Hopefully this will make sense. In short, what I have is a pt with multiple encounters. Starting with the first encounter (and is always included) and then including the next encounter if within 4 hrs.
If the next encounter does not meet criteria then all other observations will not be included in the output-
The code below shows the problem. It outputs rows 1,2, and 4. I want rows 1&2 but not 4.
Any tips appreciated on this
TIA
With Base as
(select 123 as ID, 12345 as enc_id, TO_DATE('2019-07-01 13:27:18', 'YYYY-MM-DD HH24:MI:SS') as dt from dual union
select 123 as ID, 12346 as enc_id, TO_DATE('2019-07-01 16:27:18', 'YYYY-MM-DD HH24:MI:SS') as dt from dual union
select 123 as ID, 12347 as enc_id, TO_DATE('2019-07-02 16:27:18', 'YYYY-MM-DD HH24:MI:SS') as dt from dual union
select 123 as ID, 12348 as enc_id, TO_DATE('2019-07-02 18:27:18', 'YYYY-MM-DD HH24:MI:SS') as dt from dual)
select * from (select ID,ENC_ID,dt,row_number() over (partition by ID order by DT) RK,
lag(dt) over (partition by ID order by dt) prev_dt,
(DT-lag(dt) over (partition by ID order by dt))*24 as time_dif_hrs from base) where RK=1 or TIME_DIF_HRS<4
You can use another analytical function sum as follows:
Select ID,ENC_ID,dt from
(select ID,ENC_ID,dt,rk,
sum(case when (date - prev_date)* 24 < 4 then 0 else 1 end)
over( partition by ID order by DT) as cond_met_running
from (select ID,ENC_ID,dt,
row_number() over (partition by ID order by DT) RK,
lag(dt) over (partition by ID order by dt) prev_dt
from base)
)
) Where rk = 1 or cond_met_running = 0
Related
I'm trying to build a query for the following scenario,
Group records by license ID and get min and max dates
For a given license ID, if there are two earliest start dates, then start date of the particular ID has to be updated as latest start date in that grouping.
Since I'm new to sql, I need help to satisfy condition 2. Any help is greatly appreciated. Thanks
Actual data
LicenseID
StartDate
EndDate
100
4/3/2000
3/1/2013
100
4/3/2000
2/2/2017
100
3/1/2013
1/23/2015
100
1/23/2015
2/2/2017
100
2/2/2017
2/9/2018
100
2/2/2017
12/18/2018
100
12/18/2018
2/16/2021
Expected output
LicenseID
StartDate
EndDate
100
12/18/2018
2/16/2021
Here's one option; read comments within code.
Sample data:
SQL> with test (id, start_date, end_date) as
2 (select 100, date '2000-04-03', date '2013-03-01' from dual union all
3 select 100, date '2000-04-03', date '2017-02-02' from dual union all
4 select 100, date '2018-12-18', date '2021-02-16' from dual
5 ),
Query begins here:
6 -- rank start dates per each ID
7 temp as
8 (select id,
9 min(start_date) over (partition by id) min_sd,
10 max(start_date) over (partition by id) max_sd,
11 rank() over (partition by id order by start_date) rnk_sd,
12 --
13 max(end_date) over (partition by id) max_ed
14 from test
15 ),
16 -- count number of the 1st start dates
17 temp2 as
18 (select id,
19 sum(case when rnk_sd = 1 then 1 else 0 end) cnt_sd
20 from temp
21 group by id
22 )
23 -- if number of the 1st start dates is 1, take MIN_SD. Otherwise, take MAX_SD
24 select distinct
25 b.id,
26 case when b.cnt_sd = 1 then a.min_sd else a.max_sd end start_date,
27 a.max_ed end_date
28 from temp2 b join temp a on a.id = b.id;
Result:
ID START_DATE END_DATE
---------- ---------- ----------
100 12/18/2018 02/16/2021
SQL>
This can filter them:
WITH sample_data AS
(
SELECT 100 AS LicenseID, TO_DATE('04/03/2000','MM/DD/YYYY') AS StartDate, TO_DATE('03/01/2013','MM/DD/YYYY') AS EndDate FROM DUAL UNION ALL
SELECT 100, TO_DATE('04/03/2000','MM/DD/YYYY'), TO_DATE('02/02/2017','MM/DD/YYYY') FROM DUAL UNION ALL
SELECT 100, TO_DATE('03/01/2013','MM/DD/YYYY'), TO_DATE('01/23/2015','MM/DD/YYYY') FROM DUAL UNION ALL
SELECT 100, TO_DATE('01/23/2015','MM/DD/YYYY'), TO_DATE('02/02/2017','MM/DD/YYYY') FROM DUAL UNION ALL
SELECT 100, TO_DATE('02/02/2017','MM/DD/YYYY'), TO_DATE('02/09/2018','MM/DD/YYYY') FROM DUAL UNION ALL
SELECT 100, TO_DATE('02/02/2017','MM/DD/YYYY'), TO_DATE('12/18/2018','MM/DD/YYYY') FROM DUAL UNION ALL
SELECT 100, TO_DATE('12/18/2018','MM/DD/YYYY'), TO_DATE('02/16/2021','MM/DD/YYYY') FROM DUAL
)
SELECT dat.licenseID, CASE WHEN dups.licenseID IS NOT NULL THEN MAX(StartDate)
ELSE MIN(StartDate)
END,
CASE WHEN dups.licenseID IS NOT NULL THEN MAX(EndDate)
ELSE MIN(EndDate)
END
FROM sample_data dat
LEFT OUTER JOIN (SELECT COUNT(1), sd.LicenseID
FROM sample_data sd
INNER JOIN (SELECT MIN(StartDate) AS StartDate, LicenseID
FROM sample_data
GROUP BY LicenseID) mins
ON sd.LicenseID = mins.LicenseID AND sd.startDate = mins.StartDate
GROUP BY sd.LicenseID
HAVING COUNT(1) > 1) dups
ON dups.LicenseID = dat.licenseID
GROUP BY dat.licenseID, dups.licenseID;
You can use:
SELECT licenseid,
MAX(startdate) AS startdate,
MAX(enddate) KEEP (DENSE_RANK LAST ORDER BY startdate) AS enddate
FROM table_name
GROUP BY licenseid
HAVING COUNT(*) KEEP (DENSE_RANK FIRST ORDER BY startdate) > 1;
or:
SELECT licenseid,
max_startdate AS startdate,
max_enddate As enddate
FROM (
SELECT licenseid,
RANK()
OVER (PARTITION BY licenseid ORDER BY startdate) AS rnk,
ROW_NUMBER()
OVER (PARTITION BY licenseid, startdate ORDER BY enddate) AS rn,
MAX(startdate)
OVER (PARTITION BY licenseid) AS max_startdate,
MAX(enddate)
KEEP (DENSE_RANK LAST ORDER BY startdate)
OVER (PARTITION BY licenseid) AS max_enddate
FROM table_name t
)
WHERE rnk = 1
AND rn = 2;
Which, for the sample data:
CREATE TABLE table_name (licenseid, startdate, enddate) AS
SELECT 100, DATE'2000-04-03', DATE'2013-03-01' FROM DUAL UNION ALL
SELECT 100, DATE'2000-04-03', DATE'2017-02-02' FROM DUAL UNION ALL
SELECT 100, DATE'2013-03-01', DATE'2015-01-23' FROM DUAL UNION ALL
SELECT 100, DATE'2015-01-23', DATE'2017-02-02' FROM DUAL UNION ALL
SELECT 100, DATE'2017-02-02', DATE'2018-02-09' FROM DUAL UNION ALL
SELECT 100, DATE'2018-02-02', DATE'2018-12-18' FROM DUAL UNION ALL
SELECT 100, DATE'2018-12-18', DATE'2021-02-16' FROM DUAL;
Both output:
LICENSEID
STARTDATE
ENDDATE
100
2018-12-18 00:00:00
2021-02-16 00:00:00
If you do want to perform an UPDATE of that second row then:
MERGE INTO table_name dst
USING (
SELECT ROWID AS rid,
max_startdate,
max_enddate
FROM (
SELECT RANK()
OVER (PARTITION BY licenseid ORDER BY startdate) AS rnk,
ROW_NUMBER()
OVER (PARTITION BY licenseid, startdate ORDER BY enddate) AS rn,
MAX(startdate)
OVER (PARTITION BY licenseid) AS max_startdate,
MAX(enddate)
KEEP (DENSE_RANK LAST ORDER BY startdate)
OVER (PARTITION BY licenseid) AS max_enddate
FROM table_name t
)
WHERE rnk = 1
AND rn = 2
)src
ON (src.rid = dst.ROWID)
WHEN MATCHED THEN
UPDATE
SET startdate = src.max_startdate,
enddate = src.max_enddate;
db<>fiddle here
The problem I am facing is how to find distinct time periods from multiple time periods with overlap in Teradata ANSI SQL.
For example, the attached tables contain multiple overlapping time periods, how can I combine those time periods into 3 unique time periods in Teradata SQL???
I think I can do it in python with the loop function, but not sure how to do it in SQL
ID
Start Date
End Date
001
2005-01-01
2006-01-01
001
2005-01-01
2007-01-01
001
2008-01-01
2008-06-01
001
2008-04-01
2008-12-01
001
2010-01-01
2010-05-01
001
2010-04-01
2010-12-01
001
2010-11-01
2012-01-01
My expected result is:
ID
start_Date
end_date
001
2005-01-01
2007-01-01
001
2008-01-01
2008-12-01
001
2010-01-01
2012-01-01
From Oracle 12, you can use MATCH_RECOGNIZE to perform a row-by-row comparison:
SELECT *
FROM table_name
MATCH_RECOGNIZE(
PARTITION BY id
ORDER BY start_date
MEASURES
FIRST(start_date) AS start_date,
MAX(end_date) AS end_date
ONE ROW PER MATCH
PATTERN (overlapping_ranges* last_range)
DEFINE overlapping_ranges AS NEXT(start_date) <= MAX(end_date)
)
Which, for the sample data:
CREATE TABLE table_name (ID, Start_Date, End_Date) AS
SELECT '001', DATE '2005-01-01', DATE '2006-01-01' FROM DUAL UNION ALL
SELECT '001', DATE '2005-01-01', DATE '2007-01-01' FROM DUAL UNION ALL
SELECT '001', DATE '2008-01-01', DATE '2008-06-01' FROM DUAL UNION ALL
SELECT '001', DATE '2008-04-01', DATE '2008-12-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-01-01', DATE '2010-05-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-04-01', DATE '2010-12-01' FROM DUAL UNION ALL
SELECT '001', DATE '2010-11-01', DATE '2012-01-01' FROM DUAL;
Outputs:
ID
START_DATE
END_DATE
001
2005-01-01 00:00:00
2007-01-01 00:00:00
001
2008-01-01 00:00:00
2008-12-01 00:00:00
001
2010-01-01 00:00:00
2012-01-01 00:00:00
db<>fiddle here
Update: Alternative query
SELECT id,
start_date,
end_date
FROM (
SELECT id,
dt,
SUM(cnt) OVER (PARTITION BY id ORDER BY dt) AS grp,
cnt
FROM (
SELECT ID,
dt,
SUM(type) OVER (PARTITION BY id ORDER BY dt, ROWNUM) * type AS cnt
FROM table_name
UNPIVOT (dt FOR type IN (start_date AS 1, end_date AS -1))
)
WHERE cnt IN (1,0)
)
PIVOT (MAX(dt) FOR cnt IN (1 AS start_date, 0 AS end_date))
Or, an equivalent that does not use UNPIVOT, PIVOT or ROWNUM and works in both Oracle and PostgreSQL:
SELECT id,
MAX(CASE cnt WHEN 1 THEN dt END) AS start_date,
MAX(CASE cnt WHEN 0 THEN dt END) AS end_date
FROM (
SELECT id,
dt,
SUM(cnt) OVER (PARTITION BY id ORDER BY dt) AS grp,
cnt
FROM (
SELECT ID,
dt,
SUM(type) OVER (PARTITION BY id ORDER BY dt, rn) * type AS cnt
FROM (
SELECT r.*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY dt ASC, type DESC) AS rn
FROM (
SELECT id, 1 AS type, start_date AS dt FROM table_name
UNION ALL
SELECT id, -1 AS type, end_date AS dt FROM table_name
) r
) p
) s
WHERE cnt IN (1,0)
) t
GROUP BY id, grp
Update 2: Another Alternative
SELECT id,
MIN(start_date) AS start_date,
MAX(end_Date) AS end_date
FROM (
SELECT t.*,
SUM(CASE WHEN start_date <= prev_max THEN 0 ELSE 1 END)
OVER (PARTITION BY id ORDER BY start_date) AS grp
FROM (
SELECT t.*,
MAX(end_date) OVER (
PARTITION BY id ORDER BY start_date
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
) AS prev_max
FROM table_name t
) t
) t
GROUP BY id, grp
db<>fiddle Oracle PostgreSQL
This is a gaps and islands problem. Try this:
with u as
(select ID, start_date, end_date,
case
when start_date <= lag(end_date) over(partition by ID order by start_date, end_date) then 0
else 1 end as grp
from table_name),
v as
(select ID, start_date, end_date,
sum(grp) over(partition by ID order by start_date, end_date) as island
from u)
select ID, min(start_date) as start_Date, max(end_date) as end_date
from v
group by ID, island;
Fiddle
Basically you can identify "islands" by comparing start_date of current row to end_date of previous row (ordered by start_date, end_date), if it precedes it then it's the same island. Then you can do a rolling sum() to get the island numbers. Finally select min(start_date) and max(end_date) from each island to get the desired output.
This may work ,with little bit of change in function , I tried it in Dbeaver :
select ID,Start_Date,End_Date
from
(
select t.*,
dense_rank () over(partition by extract (year from Start_Date) order BY End_Date desc) drnk
from testing_123 t
) temp
where temp.drnk = 1
ORDER BY Start_Date;
Try this
WITH a as (
SELECT
ID,
LEFT(Start_Date, 4) as Year,
MIN(Start_Date) as New_Start_Date
FROM
TAB1
GROUP BY
ID,
LEFT(Start_Date, 4)
), b as (
SELECT
a.ID,
Year,
New_Start_Date,
End_Date
FROM
a
LEFT JOIN
TAB1
ON LEFT(a.New_Start_Date, 4) = LEFT(TAB1.Start_Date, 4)
)
select
ID,
New_Start_Date as Start_Date,
MAX(End_Date)
from
b
GROUP BY
ID,
New_Start_Date;
Example: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=97f91b68c635aebfb752538cdd752ace
I want to get a value from the previous row that matches a certain condition.
For example: here I want for each row to get the timestamp from the last event = 1.
I feel I can do it without joins with LAG and PARTITION BY with CASE but I am not able to crack it.
Please help.
Here is one approach using analytic functions:
WITH cte AS (
SELECT *, COUNT(CASE WHEN event = 1 THEN 1 END) OVER
(PARTITION BY customer_id ORDER BY ts) cnt
FROM yourTable
)
SELECT ts, customer_id, event,
MAX(CASE WHEN event = 1 THEN ts END) OVER
(PARTITION BY customer_id, cnt) AS desired_result
FROM cte
ORDER BY customer_id, ts;
Demo
We can articulate your problem by saying that your want the desired_result column to contain the most recent timestamp value when the event was 1. The count (cnt) in the CTE above computes a pseudo group of records for each time the event is 1. Then we simply do a conditional aggregation over customer and pseudo group to find the timestamp value.
One more approach with "one query":
with data as
(
select sysdate - 0.29 ts, 111 customer_id, 1 event from dual union all
select sysdate - 0.28 ts, 111 customer_id, 2 event from dual union all
select sysdate - 0.27 ts, 111 customer_id, 3 event from dual union all
select sysdate - 0.26 ts, 111 customer_id, 1 event from dual union all
select sysdate - 0.25 ts, 111 customer_id, 1 event from dual union all
select sysdate - 0.24 ts, 111 customer_id, 2 event from dual union all
select sysdate - 0.23 ts, 111 customer_id, 1 event from dual union all
select sysdate - 0.22 ts, 111 customer_id, 1 event from dual
)
select
ts, event,
last_value(case when event=1 then ts end) ignore nulls
over (partition by customer_id order by ts) desired_result,
max(case when event=1 then ts end)
over (partition by customer_id order by ts) desired_result_2
from data
order by ts
Edit: As suggested by MatBailie the max(case...) works as well and is a more general approach. The "last_value ... ignore nulls" is Oracle specific.
I want to be able to increment a value when a "separator row" is reached.
As an example; zero is the separator row.
GROUP_ID
--------
1
1
1
0
1
1
0
1
1
1
What I want is:
GROUP_ID
--------
1
1
1
0
2
2
0
3
3
3
Of course outlier scenarios like starting with zero, or consecutive zeros can exist also. I though NTILE might be useful as it has a similar idea of creating buckets.
select START_DATE, END_DATE, DTEDIF, GROUP_ID
FROM (select START_DATE,
END_DATE,
DTEDIF,
(CASE
WHEN DTEDIF <= 24 THEN
1
ELSE
0
END) GROUP_ID
FROM (select START_DATE,
END_DATE,
MONTHS_BETWEEN(END_DATE, START_DATE) as DTEDIF
from (select start_date,
LEAD(start_date, 1) OVER(order by start_date) as END_DATE
from (select sysdate start_date
from dual
union all
select add_months(sysdate, 12) dt
from dual --12
union all
select add_months(sysdate, 22) dt
from dual --10
union all
select add_months(sysdate, 40) dt
from dual --18
union all
select add_months(sysdate, 68) dt
from dual --28
union all
select add_months(sysdate, 70) dt
from dual --18
union all
select add_months(sysdate, 88) dt
from dual --18
union all
select add_months(sysdate, 118) dt
from dual --30
))));
Here is one method: Count the number of zeros before each row:
select t.*,
(case when group_id = 0 then group_id
else sum(case when group_id = 0 then 1 else 0 end) over (order by <ordering column>)
end) as new_group_id
from t;
SQL tables represent unordered sets. So your question requires a separate column to represent the ordering of the data.
You can achieve this by using combination of lag and sum analytical function.
I consider that there is ordering column available with name id.
Select group_id,
Sum(case when coalesce(lag_group_id,0) then 1 else 0 end)
over(order by id) as your_result from
(Select id,
group_id,
lag(group_id) over (order by id) as lag_group_id
from your_table)
Cheers!!
I have a table per_all_Assignments_f with date_from and date_to and following column structure :
PERSON_ID DATE_FROM DATE_TO GRADE
--------- ------------ ----------- -----
12 01-Jan-2018 28-Feb-2018 c
12 01-Mar-2018 29-Mar-2018 a
12 30-Mar-2018 31-dec-4712 b
13 01-jan-2018 31-dec-4712 c
In the above table, I have to retrieve the latest grade change i.e. for person_id '12', I have to retrieve both record rows : 30-mar-2018 to 31 dec 4712 being the latest and one prior row. What function can i use for this ?
solved by :
SELECT person_id,
asg.grade_id,
lag(asg.grade_id) Over (Partition By person_ID Order By start_date) as prev_ppg_line1,
lag(start_date) Over (Partition By person_ID Order By start_date)
as prev_ppg_effective_start_date,
start_date,
row_Number() Over (Partition By person_ID Order By effective_start_date) as rn
FROM asg_table asg
WHERE person_id = 12;
This query will fetch 3 rows with all the previous changes. I want to fetch the latest change only without using max on effective start date
You can use row_number and lead analytic functions together inside the subquery as :
select person_id, date_From, date_to, grade
from
(
with per_all_Assignments_f(person_id, date_From, date_to, grade) as
(
select 12,date'2018-01-01',date'2018-02-28','c' from dual union all
select 12,date'2018-03-01',date'2018-03-29','a' from dual union all
select 12,date'2018-03-30',date'4172-12-31','b' from dual union all
select 13,date'2018-01-01',date'4172-12-31','c' from dual
)
select t.*,
lead(grade) over (order by date_From desc) as ld,
row_number() over (order by date_From desc) as rn
from per_all_Assignments_f t
)
where rn <= 2
and grade != ld
order by rn desc;
PERSON_ID DATE_FROM DATE_TO GRADE
---------- ----------- ---------- -------
12 01.03.2018 29.03.2018 a
12 30.03.2018 31.12.4172 b
Rextester Demo
Seems like you just want all with a row_number() of 1 or 2 partitioned by the person and ordered by the beginning descending.
SELECT person_id,
date_from,
date_to,
grade
FROM (SELECT person_id,
date_from,
date_to,
grade,
row_number() OVER (PARTITION BY person_id
ORDER BY date_from DESC) rn
FROM per_all_assignments_f t) x
WHERE rn IN (1, 2)
ORDER BY person_id ASC,
date_from DESC;