How do I turn variables into values for a new variable? (part 2 with a twist)

Hello_there — Fri, 24 Mar 2023 15:01:34 GMT

Hi, I posted a question earlier that was answered by @PaigeMiller , and I got a solution.

However, this new data set has a twist. Some variables have letters attached to them.

The purpose of this is that I was given raw data, and I'm creating an output that condenses information into one variable.

There are 4 subjects, and the "x" represents the value that i want presented in a new variable called CRIT.

So if zzinc3="x", then CRIT=I3.

If zzinc2a="x" and zzexc2b="x" and zzexc3="x" then CRIT=I2A, E2B , E3 ** new twist**

New data with a twist:

data have2; 
infile datalines dsd dlm=",";
	input subject $ zzinc1 $ zzinc2a $ zzinc2b $ zzinc2c $ zzinc3 $ zzexc1 $ zzexc2a $ zzexc2b $ zzexc3 $;
datalines;
001, x, x, , x, , , x, , 
002, , x, , , x, , , ,
003, , x, , x, , , , ,
004, , , , , x, , , ,x
;
run;

desired output:

subject CRIT

001 I1, I2A, I2C, E2A

002 I2A, I3

003 I2A, I2C,

004 I3, E3

@PaigeMiller 's solution below for part1

data want;
    set have;
    length crit $ 200;
    array zzi zzinc:;
    array zze zzexc:;
    crit=' ';
    do i=1 to dim(zzi);
        if zzi(i)='x' then crit=cats(crit,', I',i);
    end;
    do i=1 to dim(zze);
        if zze(i)='x' then crit=cats(crit,', E',i);
    end;
    if crit=:',' then crit=substr(crit,2);
    drop i;
    keep subject crit;
run;

link to part1: https://communities.sas.com/t5/SAS-Programming/How-do-I-turn-variables-into-values-for-a-new-variable/m-p/866127#M342036

Re: How do I turn variables into values for a new variable? (part 2 with a twist)

PaigeMiller — Fri, 24 Mar 2023 15:07:50 GMT

data want2;
    set have2;
    length crit $ 200;
    array zzi zzinc:;
    array zze zzexc:;
    crit=' ';
    do i=1 to dim(zzi);
        if zzi(i)='x' then do;
            varname=vname(zzi(i)); 
            crit=cats(crit,', I',substr(varname,6));
        end;
    end;
    do i=1 to dim(zze);
        if zze(i)='x' then do;
            varname=vname(zze(i)); 
            crit=cats(crit,', E',substr(varname,6));
        end;
    end;
    if crit=:',' then crit=substr(crit,2);
    drop i varname;
    keep subject crit;
run;

Re: How do I turn variables into values for a new variable? (part 2 with a twist)

Hello_there — Fri, 24 Mar 2023 15:14:54 GMT

Thanks, PaigeMiller. Very helpful!

topic Re: How do I turn variables into values for a new variable? (part 2 with a twist) in SAS Programming

How do I turn variables into values for a new variable? (part 2 with a twist)

Re: How do I turn variables into values for a new variable? (part 2 with a twist)

Re: How do I turn variables into values for a new variable? (part 2 with a twist)