I have the following table,
id status price date
2 complete 10 2020-01-01 10:10:10
2 complete 20 2020-02-02 10:10:10
2 complete 10 2020-03-03 10:10:10
3 complete 10 2020-04-04 10:10:10
4 complete 10 2020-05-05 10:10:10
Required output,
id status_count price ratio
2 0 0 0
2 1 10 0
2 2 30 0.33
I am looking to add the price for previous row. Row 1 is 0 because it has no previous row value.
Find ratio ie 10/30=0.33
You can use analytical function ROW_NUMBER and SUM as follows:
SELECT
id,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id ORDER BY date), 0) - price as price
FROM yourTable;
DB<>Fiddle demo
I think you want something like this:
SELECT
id,
COUNT(*) OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id
ORDER BY date ROWS BETWEEN
UNBOUNDED PRECEDING AND 1 PRECEDING), 0) price
FROM yourTable;
Demo
Please also check another method:
with cte
as(*,ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
SUM(price) OVER (PARTITION BY id ORDER BY date) ss from yourTable)
select id,status_count,isnull(ss,0)-price price
from cte
Related
I have a table with 3 columns id, start_date, end_date
Some of the values are as follows:
1 2018-01-01 2030-01-01
1 2017-10-01 2018-10-01
1 2019-01-01 2020-01-01
1 2015-01-01 2016-01-01
2 2010-01-01 2011-02-01
2 2010-10-01 2010-12-01
2 2008-01-01 2009-01-01
I have the above kind of data set where I have to filter out overlap date range by keeping maximum datarange and keep the other date range which is not overlapping for a particular id.
Hence desired output should be:
1 2018-01-01 2030-01-01
1 2015-01-01 2016-01-01
2 2010-01-01 2011-02-01
2 2008-01-01 2009-01-01
I am unable to find the right way to code in impala. Can someone please help me.
I have tried like,
with cte as(
select a*, row_number() over(partition by id order by datediff(end_date , start_date) desc) as flag from mytable a) select * from cte where flag=1
but this will remove other date range which is not overlapping. Please help.
use row number with countItem for each id
with cte as(
select *,
row_number() over(partition by id order by id) as seq,
count(*) over(partition by id order by id) as countItem
from mytable
)
select id,start_date,end_date
from cte
where seq = 1 or seq = countItem
or without cte
select id,start_date,end_date
from
(select *,
row_number() over(partition by id order by id) as seq,
count(*) over(partition by id order by id) as countItem
from mytable) t
where seq = 1 or seq = countItem
demo in db<>fiddle
You can use a cumulative max to see if there is any overlap with preceding rows. If there is not, then you have the first row of a new group (row in the result set).
A cumulative sum of the starts assigns each row in the source to a group. Then aggregate:
select id, min(start_date), max(end_date)
from (select t.*,
sum(case when prev_end_date >= start_date then 0 else 1 end) over
(partition by id
order by start_date
rows between unbounded preceding and current row
) 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_end_date
from t
) t
) t
group by id, grp;
I have the following table to store history for entities:
Date Id State
-------------------------------------
2017-10-10 1 0
2017-10-12 1 4
2018-5-30 1 8
2019-4-1 2 0
2018-3-6 2 4
2018-3-7 2 0
I want to get last entry for each Id in one week period e.g.
Date Id State
-------------------------------------
2017-10-12 1 4
2018-5-30 1 8
2019-4-1 2 0
2018-3-7 2 0
I'd try to use Partition by:
select
ID
,Date
,State
,DatePart(week,Date) as weekNumber
from TableA
where Date = (
select max(Date) over (Partition by Id Order by DatePart(week, Date) Desc)
)
order by ID
but it still gives me more than one result per week.
You can use ROW_NUMBER():
SELECT a.*
FROM (SELECT a.*, ROW_NUMBER() OVER (PARTITION BY a.id, DATEPART(WK, a.Date) ORDER BY a.Date DESC) AS Seq
FROM tablea a
) a
WHERE seq = 1
ORDER BY id, Date;
suppose I have the following data frame in Reradata SQL.
How can I get the variation between the highest and lowest date, at user level? Regards
Initial table
user date price
1 1-1 10
1 2-1 20
1 3-1 30
2 1-1 12
2 2-1 22
2 3-1 32
3 1-1 13
3 2-1 23
3 3-1 33
Final table
user var_price
1 30/10-1
2 32/12-1
3 33/13-1
Try this-
SELECT B.[user],
CAST(SUM(B.max_price) AS VARCHAR)+'/'+CAST(SUM(B.min_price) AS VARCHAR)+ '-1' var_price,
SUM(B.max_price)/SUM(B.min_price) -1 calculated_var_price
FROM
(
SELECT * FROM
(
SELECT [user],0 max_price,price min_price,ROW_NUMBER() OVER (PARTITION BY [user] ORDER BY DATE) RN
FROM your_table
)A WHERE RN = 1
UNION ALL
SELECT * FROM
(
SELECT [user],price max_price,0 min_price, ROW_NUMBER() OVER (PARTITION BY [user] ORDER BY DATE DESC) RN
FROM your_table
)A WHERE RN = 1
)B
GROUP BY B.[user]
Output is-
user var_price calculated_var_price
1 30/10-1 2
2 32/12-1 1
3 33/13-1 1
Is this what you want?
select user, max(price) / min(price) - 1
from t
group by user;
Your values are monotonically increasing, so max() and min() seems like the simplest solution.
EDIT:
You can use window functions:
select user, max(last_price) / max(first_price) - 1
from (select t.*,
first_value(price) over (partition by user order by date rows between unbounded preceding and current_row) as first_price,
first_value(price) over (partition by user order by date desc rows between unbounded preceding and current_row) as last_price
from t
) t
group by user;
select user
,price as first_price
,last_value(price)
over (paritition by user
order by date
rows between unbounded preceding and unbounded following) as last_price
from mytab
qualify
row_number() -- lowest date only
over (paritition by user
order by date) = 1
This returns the row with the lowest date and adds the price of the latest date
This question already has an answer here:
SQL: Gaps and Islands, Grouped dates
(1 answer)
Closed 5 years ago.
I have the following dataset:
enter image description here
Here is script for this data:
;with dataset AS (
select 'EMP01' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-07' AS DATE) AS CUT_DATE
UNION
select 'EMP01' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-15' AS DATE) AS CUT_DATE
UNION
select 'EMP02' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-09' AS DATE) AS CUT_DATE
)
select *
from dataset
I need to divide these periods (PERIOD_START and PERIOD_END) by CUT_DATE (exclude cut dates from that periods) The number of cut dates could be any (3,5,8 etc).
Expecting result for the dataset above is:
If your version of SQL Server supports LAG, you can use this.
SELECT EMPLOYEE_ID,
ITEM_TYPE,
MIN(APPLY_DATE) AS STARTDATE,
MAX(APPLY_DATE) AS ENDDATE
FROM
(SELECT T.*,
SUM(CASE WHEN PREV_TYPE=ITEM_TYPE THEN 0 ELSE 1 END)
OVER(PARTITION BY EMPLOYEE_ID ORDER BY APPLY_DATE) AS GRP
FROM (SELECT D.*,
LAG(ITEM_TYPE) OVER(PARTITION BY EMPLOYEE_ID ORDER BY APPLY_DATE) AS PREV_TYPE
FROM DATA D
) T
) T
WHERE ITEM_TYPE IN ('Sickness','Vacation')
GROUP BY EMPLOYEE_ID,ITEM_TYPE,GRP
The logic is to get the previous row's item_type (based on ascending order of apply_date) and compare it with the current row's value. If they are equal, they belong to the same group. Else you start a new group. This is done in the sum window function. After groups are assigned, you just need to get the max and min date for an employee_id,item_type.
Sample Demo
You would use the LAG function.
If you order by something, the LAG function gives the previous value;
a full description can be found at: http://www.sqlservercentral.com/articles/T-SQL/106783/
Take a look at vkp's answer for a full query
This is another way if way if lag is supported.
Rextester Sample
with tbl as
(select d.*
,case when (item_type = lag(item_type) over (partition by employee_id order by apply_date))
then 0
else 1
end grp_tmp
from DATA2 d
where
item_type <> 'Worked'
)
,tbl2 as
(select t.*
,sum(grp_tmp) over (order by employee_id,apply_date
rows between unbounded preceding and current row
)
as grp
from tbl t
)
select
EMPLOYEE_ID
,ITEM_TYPE
,(CONVERT(VARCHAR(24),min(apply_date),103)
+' - '
+CONVERT(VARCHAR(24),max(apply_date),103)
) as range
from tbl2
group by EMPLOYEE_ID,
ITEM_TYPE
,grp
order by
employee_id
,min(apply_date);
Output
+-------------+-----------+-------------------------+
| EMPLOYEE_ID | ITEM_TYPE | range |
+-------------+-----------+-------------------------+
| 1 | Sickness | 23/05/2017 - 24/05/2017 |
| 1 | Vacation | 26/05/2017 - 29/05/2017 |
| 1 | Sickness | 01/06/2017 - 01/06/2017 |
| 2 | Sickness | 25/05/2017 - 30/05/2017 |
+-------------+-----------+-------------------------+
I am trying to perform a window function on a data-set in Redshift using days an an interval for the preceding rows.
Example data:
date ID score
3/1/2017 123 1
3/1/2017 555 1
3/2/2017 123 1
3/3/2017 555 3
3/5/2017 555 2
SQL window function for avg score from the last 3 scores:
select
date,
id,
avg(score) over
(partition by id order by date rows
between preceding 3 and
current row) LAST_3_SCORES_AVG,
from DATASET
Result:
date ID LAST_3_SCORES_AVG
3/1/2017 123 1
3/1/2017 555 1
3/2/2017 123 1
3/3/2017 555 2
3/5/2017 555 2
Problem is that I would like the average score from the last 3 DAYS (moving average) and not the last three tests. I have gone over the Redshift and Postgre Documentation and can't seem to find any way of doing it.
Desired Result:
date ID 3_DAY_AVG
3/1/2017 123 1
3/1/2017 555 1
3/2/2017 123 1
3/3/2017 555 2
3/5/2017 555 2.5
Any direction would be appreciated.
You can use lag() and explicitly calculate the average.
select t.*,
(score +
(case when lag(date, 1) over (partition by id order by date) >=
date - interval '2 day'
then lag(score, 1) over (partition by id order by date)
else 0
end) +
(case when lag(date, 2) over (partition by id order by date) >=
date - interval '2 day'
then lag(score, 2) over (partition by id order by date)
else 0
end)
)
) /
(1 +
(case when lag(date, 1) over (partition by id order by date) >=
date - interval '2 day'
then 1
else 0
end) +
(case when lag(date, 2) over (partition by id order by date) >=
date - interval '2 day'
then 1
else 0
end)
)
from dataset t;
The following approach could be used instead of the RANGE window option in a lot of (or all) cases.
You can introduce "expiry" for each of the input records. The expiry record would negate the original one, so when you aggregate all preceding records, only the ones in the desired range will be considered.
AVG is a bit harder as it doesn't have a direct opposite, so we need to think of it as SUM/COUNT and negate both.
SELECT id, date, running_avg_score
FROM
(
SELECT id, date, n,
SUM(score) OVER (PARTITION BY id ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
/ NULLIF(SUM(n) OVER (PARTITION BY id ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW), 0) as running_avg_score
FROM
(
SELECT date, id, score, 1 as n
FROM DATASET
UNION ALL
-- expiry and negate
SELECT DATEADD(DAY, 3, date), id, -1 * score, -1
FROM DATASET
)
) a
WHERE a.n = 1