE 0
END as overlap_start_time,
CASE
WHEN follower_table.end_time < lead_table.end_time THEN follower_table.end_time
WHEN follower_table.end_time > lead_table.end_time THEN lead_table.end_time
ELSE 0
END as overlap_end_time
Then created an outer query to lookup the temp table just created to find the difference between start time and end time for each lead and follower combination in seconds
select temp_table.id,
temp_table.overlap_start_time,
temp_table.overlap_end_time,
DATEDIFF_BIG(second,
temp_table.overlap_start_time,
temp_table.overlap_end_time) as overlap_time FROM temp_table
Wondering how to select from a table:
FIELDID personID purchaseID dateofPurchase
--------------------------------------------------
2 13 147 2014-03-21 00:00:00
3 15 165 2015-03-23 00:00:00
4 13 456 2018-03-24 00:00:00
5 1 133 2018-03-21 00:00:00
6 23 123 2013-03-22 00:00:00
7 25 456 2013-03-21 00:00:00
8 25 456 2013-03-23 00:00:00
9 22 456 2013-03-28 00:00:00
10 25 589 2013-03-21 00:00:00
11 82 147 1991-10-22 00:00:00
12 82 453 2003-03-22 00:00:00
I'd like to get a result table of two columns: weekday and the number of purchases of each weekday, but only count the distinct days of purchases if done by the same person on the same day - for example since personID 25 purchased two things on 2013-03-21, that should only count as one 'thursday' instead of 2.
Basically, if the personID and the dateofPurchase are the same for more than one row, only count it once is what I want.
Here is what I have currently: It does everything correctly except it will count the above scenario under the thursday twice, when I would only want to add one:
SELECT v.wkday as day, COUNT(*) as 'absences'
FROM dbo.AttendanceRecord pr CROSS APPLY
(VALUES (CASE WHEN DATEPART(WEEKDAY, date) IN (1, 7)
THEN 'Weekend'
ELSE DATENAME(WEEKDAY, date)
END)
) v(wkday)
GROUP BY v.wkday;
to clarify:
If an item is purchased for at least one puchaseID on a specific day they will be counted as purchased for that day, and do not need to be counted again for each new purchase ID on that day.
I think you want to count distinct persons, so that would be:
COUNT(DISTINCT personid) as absences
Note that single quotes are not appropriate around column aliases. If you need to escape them, use square braces.
EDIT:
If you want to count distinct person-days, then you can use:
COUNT(DISTINCT CONCAT(personid, ':', dateofpurchase) as absences
I am trying to calculate the churn rate from a data that has customer_id, group, date. The aggregation is going to be by id, group and date. The churn formula is (customers in previous cohort - customers in last cohort)/customers in previous cohort
customers in previous cohort refers to cohorts in before 28 days
customers in last cohort refers to cohorts in last 28 days
I am not sure how to aggregate them by date range to calculate the churn.
Here is sample data that I copied from SQL Group by Date Range:
Date Group Customer_id
2014-03-01 A 1
2014-04-02 A 2
2014-04-03 A 3
2014-05-04 A 3
2014-05-05 A 6
2015-08-06 A 1
2015-08-07 A 2
2014-08-29 XXXX 2
2014-08-09 XXXX 3
2014-08-10 BB 4
2014-08-11 CCC 3
2015-08-12 CCC 2
2015-03-13 CCC 3
2014-04-14 CCC 5
2014-04-19 CCC 4
2014-08-16 CCC 5
2014-08-17 CCC 3
2014-08-18 XXXX 2
2015-01-10 XXXX 3
2015-01-20 XXXX 4
2014-08-21 XXXX 5
2014-08-22 XXXX 2
2014-01-23 XXXX 3
2014-08-24 XXXX 2
2014-02-25 XXXX 3
2014-08-26 XXXX 2
2014-06-27 XXXX 4
2014-08-28 XXXX 1
2014-08-29 XXXX 1
2015-08-30 XXXX 2
2015-09-31 XXXX 3
The goal is to calculate the churn rate every 28 days in between 2014 and 2015 by the formula given above. So, it is going to be aggregating the data by rolling it by 28 days and calculating the churn by the formula.
Here is what I tried to aggregate the data by date range:
SELECT COUNT(distinct customer_id) AS count_ids, Group,
DATE_SUB(CAST(Date AS DATE), INTERVAL 56 DAY) AS Date_min,
DATE_SUB(CURRENT_DATE, INTERVAL 28 DAY) AS Date_max
FROM churn_agg
GROUP BY count_ids, Group, Date_min, Date_max
Hope someone will help me with aggregation and churn calculation. I want to simply deduct the aggregated count_ids to deduct it from the next aggregated count_ids which is after 28 days. So this is going to be successive deduction of the same column value (count_ids). I am not sure if I have to use rolling window or simple aggregation to find the churn.
As corrected by #jarlh, it's not 2015-09-31 but 2015-09-30
You can use this to create 28 days calendar:
create table daysby28 (i int, _Date date);
insert into daysby28 (i, _Date)
SELECT i, cast('01-01-2014'as date) + i*INTERVAL '28 day'
from generate_series(0,50) i
order by 1;
After you use #jarlh churn_agg table creation he sent with the fiddle, with this query, you get what you want:
with cte as
(
select count(Customer) as TotalCustomer, Cohort, CohortDateStart From
(
select distinct a.Customer_id as Customer, b.i as Cohort, b._Date as CohortDateStart
from churn_agg a left join daysby28 b on a._Date >= b._Date and a._Date < b._Date + INTERVAL '28 day'
) a
group by Cohort, CohortDateStart
)
select a.CohortDateStart,
1.0*(b.TotalCustomer - a.TotalCustomer)/(1.0*b.TotalCustomer) as Churn from cte a
left join cte b on a.cohort > b.cohort
and not exists(select 1 from cte c where c.cohort > b.cohort and c.cohort < a.cohort)
order by 1
The fiddle of all together is here
Is there a way to find the solution so that I need for 2 days, there are 2 UD's because there are June 24 2 times and for the rest there are single days.
I am showing the expected output here:
Primary key UD Date
-------------------------------------------
1 123 2015-06-24 00:00:00.000
6 456 2015-06-24 00:00:00.000
2 123 2015-06-25 00:00:00.000
3 658 2015-06-26 00:00:00.000
4 598 2015-06-27 00:00:00.000
5 156 2015-06-28 00:00:00.000
No of times Number of days
-----------------------------
4 1
2 2
The logic is 4 users are there who used the application on 1 day and there are 2 userd who used the application on 2 days
You can use two levels of aggregation:
select cnt, count(*)
from (select date, count(*) as cnt
from t
group by date
) d
group by cnt
order by cnt desc;
I have a database with the following data:
Group ID Time
1 1 16:00:00
1 2 16:02:00
1 3 16:03:00
2 4 16:09:00
2 5 16:10:00
2 6 16:14:00
I am trying to find the difference in times between the consecutive rows within each group. Using LAG() and DATEDIFF() (ie. https://stackoverflow.com/a/43055820), right now I have the following result set:
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 00:06:00
2 5 00:01:00
2 6 00:04:00
However I need the difference to reset when a new group is reached, as in below. Can anyone advise?
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 NULL
2 5 00:01:00
2 6 00:04:00
The code would look something like:
select t.*,
datediff(second, lag(time) over (partition by group order by id), time)
from t;
This returns the difference as a number of seconds, but you seem to know how to convert that to a time representation. You also seem to know that group is not acceptable as a column name, because it is a SQL keyword.
Based on the question, you have put group in the order by clause of the lag(), not the partition by.