How to count continuous years

589626

I have a table with the following data:

ID YEAR
1 2009
1 2008
1 2007
1 2006
1 2005
1 2004
1 2002
1 2001
2 2009
2 2008
2 2005

I want to be able to count the number of records from the current year back, but if there is a break I want to stop the count, so it is only counting continuous years that are being counted. So for the above data I would want the following returned:

ID COUNT
1 6
2 2

I cannot figure out how to do this using either a cursor or Analytical functions. Any ideas would be greatly appreciated.

Thanks,
Andrew

Frank Kulash

Hi, Andrew,

Analytic fucntions are great for that, but not as easy as you might wish.

Using the analytic LAG (or LEAD) function, you can see iwhat the difference is between a year and the last year for the same id:

LAG (year) OVER (PARTITION BY id ORDER BY year DESC) 
- year

Using a CASE staemnet, you can mark each row where the difference was 1 as 0, and all the rows where the difference was more than 1 as 1.
Then, using the analytic SUM function, you can add up all those numbers through the present row

SUM (dif) OVER (PARTITION BY id ORDER BY year DESC) AS grp

Since analytic functions can't be nested, this usually involves two sub-queries.

WITH   got_new_grp  AS
(
	SELECT	id
	,	year
	,	CASE
			WHEN	LAG (year) OVER ( PARTITION BY	id
				    	   	  ORDER BY  	year	DESC
						) - year > 1
			THEN	1
		END	AS new_grp
	FROM	table_x
)
,	got_grp	AS
(
	SELECT	got_new_grp.*
	,	COUNT (new_grp) OVER ( PARTITION BY	id
		      		       ORDER BY	 	year	DESC
				     )   AS grp
	FROM	got_new_grp
)
SELECT	id
,	COUNT (*)	AS cnt
FROM	got_grp
WHERE	grp	= 0
GROUP BY	id
ORDER BY	id;

In this solution, ecery id can have a different starting point. For example, if you delete the row with id=2 and year=2009, then the count for id=2 is 1, not 0.

You might have a special case, where the combination of id and year is unique, and the difference is always 1 or more. If so, there's a cute trick to avoid one of the sub-queries.

Edited by: Frank Kulash on Feb 25, 2009 1:48 PM

Frank Kulash

Hi,

Here's a solution for the special case (no duplicate years, difference is always 1 or more):

WITH	got_grp	AS
(
	SELECT	id
	,	year
	,	(	MAX (year) OVER (PARTITION BY id)
		-	year
		) - ROW_NUMBER () OVER (PARTITION BY id ORDER BY year DESC)
		        AS  grp
	FROM	table_x
)
SELECT	id
,	COUNT (*)	AS cnt
FROM	got_grp
WHERE	grp	= -1
GROUP BY	id
ORDER BY	id;

589626

Works like a charm, thanks so much.

Andrew

Aketi Jyuuzou

I used sense of TabibitoZan B-)

with YearT as(
select 1 as ID,2009 as YEAR from dual union
select 1,2008 from dual union
select 1,2007 from dual union
select 1,2006 from dual union
select 1,2005 from dual union
select 1,2004 from dual union
select 1,2002 from dual union
select 1,2001 from dual union
select 2,2009 from dual union
select 2,2008 from dual union
select 2,2005 from dual)
select ID,
count(*) Keep(Dense_Rank Last order by distance) as cnt
from (select ID,
      Year+Row_Number() over(partition by ID order by YEAR desc) as distance
        from YearT)
group by ID;

ID  CNT
--  ---
 1    6
 2    2

user503699

Aketi,

Hats off to you and your "sense of TabibitoZan" (whatever that means)...
Wish I can learn to "THINK" like this.

Aketi Jyuuzou

I mentions "Tabibitozan" in below threads ;-)
584668
450745

"Tabibitozan" is one of math problem.
I do not know what "Tabibitozan" is called in English.

589626

That's some sweet code. Thanks for all the advice guys.