7 Replies Latest reply on Feb 26, 2009 7:46 PM by 589626

# How to count continuous years

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
• ###### 1. Re: How to count continuous years
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
• ###### 2. Re: How to count continuous years
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;``````
• ###### 3. Re: How to count continuous years
Works like a charm, thanks so much.

Andrew
• ###### 4. Re: How to count continuous years
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``````
• ###### 5. Re: How to count continuous years
Aketi,

Hats off to you and your "sense of TabibitoZan" (whatever that means)...
Wish I can learn to "THINK" like this.
• ###### 6. Re: How to count continuous years
I mentions "Tabibitozan" in below threads ;-)
Sum over group
Group by preserving the order

"Tabibitozan" is one of math problem.
I do not know what "Tabibitozan" is called in English.
• ###### 7. Re: How to count continuous years
That's some sweet code. Thanks for all the advice guys.