I'm trying to add a column on this table and stuck for a little while
ID
Category 1
Date
Data1
A
1
2022-05-30
21
B
2
2022-05-21
15
A
2
2022-05-02
33
A
1
2022-02-11
3
B
2
2022-05-01
19
A
1
2022-05-15
null
A
1
2022-05-20
11
A
2
2022-04-20
22
to
ID
Category 1
Date
Data1
Picked_Data
A
1
2022-05-30
21
11
B
2
2022-05-21
15
19
A
2
2022-05-02
33
22
A
1
2022-02-11
3
some number or null
B
2
2022-05-01
19
some number or null
A
1
2022-05-15
null
some number or null
A
1
2022-05-20
11
some number or null
A
2
2022-04-20
22
some number or null
The logic is to partition by Category1 and ID then pick the latest none null value within the past 28 days. If there is no data exist, it'll be null
For the first row, ID = A and Category 1, it will pick 7th row as they are in the same category, ID and the date difference is <= 28. It skipped row 4th and 6th as the date is too far back and null value.
I've tried querying this by
select first_value(Data1) over (partition bty Category1 order by case when Data1 is not null and Date between Date - Inteverval 28 DAY and Date then 1 else 2) as Picked_Data
but it's picking incorrect rows,my guess is this query
Date between Date - Inteverval 28 DAY and Date
is not picking the correct date.. could anyone give me advise/suggestion how I could twick this query?
Consider below approach
select *,
first_value(data1 ignore nulls) over past_28_days as picked_data
from your_table
window past_28_days as (
partition by id, category_1
order by unix_date(date)
range between 29 preceding and 1 preceding
)
if applied to sample data in your question - output is
Consider below approach:
with sample_data as (
select 'A' as ID, 1 as category_1, date('2022-05-30') as date, 21 as data1,
union all select 'B' as ID, 2 as category_1, date('2022-05-21') as date, 15 as data1,
union all select 'A' as ID, 2 as category_1, date('2022-05-02') as date, 33 as data1,
union all select 'A' as ID, 1 as category_1, date('2022-02-11') as date, 3 as data1,
union all select 'B' as ID, 2 as category_1, date('2022-05-01') as date, 19 as data1,
union all select 'A' as ID, 1 as category_1, date('2022-05-15') as date, NULL as data1,
union all select 'A' as ID, 1 as category_1, date('2022-05-20') as date, 11 as data1,
union all select 'A' as ID, 2 as category_1, date('2022-04-20') as date, 22 as data1,
),
with_next_data as (
select *,
lag(date) over (partition by ID,category_1 order by date) as next_date,
lag(data1) over (partition by ID,category_1 order by date) as next_data,
from sample_data
)
select
id,
category_1,
date,
data1,
if(date_diff(date, next_date,day) <= 28, next_data, null) as picked_data
from with_next_data
Output:
Related
I was wondering if the following case is possible within BigQuery.
There are 2 tables of intervals. The intervals in a single table do not overlap with other intervals in the same table. The intervals however can overlap with intervals in the other table.
I want to merge the intervals, but also divide the intervals into multiple intervals if they overlap. So if for example the interval is in table A from 5/8/2020 - 5/9/2020 and there is an interval in B 18/8/2020 - 1/9/2020, then I want to split the interval as 5/8/2020 - 18/8/2020 (in A), 18/8/2020 - 1/9/2020 (in A and B) and 1/9/2020 - 5/9/2020 (in A).
A more extensive example: We have a table with intervals where people eat Apples
ID
StartDate
EndDate
1
01/01/19
01/04/19
2
01/01/19
03/01/19
And a table with intervals where people eat Bananas
ID
StartDate
EndDate
1
15/12/18
12/01/19
1
01/02/19
17/02/19
1
15/03/19
15/04/19
2
01/06/19
01/07/19
And now we want to combine those intervals and classify the intervals as either, apple eaters, banana eaters, or apple and banana eaters.
ID
StartDate
EndDate
type
1
15/12/18
01/01/19
B
1
01/01/19
12/01/19
AB
1
12/01/19
01/02/19
A
1
01/02/19
17/02/19
AB
1
17/02/19
15/03/19
A
1
15/03/19
01/04/19
AB
1
01/04/19
15/04/19
B
2
01/01/19
03/01/19
A
2
01/06/19
01/07/19
B
Is it possible to solve this with bigQuery?
Consider below query :
WITH stacked AS (
SELECT ID, date, STRING_AGG(type, '' ORDER BY type) type FROM (
SELECT *, 'A' type FROM Apples
UNION ALL
SELECT *, 'B' type FROM Bananas
), UNNEST (GENERATE_DATE_ARRAY(PARSE_DATE('%d/%m/%y', StartDate), PARSE_DATE('%d/%m/%y', EndDate), INTERVAL 1 DAY)) date
GROUP BY 1, 2
),
partitioned AS (
SELECT ID, date, type,
COUNTIF(flag) OVER w AS div,
type = 'AB' AND LEAD(type) OVER w <>'AB' in_AB,
type <> 'AB' AND LAG(type) OVER w = 'AB' out_AB,
type <> 'AB' AND LEAD(type, 1, 'A') OVER w <> 'AB' bw_AB,
FROM (
SELECT ID, date, type, type <> LAG(type) OVER (PARTITION BY ID ORDER BY date) AS flag
FROM stacked
)
WINDOW w AS (PARTITION BY ID ORDER BY date)
)
SELECT ID,
MIN(IF(out_AB, date - 1, date)) StartDate,
MAX(IF(in_AB or bw_AB, date, date + 1)) EndDate,
ANY_VALUE(type) type
FROM partitioned
GROUP BY ID, div
ORDER BY 1, 2;
With sample tables:
CREATE TEMP TABLE Apples AS
select 1 ID, '01/01/19' StartDate, '01/04/19'EndDate union all
select 2, '01/01/19', '03/01/19';
CREATE TEMP TABLE Bananas AS
select 1 ID, '15/12/18' StartDate, '12/01/19' EndDate union all
select 1, '01/02/19', '17/02/19' union all
select 1, '15/03/19', '15/04/19' union all
select 2, '01/06/19', '01/07/19';
Trying to get userid recent aggregate value for session_id.
(session_id 3 has two records, recent agg value is 80.00
session_id 4 has four records, recent agg value is 95.00
session_id 6 has three records, recent agg value is 72.00
Table:session_agg
id session_id userid agg date
-- ---------- ------ ----- -------
1 3 11 60.00 1573561586
4 3 11 80.00 1573561586
6 4 11 35.00 1573561749
7 4 11 50.00 1573561751
8 4 11 70.00 1573561912
10 4 11 95.00 1573561921
11 6 14 40.00 1573561945
12 6 14 67.00 1573561967
13 6 14 72.00 1573561978
select id, session_id, userid, agg, date from session_agg
WHERE date IN (select MAX(date) from session_agg GROUP BY session_id) AND
userid = 11
If you want to stick with your current approach, then you need to correlate the session_id in the subquery which checks for the max date for each session:
SELECT id, session_id, userid, add, date
FROM session_agg sa1
WHERE
date = (SELECT MAX(date) FROM session_agg sa2 WHERE sa2.session_id = sa1.session_id) AND
userid = 11;
But, if your version of SQL supports analytic functions, ROW_NUMBER is an easier way to do this:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY session_id ORDER BY date DESC) rn
FROM session_agg
)
SELECT id, session_id, userid, add, date
FROM cte
WHERE rn = 1;
I have a oracle table with the following format,
For eg:
JLID Dcode SID TDT QTY
8295783 3119255 9842 3/5/2018 14
8269771 3119255 9842 3/6/2018 11
8302211 3119255 1126 3/1/2018 19
Here I have different SID for the same Dcode, now I need to get the SID with the maximum Qty. (i.e) for SID 9842 - (14+11)=25, for SID 1126 it is 19, then the results should be on SID 9842. So, our query should returns the following results
JLID Dcode START_DT END_DT SID
111 3119255 3/1/2018 3/31/2018 12:00 9842
Startdate and enddate should be calculated from TDT (i.e) start date is the first date of the month and the end date is the last date of the month
Can anyone please suggest me some ideas to do it.
It might be as simple as this:
SELECT Dcode, start_date, end_date, SID FROM (
SELECT Dcode, SID, TRUNC(start_date, 'MONTH') AS start_date
, LAST_DAY(end_date) AS end_date
, ROW_NUMBER() OVER ( PARTITION BY Dcode ORDER BY total_qty DESC ) AS rn
FROM (
SELECT Dcode, SID, MIN(TDT) AS start_date, MAX(TDT) AS end_date
, SUM(QTY) AS total_qty
FROM mytable
GROUP BY Dcode, SID
)
) WHERE rn = 1
In the inner most subquery I aggregation to get the range of dates and total quantity for particular values of Dcode and SID. Then I use an anaylitic (window) function to get the row for which total quantity is the greatest. (You would want to use RANK() in place of ROW_NUMBER() in the event you want to return more than one value of SID with the same quantity.)
Here's one option which doesn't contain JLID = 111 in the final result as I have no idea where you took it from.
SQL> with test (jlid, dcode, sid, tdt, qty) as
2 (select 8295783, 3119255, 9842, date '2018-03-05', 14 from dual union
3 select 8269771, 3119255, 9842, date '2018-08-22', 11 from dual union
4 select 8302211, 3119255, 1126, date '2018-03-01', 19 from dual union
5 --
6 select 1234567, 1112223, 1000, date '2018-06-16', 88 from dual
7 )
8 select dcode,
9 min (trunc (tdt, 'mm')) start_dt, --> MIN
10 max (last_day (tdt)) end_dt, --> MAX
11 sid
12 from (select dcode,
13 sid,
14 tdt,
15 sqty,
16 rank () over (partition by dcode order by sqty desc) rnk
17 from (select dcode,
18 sid,
19 tdt,
20 sum (qty) over (partition by dcode, sid) sqty
21 from test))
22 where rnk = 1
23 group by dcode, sid; --> GROUP BY
DCODE START_DT END_DT SID
---------- ---------------- ---------------- ----------
1112223 01.06.2018 00:00 30.06.2018 00:00 1000
3119255 01.03.2018 00:00 31.08.2018 00:00 9842
SQL>
Suppose I have my table like:
uid day_used_app
--- -------------
1 2012-04-28
1 2012-04-29
1 2012-04-30
2 2012-04-29
2 2012-04-30
2 2012-05-01
2 2012-05-21
2 2012-05-22
Suppose I want the number of unique users who returned to the app at least 2 different days in the last 7 days (from 2012-05-03).
So as an example to retrieve the number of users who have used the application on at least 2 different days in the past 7 days:
select count(distinct case when num_different_days_on_app >= 2
then uid else null end) as users_return_2_or_more_days
from (
select uid,
count(distinct day_used_app) as num_different_days_on_app
from table
where day_used_app between current_date() - 7 and current_date()
group by 1
)
This gives me:
users_return_2_or_more_days
---------------------------
2
The question I have is:
What if I want to do this for every day up to now so that my table looks like this, where the second field equals the number of unique users who returned 2 or more different days within a week prior to the date in the first field.
date users_return_2_or_more_days
-------- ---------------------------
2012-04-28 2
2012-04-29 2
2012-04-30 3
2012-05-01 4
2012-05-02 4
2012-05-03 3
Would this help?
WITH
-- your original input, don't use in "real" query ...
input(uid,day_used_app) AS (
SELECT 1,DATE '2012-04-28'
UNION ALL SELECT 1,DATE '2012-04-29'
UNION ALL SELECT 1,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-04-29'
UNION ALL SELECT 2,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-05-01'
UNION ALL SELECT 2,DATE '2012-05-21'
UNION ALL SELECT 2,DATE '2012-05-22'
)
-- end of input, start "real" query here, replace ',' with 'WITH'
,
one_week_b4 AS (
SELECT
uid
, day_used_app
, day_used_app -7 AS day_used_1week_b4
FROM input
)
SELECT
one_week_b4.uid
, one_week_b4.day_used_app
, count(*) AS users_return_2_or_more_days
FROM one_week_b4
JOIN input
ON input.day_used_app BETWEEN one_week_b4.day_used_1week_b4 AND one_week_b4.day_used_app
GROUP BY
one_week_b4.uid
, one_week_b4.day_used_app
HAVING count(*) >= 2
ORDER BY 1;
Output is:
uid|day_used_app|users_return_2_or_more_days
1|2012-04-29 | 3
1|2012-04-30 | 5
2|2012-04-29 | 3
2|2012-04-30 | 5
2|2012-05-01 | 6
2|2012-05-22 | 2
Does that help your needs?
Marco the Sane ...
SELECT DISTINCT
t1.day_used_app,
(
SELECT SUM(CASE WHEN t.num_visits >= 2 THEN 1 ELSE 0 END)
FROM
(
SELECT uid,
COUNT(DISTINCT day_used_app) AS num_visits
FROM table
WHERE day_used_app BETWEEN t1.day_used_app - 7 AND t1.day_used_app
GROUP BY uid
) t
) AS users_return_2_or_more_days
FROM table t1
Say I've got a table with two columns (date and price). If I select over a range of dates, then is there a way to count the number of price changes over time?
For instance:
Date | Price
22-Oct-11 | 3.20
23-Oct-11 | 3.40
24-Oct-11 | 3.40
25-Oct-11 | 3.50
26-Oct-11 | 3.40
27-Oct-11 | 3.20
28-Oct-11 | 3.20
In this case, I would like it to return a count of 4 price changes.
Thanks in advance.
You can use the analytic functions LEAD and LAG to access to prior and next row of a result set and then use that to see if there are changes.
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select date '2011-10-22' dt, 3.2 price from dual union all
3 select date '2011-10-23', 3.4 from dual union all
4 select date '2011-10-24', 3.4 from dual union all
5 select date '2011-10-25', 3.5 from dual union all
6 select date '2011-10-26', 3.4 from dual union all
7 select date '2011-10-27', 3.2 from dual union all
8 select date '2011-10-28', 3.2 from dual
9 )
10 select sum(is_change)
11 from (
12 select dt,
13 price,
14 lag(price) over (order by dt) prior_price,
15 (case when lag(price) over (order by dt) != price
16 then 1
17 else 0
18 end) is_change
19* from t)
SQL> /
SUM(IS_CHANGE)
--------------
4
Try this
select count(*)
from
(select date,price from table where date between X and Y
group by date,price )
Depending on the Oracle version use either analytical functions (see answer from Justin Cave) or this
SELECT
SUM (CASE WHEN PREVPRICE != PRICE THEN 1 ELSE 0 END) CNTCHANGES
FROM
(
SELECT
C.DATE,
C.PRICE,
MAX ( D.PRICE ) PREVPRICE
FROM
(
SELECT
A.Date,
A.Price,
(SELECT MAX (B.DATE) FROM MyTable B WHERE B.DATE < A.DATE) PrevDate
FROM MyTable A
WHERE A.DATE BETWEEN YourStartDate AND YourEndDate
) C
INNER JOIN MyTable D ON D.DATE = C.PREVDATE
GROUP BY C.DATE, C.PRICE
)