Windowing function puzzle

• Aaron N. Cutshall

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  March 2, 2017 at 4:01 pm

  #341278

  I have a complex query that produces data for multiple sets.  I need to ensure that for each set of criteria that I get data for each set.  The following query works, but I'd rather use a windowing function to avoid the double execution of the complex query:
  

  select a.*

  from dbo.ComplexQuery a

  inner join (select ColumnA, ColumnB, ColumnC, count(distinct ColumnD) as ColumnDCount

  from dbo.ComplexQuery

  group by ColumnA, ColumnB, ColumnC

  ) b

  on b.ColumnA = a.ColumnA

  and b.ColumnB = a.ColumnB

  and b.ColumnC = a.ColumnC

  and b.ColumnDCount = 2

  order by a.ColumnA, a.ColumnB, a.ColumnC, a.ColumnD

  This is about as simple of a query that I could produce which replicates the scenario I have.  To complicate things, it's possible to have multiple records of each ColumnD value with some in one group, some in the other, but we're interested in those in both groups only.  I'll have to work on some sample data, but I was hoping in the meantime someone had an idea.

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• Eddie Wuerch

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  March 2, 2017 at 8:28 pm

  #1931399

  You've basically hand-coded a DENSE_RANK() ... WHERE MAX(<DENSE_RANK() value>) = 2. The sample data will help roll a query to test, but you'll likely find a similar query plan with DENSE_RANK() . Window functions kick off a second trip through the data to prepare the results (for each distinct PARTITION BY clause), which are joined back to the base query results to produce the final output.

  Eddie Wuerch
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• sgmunson

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  March 2, 2017 at 8:32 pm

  #1931400

  Why not use a windowed version of COUNT?

  SELECT a.*
FROM dbo.ComplexQuery AS a
WHERE COUNT(DISTINCT ColumnD) OVER(PARTITION BY ColumnA, ColumnB, ColumnC) = 2

  If it won't let you use that in the WHERE clause, you can always just SELECT it in a CTE and then use the WHERE against that new column name in the final SELECT.

  Steve (aka sgmunson) 🙂 🙂 🙂
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• Aaron N. Cutshall

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  March 3, 2017 at 5:27 am

  #1931462

  Eddie Wuerch - Thursday, March 2, 2017 8:27 PM

  You've basically hand-coded a DENSE_RANK() ... WHERE MAX(<DENSE_RANK() value>) = 2. The sample data will help roll a query to test, but you'll likely find a similar query plan with DENSE_RANK() . Window functions kick off a second trip through the data to prepare the results (for each distinct PARTITION BY clause), which are joined back to the base query results to produce the final output.

  Thanks Eddie, but I already tried the DENSE_RANK() function.  Here's an example of my situation:

  ColumnA	ColumnB	ColumnC	ColumnD	Dense_Rank	What I Need
  123	ABC	987	A	1	2
  123	ABC	987	B	2	2
  456	DEF	654	A	1	1
  789	GHI	321	A	1	2
  789	GHI	321	B	2	2

  I hope this is clear enough.  Essentially, I need to ensure that I get both of the records as identified in ColumnD.  In the above case, I should get the records with Column A of 123 and 789 but NOT the record of 456 because it didn't have both and A and B values for ColumnD.  If it weren't for the need to return all columns of each record (I didn't display them all), I could simply deal with a GROUP BY.  Having the total row count in the column "What I Need" allows me to properly filter out incomplete sets.

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• Aaron N. Cutshall

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  March 3, 2017 at 5:28 am

  #1931464

  sgmunson - Thursday, March 2, 2017 8:32 PM

  Why not use a windowed version of COUNT?

  SELECT a.*
FROM dbo.ComplexQuery AS a
WHERE COUNT(DISTINCT ColumnD) OVER(PARTITION BY ColumnA, ColumnB, ColumnC) = 2

  If it won't let you use that in the WHERE clause, you can always just SELECT it in a CTE and then use the WHERE against that new column name in the final SELECT.

  Unfortunately, the DISTINCT keyword is not allowed in a window function with OVER() and without OVER(), it's no longer a window function!  This would have been the ideal solution!!

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• Aaron N. Cutshall

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  March 3, 2017 at 5:45 am

  #1931467

  Actually, let me clarify it a bit more and simplify the situation more succinctly:
  

  ColumnA	ColumnB	OtherData1	OtherData2	Dense_Rank	What I Need
  123	A	ABC	987	1	2
  123	B	BCD	876	2	2
  456	A	DEF	654	1	1
  789	A	GHI	321	1	2
  789	B	HIJ	210	2	2
  789	B	IJK	312	3	2

  The query that works is:
  select a.ColumnA, a.ColumnB, a.OtherData1, a.OtherData2

  from dbo.ComplexQuery a

  inner join (select ColumnA, count(distinct ColumnB) as ColumnBCount

  from dbo.ComplexQuery

  group by ColumnA

  ) b

  on b.ColumnA = a.ColumnA

  and b.ColumnBCount = 2

  order by a.ColumnA, a.ColumnB

  So, while the above works, it's not very efficient.  I had hoped to use something as suggested earlier with a COUNT(DISTINCT ColumnB) OVER(PARTITION BY ColumnA) but the DISTINCT is definitely not allowed.  Thoughts?

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• Mark Cowne

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  March 3, 2017 at 7:03 am

  #1931491

  You can use

  
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1

  instead of 

  
COUNT(DISTINCT ColumnB) OVER(PARTITION BY ColumnA)


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• Jeff Atherton

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  March 3, 2017 at 7:12 am

  #1931494

  Window functions are only allowed in the SELECT and ORDER BY clause. So if you really wanted to use one of them then the best option would be to put it in a CTE.  If you're just looking for alternative ways to write your select stmt then here's one option:

  
SELECT *
FROM ComplexQuery
WHERE ColumnA IN
(
SELECT ColumnA
FROM ComplexQuery
GROUP BY ColumnA
HAVING count(distinct ColumnB) = 2
)

• Aaron N. Cutshall

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  March 3, 2017 at 8:33 am

  #1931526

  Mark Cowne - Friday, March 3, 2017 7:03 AM

  You can use

  
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1

  instead of 

  
COUNT(DISTINCT ColumnB) OVER(PARTITION BY ColumnA)


  This works when the situation only calls for 2 values for ColumnB (which I showed in the examples).  What about where there are 3, 4, or more?

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• Aaron N. Cutshall

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  March 3, 2017 at 8:36 am

  #1931527

  Jeff Atherton - Friday, March 3, 2017 7:12 AM

  Window functions are only allowed in the SELECT and ORDER BY clause. So if you really wanted to use one of them then the best option would be to put it in a CTE.  If you're just looking for alternative ways to write your select stmt then here's one option:

  
SELECT *
FROM ComplexQuery
WHERE ColumnA IN
(
SELECT ColumnA
FROM ComplexQuery
GROUP BY ColumnA
HAVING count(distinct ColumnB) = 2
)

  This would also work, but it requires me to still have the complex query in there twice.  Unfortunately, I can't make use of a CTE in my particular scenario as I only gave a simplified portion of my overall query structure.  Even then, referencing a CTE twice still causes it to execute twice.  Window functions, as I understand it, execute it once but will navigate the result set multiple times as needed.

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• Mark Cowne

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  March 3, 2017 at 8:49 am

  #1931538

  Aaron N. Cutshall - Friday, March 3, 2017 8:33 AM

  Mark Cowne - Friday, March 3, 2017 7:03 AM

  You can use

  
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1

  instead of 

  
COUNT(DISTINCT ColumnB) OVER(PARTITION BY ColumnA)


  This works when the situation only calls for 2 values for ColumnB (which I showed in the examples).  What about where there are 3, 4, or more?

  Can you post some sample data.

  This appears to work
  
DECLARE @mytable TABLE(ColumnA INT, ColumnB CHAR(1))
INSERT INTO @mytable(columnA, ColumnB)
VALUES(123,'A'),(123,'B'),(456,'A'),(789,'A'),(789,'B'),(789,'B'),(888,'A'),(888,'B'),(888,'B'),(888,'D'),(888,'C'),(888,'C');

  SELECT ColumnA, ColumnB,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1 AS DistinctCount
FROM @mytable
ORDER BY ColumnA;


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• Jeff Atherton

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  March 3, 2017 at 9:01 am

  #1931546

  Aaron, so is this a performance issue? How much data are we talking about here? How long does the query take to run with the ComplexQuery in there twice? How often will this query need to be run everyday? Do the users need up to the minute results or can you run some ETL overnight to a new table just for reporting?

• Aaron N. Cutshall

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  March 3, 2017 at 10:08 am

  #1931572

  Mark Cowne - Friday, March 3, 2017 8:49 AM

  Can you post some sample data.

  This appears to work
  
DECLARE @mytable TABLE(ColumnA INT, ColumnB CHAR(1))
INSERT INTO @mytable(columnA, ColumnB)
VALUES(123,'A'),(123,'B'),(456,'A'),(789,'A'),(789,'B'),(789,'B'),(888,'A'),(888,'B'),(888,'B'),(888,'D'),(888,'C'),(888,'C');

  SELECT ColumnA, ColumnB,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1 AS DistinctCount
FROM @mytable
ORDER BY ColumnA;


  Mark,

  This is PERFECT!!  WOW!!  I'm going to have to study this one to make sure I truly understand it.  Here's your revised query with additional data:
  DECLARE @mytable TABLE(ColumnA INT, ColumnB CHAR(1), OtherData1 CHAR(3), OtherData2 INT)
INSERT INTO @mytable(columnA, ColumnB, OtherData1, OtherData2)
VALUES    (123,'A','ABC',987),
        (123,'B','BCD',876),
        (123,'C','CDE',765),
        (456,'A','DEF',654),
        (789,'A','GHI',321),
        (789,'B','HIJ',210),
        (789,'B','IJK',312),
        (258,'A','FDE',753),
        (258,'B','THE',951),
        (258,'C','PKH',873),
        (246,'A','LDJ',983),
        (246,'B','WKS',893),
        (246,'C','FUW',373),
        (246,'C','WLC',583),
        (246,'D','GKD',353);

  SELECT ColumnA, ColumnB, OtherData1, OtherData2,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) AS DenseRankOnly,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) AS DenseRankDesc,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1 AS DistinctCount
FROM @mytable
ORDER BY ColumnA, ColumnB;

  It certainly presents the results just as I needed it in all situations.  I'm totally impressed!!

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• Mark Cowne

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  March 3, 2017 at 10:57 am

  #1931580

  Aaron N. Cutshall - Friday, March 3, 2017 10:08 AM

  Mark Cowne - Friday, March 3, 2017 8:49 AM

  Can you post some sample data.

  This appears to work
  
DECLARE @mytable TABLE(ColumnA INT, ColumnB CHAR(1))
INSERT INTO @mytable(columnA, ColumnB)
VALUES(123,'A'),(123,'B'),(456,'A'),(789,'A'),(789,'B'),(789,'B'),(888,'A'),(888,'B'),(888,'B'),(888,'D'),(888,'C'),(888,'C');

  SELECT ColumnA, ColumnB,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1 AS DistinctCount
FROM @mytable
ORDER BY ColumnA;


  Mark,

  This is PERFECT!!  WOW!!  I'm going to have to study this one to make sure I truly understand it.  Here's your revised query with additional data:
  DECLARE @mytable TABLE(ColumnA INT, ColumnB CHAR(1), OtherData1 CHAR(3), OtherData2 INT)
INSERT INTO @mytable(columnA, ColumnB, OtherData1, OtherData2)
VALUES    (123,'A','ABC',987),
        (123,'B','BCD',876),
        (123,'C','CDE',765),
        (456,'A','DEF',654),
        (789,'A','GHI',321),
        (789,'B','HIJ',210),
        (789,'B','IJK',312),
        (258,'A','FDE',753),
        (258,'B','THE',951),
        (258,'C','PKH',873),
        (246,'A','LDJ',983),
        (246,'B','WKS',893),
        (246,'C','FUW',373),
        (246,'C','WLC',583),
        (246,'D','GKD',353);

  SELECT ColumnA, ColumnB, OtherData1, OtherData2,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) AS DenseRankOnly,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) AS DenseRankDesc,
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) +
DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB DESC) - 1 AS DistinctCount
FROM @mytable
ORDER BY ColumnA, ColumnB;

  It certainly presents the results just as I needed it in all situations.  I'm totally impressed!!

  You're  welcome.

  ____________________________________________________

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  How to get the best help on a forum

  http://www.sqlservercentral.com/articles/Best+Practices/61537

• Luis Cazares

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  March 3, 2017 at 12:10 pm

  #1931602

  Here's an alternative. I'm not sure which would perform better as both have issues that can slow them down.
  
WITH CTE AS(
  SELECT ColumnA, ColumnB, OtherData1, OtherData2,
  DENSE_RANK() OVER(PARTITION BY ColumnA ORDER BY ColumnB) AS DenseRank
  FROM @mytable
)
SELECT ColumnA, ColumnB, OtherData1, OtherData2,
  MAX(DenseRank) OVER( PARTITION BY ColumnA) DistinctCount
FROM CTE
ORDER BY ColumnA, ColumnB;


  Luis C.
  General Disclaimer:
  Are you seriously taking the advice and code from someone from the internet without testing it? Do you at least understand it? Or can it easily kill your server?

  How to post data/code on a forum to get the best help: Option 1 / Option 2