Solved

Using TSQL to identify sequencing of groups

Posted on 2016-08-23
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44 Views
Last Modified: 2016-08-30
I've got an MS-SQL table
create table test(
  name varchar(100),
  sequence integer,
  position integer);

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that contains:
insert into test values('A', null, 1);
insert into test values('A', null, 2);
insert into test values('A', null, 3);
insert into test values('B', null, 4);
insert into test values('B', null, 10);
insert into test values('A', null, 11);
insert into test values('A', null, 15);
insert into test values('B', null, 16);
insert into test values('C', null, 20);
insert into test values('C', null, 21);
insert into test values('A', null, 22);
insert into test values('A', null, 23);

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The field "position" is unique, no two rows contain the same value. There may however be gaps in the number list (see above)


Now what I'd like to do with TSQL is to identify the "groupings" of sequences of the "position" number, for each name.
The name "A" first sequence group is for position 1,2,3 - its sequence 2 is for position 11,15 and its sequence 3 is for position 22,23
The "B" name sequence group 1 contains position 4,10 and its sequence 2 is for position 16
And the "C" name has only got one sequence group, for position 20,21

Thus, I'd like to the result of the TSQL update be identical to if I had done the following insert to begin with:
insert into test values('A', 1, 1);
insert into test values('A', 1, 2);
insert into test values('A', 1, 3);
insert into test values('B', 1, 4);
insert into test values('B', 1, 10);
insert into test values('A', 2, 11);
insert into test values('A', 2, 15);
insert into test values('B', 2, 16);
insert into test values('C', 1, 20);
insert into test values('C', 1, 21);
insert into test values('A', 3, 22);
insert into test values('A', 3, 23);

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Any smart ideas? Otherwise I'll probably have to do this "off-line" with some kind of non-SQL programming

/Stefan
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Question by:stefanlennerbrant
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9 Comments
 
LVL 142

Assisted Solution

by:Guy Hengel [angelIII / a3]
Guy Hengel [angelIII / a3] earned 125 total points
ID: 41766815
Believe it or not, I had written an article about this kind of problems, see here:
https://www.experts-exchange.com/articles/3952/ranges-gaps-overlaps-for-number-and-date-ranges.html
0
 
LVL 142

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 41766818
the only thing you will need to add, over the article suggestion, is in the row_number() functions to add this:
row_number() over ( partition by name order by position )

once you create a "staging table" with the proper values, you can then use this other article I wrote to update the values as needed:
https://www.experts-exchange.com/articles/1517/UPDATES-with-JOIN-for-everybody.html
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LVL 69

Expert Comment

by:Éric Moreau
ID: 41766831
which version of MS SQL?
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LVL 69

Assisted Solution

by:Éric Moreau
Éric Moreau earned 125 total points
ID: 41766861
could it be:
;WITH C AS
(
  SELECT name, position
      , DENSE_RANK() OVER(PARTITION BY name ORDER BY position, name) AS grp
	  , ROW_NUMBER() OVER(ORDER BY position) AS L
	  , ROW_NUMBER() OVER(ORDER BY position) - RANK() OVER(PARTITION BY name ORDER BY position) AS T 
  FROM test 
)

SELECT name, MIN(position), MAX(position)
FROM C
GROUP BY name, T

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Author Comment

by:stefanlennerbrant
ID: 41766914
Wow, such first class answers!

I'll read those articles, lots of thanks!

And @emoreau, your suggestion kind of works out-of-the-box as well, almost - perfect.
I'll just have to get that T number to go sequentially 1,2,3,... for each "name".

/Stefan
0
 
LVL 17

Accepted Solution

by:
Pawan Kumar Khowal earned 250 total points
ID: 41769825
SELECT
              Name
            , MIN(LEVEL) StartPosition
            , MAX(LEVEL) EndPosition
FROM
(
              SELECT
                          Name
                        , LEVEL
                        , ROW_NUMBER() OVER(ORDER BY LEVEL) - ROW_NUMBER() OVER(PARTITION BY Name ORDER BY LEVEL) rnk      
              FROM THEGapPuzzle
)r
GROUP BY Name,rnk

Pawan Khowal
MSBISKILLS.COM
0
 
LVL 17

Expert Comment

by:Pawan Kumar Khowal
ID: 41770156
Well the methods given above will NOT scale when data will grow..

Use below method. It will improve the performance heavily.
SELECT Name, StartPosition, EndPosition FROM 
(
	SELECT Name,LeadValue, LagValue,position StartPosition, CASE (LEAD(name) OVER (ORDER BY ( select 1))) WHEN name THEN LEAD(position) OVER (ORDER BY ( select 1)) ELSE position END EndPosition 
	FROM
	(

		SELECT 
			 name
			,CASE (LAG(name) OVER (ORDER BY ( select 1 ))) WHEN name THEN 1 ELSE 0 END LagValue
			,CASE (LEAD(name) OVER (ORDER BY (  select 1))) WHEN name THEN 1 ELSE 0 END LeadValue
			,position
		FROM test

	) tbl2 
	WHERE tbl2.LagValue = 0 OR tbl2.LeadValue = 0
) tbl3 
WHERE LagValue=0

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Enjoy !
0
 
LVL 142

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 41770208
just to mention the LEAD/LAG version only works as from SQL 2012 on:
https://msdn.microsoft.com/en-us/library/hh231256.aspx
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Author Closing Comment

by:stefanlennerbrant
ID: 41776389
Great, spot on! Thanks!
All your suggestions work great, but I'm forced to split points in some way...
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