Make indicator to split two incidence within each individ in sas

[user40]

Hi 

I tried to make an index that assign number 1 for each sequence start in the table below. I try this with the following code: 

But the problem is that i Want to have a new unique number for index when the sequence variable (seqn) start in the beginning again. Like 1 for seqn 1 , 2, and index=2 for seqn 1 until 14 in this example? 

How can I solve this example? 

[ballardw]

I can't tell what you want the output to look like.

Note: we can't code against pictures.

It is best to provide data in the form of a data step. There are hundreds of examples on the forum for such.

[PeterClemmensen]

I think this is what you want

data have;
   do seqnr = 1, 2, 1 to 14;
      output;
   end;
run;

data want;
   set have;
   if seqnr = 1 then index + 1;
run;

[user40]

Hi

My problem that I have many individuals in this file and this indicator must be done for each individ. How can I do that?

[PeterClemmensen]

Do the individuals have IDs? As I understand it, you want to reset the index to 1 whenever seqnr is 1. Which is that my code does.

[user40]

Yes, but my dataset have first id number that are for example like this

ID, seqn
101, 1
101 ,2
101 ,3
101 ,1
101 ,2
102 ,1
102 ,2
102 ,3
102 ,4
102 ,1
102 ,2


Then I need this that the index asign 1 for ID number 1 from seqn 1 to 3 and index = 2 for 1,2 within the same id number. And it will start again on the next id-number and do it on the same way. How can i reset the index to 1 in the beginning of each id number in sas?

[PeterClemmensen]

Like this?

data have;
input ID seqn;
datalines;
101 1
101 2
101 3
101 1
101 2
102 1
102 2
102 3
102 4
102 1
102 2
;

data want;
   set have;
   by ID;
   if first.ID then index = 0;
   if seqn = 1 then index + 1;
run;

[user40]

I only got index =1 in when seqn is 1 for this solution. I need also that index =1 for all 101 when seqn go from 1-3 , and index =2 when 1-2 in the next sequence in the same id number if you understand.

[PeterClemmensen · Accepted solution]

I do understand.

Here are my results from the code above. I think that is what you describe?

ID   seqn  index
101  1     1
101  2     1
101  3     1
101  1     2
101  2     2
102  1     1
102  2     1
102  3     1
102  4     1
102  1     2
102  2     2