Answered by:
Ranking Functions Not Windowing As Expected

Hi all,I'm new to ranking functions and windowing. Please excuse the simplicity of this post  but I'm failing to grasp why I'm receiving the output I am from SQL 2008 SP2 Dev x64, SQL 2008 SP1 Std x64, and SQL 2008R2 Std x64. (This tells me I'm interpreting the function wrong!)I'm running this set of statements:
CREATE TABLE RankTest ( OrderBy1 INT NOT NULL, Value1 CHAR(1) NOT NULL ) INSERT INTO RankTest (OrderBy1, Value1) VALUES (1, 'A') INSERT INTO RankTest (OrderBy1, Value1) VALUES (2, 'A') INSERT INTO RankTest (OrderBy1, Value1) VALUES (3, 'B') INSERT INTO RankTest (OrderBy1, Value1) VALUES (4, 'B') INSERT INTO RankTest (OrderBy1, Value1) VALUES (5, 'A') INSERT INTO RankTest (OrderBy1, Value1) VALUES (6, 'A') SELECT *, ROW_NUMBER() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS rn, RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS r, DENSE_RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS dr FROM RankTest ORDER BY OrderBy1 DROP TABLE RankTest
From the SELECT, I'm expecting to see (from all of the ranking functions) is a nice series of "1, 2, 1, 2, 1, 2", like this:OrderBy1 Value1 rn r dr 1 A 1 1 1 2 A 2 2 2 3 B 1 1 1 4 B 2 2 2 5 A 1 1 1 6 A 2 2 2 OrderBy1 Value1 rn r dr 1 A 1 1 1 2 A 2 2 2 3 B 1 1 1 4 B 2 2 2 5 A 3 3 3 6 A 4 4 4 Why is the partition not working like my brain says it should? Doesn't the ORDER BY in the OVER clause mean that the ranking functions should "start over" in their counting when they see the "A" value in row 5? Or does the PARTITION BY operate "first" (to group the rows) and then the ORDER BY merely orders the contents of the partitions?
How can I modify my query to return the results I'm looking for?
Thanks!
Talk to me now on
Question
Answers

The windowing functions are working correctly, one way to get what you want is:
With cte As (SELECT *, ROW_NUMBER() OVER (ORDER BY OrderBy1)  ROW_NUMBER() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS Island RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS r, DENSE_RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS dr FROM RankTest) Select *, ROW_NUMBER() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) As rn, RANK() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) AS r, DENSE_RANK() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) AS dr From cte ORDER BY OrderBy1;
Tom
 Marked as answer by Todd McDermid Tuesday, April 26, 2011 7:01 PM
All replies

The ranking functions are working as expected. I am afraid that I am not awake on this one. To me it looks like the best solution might be to use a cursor. Additional help please?
Here is a method that is moreorless a triangular join method; however, I think this will be slower than a cursorbased version:
declare @RankTest table ( OrderBy1 INT NOT NULL, Value1 CHAR(1) NOT NULL) INSERT INTO @RankTest select 1, 'A' union all select 2, 'A' union all select 3, 'B' union all select 4, 'B' union all select 5, 'A' union all select 6, 'A' select a.orderBy1, a.value1, count(*) from @rankTest a join @rankTest b on b.value1 = a.value1 and b.orderBy1 <= a.orderBy1 and not exists ( select 0 from @rankTest c where c.value1 <> a.value1 and c.orderBy1 between b.orderBy1 and a.orderBy1 ) group by a.orderBy1, a.value1 /*  Output:  orderBy1 value1    1 A 1 2 A 2 5 A 1 6 A 2 3 B 1 4 B 2 (6 row(s) affected) */
( This is a case where the ORACLE "Lag" or "Last" analytic functions might be helpful. )

To get 1,2,1,2,1,2 like that the values for rows 5 & 6 would need to be 'C'.
This is going to get you pretty close to what you are looking for but if you really need it to reset back to 1 on row 5 you'll have to have the Value1 column be different than either of the previous sets. ROW_NUMBER() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS rn
I went through a similare problem last week and went as far as putting one ROW_NUMBER in a subselect because you can't for example do DENSE_RANK(ROW_NUMBER() OVER (PARTITION BY Value1 ORDER BY OrderBy1)) AS rn. It's not allowed.
 Aaron Nelson. 
The windowing functions are working correctly, one way to get what you want is:
With cte As (SELECT *, ROW_NUMBER() OVER (ORDER BY OrderBy1)  ROW_NUMBER() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS Island RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS r, DENSE_RANK() OVER (PARTITION BY Value1 ORDER BY OrderBy1) AS dr FROM RankTest) Select *, ROW_NUMBER() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) As rn, RANK() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) AS r, DENSE_RANK() OVER (PARTITION BY Value1, Island ORDER BY OrderBy1) AS dr From cte ORDER BY OrderBy1;
Tom
 Marked as answer by Todd McDermid Tuesday, April 26, 2011 7:01 PM


Think of it this way... The ranking is taking place before the SQL statement's ORDER BY clause. So the last line, has no bearing on any of these functions. It's only affecting the final display order.
The ranking criteria is 100% in the OVER clause of the given function, with the "PARTITION BY" columns being sorted first, followed by the "ORDER BY" columns.
So based on your example, in all 3 cases, the ranking order is...
OrderBy1 Value1 r 1 A 1 2 A 2 5 A 3 6 A 4 3 B 1 4 B 2
With [Value1] being sorted 1st followed by [OrderBy1].
Jason Long 
Todd,
> Or does the PARTITION BY operate "first" (to group the rows) and then the ORDER BY merely orders the contents of the partitions?
Bingo!!!
> How can I modify my query to return the results I'm looking for?
Tom already answered your question, so I am going to add little. You are looking to rank the rows not only by the [Value1], if not also by the island (consecutive rows) based on the values of [OrderBy1].
Island OrderBy1 Value1
1 1 A
1 2 A
1 3 B
1 4 B
2 5 A
2 6 AIf there is no gap in the [OrderBy1] sequence, then you can use:
[OrderBy1]  ROW_NUMBER() OVER(PARTITION BY Value1 ORDER BY [OrderBy1]) AS Island
otherwise you use:
ROW_NUMBER() OVER(ORDER BY [OrderBy1])  ROW_NUMBER() OVER(PARTITION BY [Value1] ORDER BY [OrderBy1]) AS Island
Then you rank the result set by:
ranking_function() OVER(PARTITION BY [Value1], Island ORDER BY [OrderBy1])
I learned this trick from Itzik BenGan, and you can read more about ranking functions in his last book about TSQL Querying.
Inside Microsoft® SQL Server® 2008: TSQL Querying
AMB

Alternatively, Plamen Ratchev has a good blog on this topic as well which I always recheck then faced islands problem
For every expert, there is an equal and opposite expert.  Becker's Law
My blog