serparate a range of string variables into multiple varialbes using array

[jenim514]

Hi!

Is it possible to change a range of variables (namepers1-namepers3)  currently a character string spearated by delimiter "/"

I was able to do it for one variable at a time...but for efficiency, is there a way to do Namepers1-namepsers3 in one datastep?

All the values in each variable are the same character string format...

data rr.namepers2006;
set sar_raw.rr2006 (keep=namepers1-namepers3) ;
array new_var(*)$ last rank grade;
i=1;
do while(scan(namepers1, i, "/") ne "");
new_var(i) =scan(namepers1, i, "/");
i+1;
end;
run;

 thank you!

[PaigeMiller]

although I'm not 100% sure I understand the problem, I believe the CATX function does what you want

but show us an example of the input, and what the output will look like for this situation

[jenim514]

current data looks like this

namepers1	namepers2	namepers3
Sample J ./0-3/ LT	Sample A ./0-1/ LT	Sample E ./0-4/ CDR

Want new colums to look like this

last1	grade1	rank1	last2	grade2	rank2	last3	grade3	rank3
Sample J.	0-3	LT	Sample A.	0-1	LT	Sample E.	0-4	CDR

[PaigeMiller]

You can use the SCAN function to separate the variables into "words" where the delimiter is a slash.

You'd want to put this into a data step loop, probably with namepers1 through namepers8 in an array.

[jenim514]

I can get the san to work....just not with a range of variables.  I'm not soing something right...

data rr.namepers2006;
set sar_raw.rr2006  (keep=namepers1-namepers3);
array new_var(*)$ name1-Name3 grade1-grade3 rank1-rank3;   *not sure if you can use array dimensions like this??;
i=1;
do while(scan(namepers1-namepers3, i, "/") ne "");
new_var(i) =scan(namepers1-namepers3, i, "/");          *trying to scan range of variables;
i+1;
end;
run;

[Reeza]

Does the number of items vary, or is it always 3?

[jenim514]

always 3

[art297]

Here is one way:

data need (keep=id _name_ i data);
  set rr2006;
  array rec(*) $ namepers1-namepers8;
  array new_var(*) $ last rank grade;
  id=_n_;
  do i=1 to dim(rec);
    if rec(i) ne '' then do;
      j=1;
      do while(scan(rec(i), j, "/") ne "");
        data =scan(rec(i), j, "/");
        _name_=vname(new_var(j));
        output;
        j+1;
      end;
    end;
  end;
run;

proc transpose data=need out=namepers2006 (drop=id _name_);
  var data;
  by id;
  id _name_ i;
run;

Art, CEO, AnalystFinder.com

[Reeza]

@art297 solution is more dynamic, but if you only have 3, the code is about the same length overall. It's what you can understand enough to change 😉

[Reeza · Accepted solution]

data want;;
infile cards dlm=',';
informat per1-per3 $20.;
input per1 per2 per3;
array name(3) $20. name1-Name3 ;
array grade(3) $20. grade1-grade3;
array rank(3) $20. rank1-rank3;   
array old_var(3) per1-per3;

do i=1 to 3;
name(i)=scan(old_var(i), 1, ".");
grade(i)=scan(old_var(i), 2, "/");
rank(i)=scan(old_var(i), 3, "/");
end;

cards;
Sample J ./0-3/ LT , Sample A ./0-1/ LT, Sample E ./0-4/ CDR
;

run;

[jenim514]

@Reeza thank you!