Related
We have DataTemp table which has the records in desc order.
select * from (
select 9,'a',3 union
select 8,'a',2 union
select 7,'b',3 union
select 6,'a',1 union
select 5,'b',2 union
select 4,'c',3 union
select 3,'c',2 union
select 2,'b',1 union
select 1,'c',1
) door (sno,id, N_th_Reocord)
sno - Auto Id.
id - code of the Doors*.
N_th_Record - for denoting the n the record.
At a time, only three* records per Door are need to store on this table. For example Door 'a' has new entry(means 4th record) then first of 'a' Door need to delete.
4th record:
select * from (
select 10,'a',4 union --- new entry
select 9,'a',3 union
select 8,'a',2 union
select 7,'b',3 union
select 6,'a',1 union -- need to delete
select 5,'b',2 union
select 4,'c',3 union
select 3,'c',2 union
select 2,'b',1 union
select 1,'c',1
) door (sno,id, N_th_Reocord)
I do following query. But I need easiest way for deleting the row. Because, we are try to reduce the time consumption of over all project.
delete from door where sno = (
select sno from (
select 10,'a',4 union
select 9,'a',3 union
select 8,'a',2 union
select 7,'b',3 union
select 6,'a',1 union
select 5,'b',2 union
select 4,'c',3 union
select 3,'c',2 union
select 2,'b',1 union
select 1,'c',1
) door (sno,id, N_th_Reocord)
where id = 'a'
order by sno desc -- For 'DataTemp' *order by* is no needed.
offset 3 rows fetch next 1 rows only
)
Note:
Three rows and three Door are given for example. Actually we work with 144 rows per 12 Doors.
Before this delete, we check lot of Business rules.
Version: SQL Server 2012
You could use ROW_NUMBER:
WITH cte AS (SELECT *,ROW_NUMBER() OVER(PARTITION BY id ORDER BY sno DESC) rn FROM t)
DELETE FROM cte WHERE rn > 3;
db<>fiddle demo
I have data like below -
Year,winning_country
2001,IND
2002,IND
2003,IND
2004,AUS
2005,AUS
2006,SA
2007,SA
2008,SA
2009,IND
2010,IND
2011,IND
2012,IND
2013,AUS
2014,AUS
2015,SA
2016,NZ
2017,SL
2018,IND
The question here is to find out the longest streak of wins for each country and desired output will be like below -
Country,no_of_wins
IND,4
AUS,2
SA,3
SL,1
NZ,1
Can someone help here.
This is a gaps and islands problem, but the simplest method is to subtract a sequence from the year. So, to get all the sequences:
select country, count(*) as streak,
min(year) as from_year, max(year) as to_year
from (select year, country,
row_number() over (partition by country order by year) as seqnum
from t
) t
group by country, (year - seqnum);
To get the longest per country, aggregate again or use window functions:
select country, streak
from (select country, count(*) as streak,
min(year) as from_year, max(year) as to_year,
row_number() over (partition by country order by count(*) desc) as seqnum_2
from (select year, country,
row_number() over (partition by country order by year) as seqnum
from t
) t
group by country, (year - seqnum)
) cy
where seqnum_2 = 1;
I prefer using row_number() to get the longest streak because it allows you to also get the years when it occurred.
Looks like an gaps-and-islands problem.
The SQL below calculates some ranking based on 2 row_number.
Then it's just a matter of grouping.
SELECT q2.Country, MAX(q2.no_of_wins) AS no_of_wins
FROM
(
SELECT q1.winning_country as Country,
COUNT(*) AS no_of_wins
FROM
(
SELECT t.Year, t.winning_country,
(ROW_NUMBER() OVER (ORDER BY t.Year ASC) -
ROW_NUMBER() OVER (PARTITION BY t.winning_country ORDER BY t.Year)) AS rnk
FROM yourtable t
) q1
GROUP BY q1.winning_country, q1.rnk
) q2
GROUP BY q2.Country
ORDER BY MAX(q2.no_of_wins) DESC
If Redshift supports analytic function, below would be the query.
with t1 as
(
select 2001 as year,'IND' as cntry from dual union
select 2002,'IND' from dual union
select 2003,'IND' from dual union
select 2004,'AUS' from dual union
select 2005,'AUS' from dual union
select 2006,'SA' from dual union
select 2007,'SA' from dual union
select 2008,'SA' from dual union
select 2009,'IND' from dual union
select 2010,'IND' from dual union
select 2011,'IND' from dual union
select 2012,'IND' from dual union
select 2013,'AUS' from dual union
select 2014,'AUS' from dual union
select 2015,'SA' from dual union
select 2016,'NZ' from dual union
select 2017,'SL' from dual union
select 2018,'IND' from dual) ,
t2 as (select year, cntry, year - row_number() over (partition by cntry order by year) as grpBy from t1 order by cntry),
t3 as (select cntry, count(grpBy) as consWins from t2 group by cntry, grpBy),
res as (select cntry, consWins, row_number() over (partition by cntry order by consWins desc) as rnk from t3)
select cntry, consWins from res where rnk=1;
Hope this helps.
Here is a solution that leverages the use of Redshift Python UDF's
There may be simpler ways to achieve the same but this is a good example of how to create a simple UDF.
create table temp_c (competition_year int ,winning_country varchar(4));
insert into temp_c (competition_year, winning_country)
values
(2001,'IND'),
(2002,'IND'),
(2003,'IND'),
(2004,'AUS'),
(2005,'AUS'),
(2006,'SA'),
(2007,'SA'),
(2008,'SA'),
(2009,'IND'),
(2010,'IND'),
(2011,'IND'),
(2012,'IND'),
(2013,'AUS'),
(2014,'AUS'),
(2015,'SA'),
(2016,'NZ'),
(2017,'SL'),
(2018,'IND')
;
create or replace function find_longest_streak(InputStr varChar)
returns integer
stable
as $$
MaxStreak=0
ThisStreak=0
ThisYearStr=''
LastYear=0
for ThisYearStr in InputStr.split(','):
if int(ThisYearStr) == LastYear + 1:
ThisStreak+=1
else:
if ThisStreak > MaxStreak:
MaxStreak=ThisStreak
ThisStreak=1
LastYear=int(ThisYearStr)
return max(MaxStreak,1)
$$ language plpythonu;
select winning_country,
find_longest_streak(listagg(competition_year,',') within group (order by competition_year))
from temp_c
group by winning_country
order by 2 desc
;
How about something like...
SELECT
winning_country,
COUNT(*)
GROUP BY winning_country
HAVING MAX(year) - MIN(year) = COUNT(year) - 1
This assumes no duplicate entries.
Creating a session abstraction do the trick:
WITH winning_changes AS (
SELECT *,
CASE WHEN LAG(winning_country) OVER (ORDER BY year) <> winning_country THEN 1 ELSE 0 END AS same_winner
FROM winners
),
sequences AS (
SELECT *,
SUM(same_winner) OVER (ORDER BY year) AS winning_session
FROM winning_changes
),
streaks AS (
SELECT winning_country AS country,
winning_session,
COUNT(*) streak
FROM sequences
GROUP BY 1,2
)
SELECT country,
MAX(streak) AS no_of_wins
FROM streaks
GROUP BY 1;
As the title says, I'm trying to add an extra column to a table which autoincrements everytime a different string in another column changes.
I would like to do this in a query.
Example:
MyCol GroupID
Cable 1
Cable 1
Foo 2
Foo 2
Foo 2
Fuzz 3
Fizz 4
Tv 5
Tv 5
The GroupID column is what I want to accomplish.
We can be sure that MyCol's strings will be the same in each subgroup (Foo will always be Foo, etc).
Thanks in advance
If I understand correctly, you can use dense_rank():
select t.*, dense_rank() over (order by col1) as groupid
from t;
You could make a temporal table with the distinct value of the MyCol and get the groupId throught the RowNumber of the temp table, and join the rownumbered result with your table.
This is a raw example in oracle:
WITH data AS
(SELECT 'Cable' MyCol FROM dual
UNION ALL
SELECT 'Cable' FROM dual
UNION ALL
SELECT 'Foo' FROM dual
UNION ALL
SELECT 'Foo' FROM dual
UNION ALL
SELECT 'Foo' FROM dual
UNION ALL
SELECT 'Fuzz' FROM dual
UNION ALL
SELECT 'Fizz' FROM dual
UNION ALL
SELECT 'Tv' FROM dual
UNION ALL
SELECT 'Tv' FROM dual
),
tablename AS
(SELECT * FROM data
),
temp AS
( SELECT DISTINCT mycol FROM tablename
),
temp2 AS
( SELECT mycol, rownum AS groupid from temp
)
SELECT tablename.mycol, temp2.groupid FROM temp2 JOIN tablename ON temp2.mycol = tablename.mycol
You could also check for a way to implement the tabibitosan method knowing that your column condition is string.
I know the question is probably badly explained, but I don't know how else to explain this. I have the following data: (ordered by date)
DATE GROUP
11-Oct-16 A
12-Oct-16 A
13-Oct-16 A
14-Oct-16 B
15-Oct-16 B
16-Oct-16 A
17-Oct-16 A
18-Oct-16 C
19-Oct-16 C
20-Oct-16 C
21-Oct-16 C
22-Oct-16 A
23-Oct-16 A
24-Oct-16 A
I want to find consecutive usage for groups. The results I want will explain this better than me:
GROUP MIN(DATE) MAX(DATE)
A 11-Oct-16 13-Oct-16
B 14-Oct-16 15-Oct-16
A 16-Oct-16 17-Oct-16
C 18-Oct-16 21-Oct-16
A 22-Oct-16 24-Oct-16
Any idea how to do this in oracle sql?
Thank you.
This can be a way:
with test("DATE","GROUP") as
(
select to_date('11-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('12-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('13-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('14-10-16', 'dd-mm-rr'),'B' from dual union all
select to_date('15-10-16', 'dd-mm-rr'),'B' from dual union all
select to_date('16-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('17-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('18-10-16', 'dd-mm-rr'),'C' from dual union all
select to_date('19-10-16', 'dd-mm-rr'),'C' from dual union all
select to_date('20-10-16', 'dd-mm-rr'),'C' from dual union all
select to_date('21-10-16', 'dd-mm-rr'),'C' from dual union all
select to_date('22-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('23-10-16', 'dd-mm-rr'),'A' from dual union all
select to_date('24-10-16', 'dd-mm-rr'),'A' from dual
)
select min("DATE"), max("DATE"), "GROUP"
from (
select "DATE",
"DATE" - row_number() over (partition by "GROUP" order by "DATE") as minDate,
"GROUP"
from test
)
group by "GROUP", minDate
order by "GROUP", minDate
The inner query builds the minimum date for a group of consecutive dates, while the external one simply aggregates by this minimum date, thus building a row for every group of consecutive dates.
As an aside, it's better to avoid using reserved words as column names.
So basically I think that the proper answer was in #mathguy comment.
Here I am just expanding and giving the proper full query.
select GROUP_NAME, grp_id, min(date_field) as starting_date, max(date_field) as ending_date
from (
select DATE_FIELD, GROUP_NAME,
row_number() over ( order by date_field) - row_number() over ( partition by group_name order by date_field ) as grp_id
from darber.my_table) innested
group by GROUP_NAME, grp_id
order by min(date_field);
Please note: I changed the field name because they were giving conflicts on my database. In this solution DATE is now DATE_FIELD and GROUP is now GROUP_NAME.
I found this to be a really smart and elegant solution.
I hope will help others.
I had the same issue with multiple members in the group. In case you are running into the same issue, replace the GROUP_NAME with "groupMember1, groupMember2, ..."
I'm building a query to show average and variance from many columns.
To get the average I use this:
SELECT *,
(SELECT AVG(t.c)
FROM (
SELECT peca_1 UNION ALL
SELECT peca_2 UNION ALL
SELECT peca_3 UNION ALL
SELECT peca_4 UNION ALL
SELECT peca_5 UNION ALL
SELECT peca_6 UNION ALL
SELECT peca_7 UNION ALL
SELECT peca_8 UNION ALL
SELECT peca_9 UNION ALL
SELECT peca_10
) t(c)
) as [media]
from Durabilidade
where cd_durabilidade = 1
The result is:
Now I need a new column with VAR(media) comparing each row with first row.
Any idea?
I think this is a case where cross apply is appropriate. I am assuming that you want the variance of the values as calculated by the var() function:
SELECT *, t.avgval as [media], t.varval
from Durabilidade d cross apply
(select avg(t.val) as avgval, var(t.val) as varval
from (select d.peca_1 union all
select d.peca_2 union all
select d.peca_3 union all
select d.peca_4 union all
select d.peca_5 union all
select d.peca_6 union all
select d.peca_7 union all
select d.peca_8 union all
select d.peca_9 union all
select d.peca_10
) t(val) -- t(val) to work
) t
where cd_durabilidade = 1
Something like this?
SELECT *,
VAR(media) AS [variance]
FROM
(
SELECT *,
(SELECT AVG(t.c)
FROM (
SELECT peca_1 UNION ALL
SELECT peca_2 UNION ALL
SELECT peca_3 UNION ALL
SELECT peca_4 UNION ALL
SELECT peca_5 UNION ALL
SELECT peca_6 UNION ALL
SELECT peca_7 UNION ALL
SELECT peca_8 UNION ALL
SELECT peca_9 UNION ALL
SELECT peca_10
) t(c)
) as [media]
from Durabilidade
where cd_durabilidade = 1
) x
GROUP BY
column1_from_durabilidade
,column2_from_durabilidade
--etc
,media