topic Re: SAS programming in SAS Programming

SAS programming

UcheOkoro — Fri, 03 Jul 2020 20:52:02 GMT

I would like to use the SAS array statement to create 3 new variables(yes or no) from 48 variables that contain just 3 drug classes, Benzos, Opioids and Anticholinergics. So the 48 variables are listed as drugflag1-drugflag48.

The new variables 1-3 are Benzo_rc, Opioid_rc and Antichol. For variable Benzo, if drugflag1- drugflag48 has Benzo then the Variable Benzo_rc="Yes" if not Benzo_rc

Data Merged_data3;
Set Merged_data3;
array drugname $DrugFlag1-DrugFlag48;
do i = 1 to 48;
IF drugname[i]= BENZO THEN benzo_rc = 1; ELSE benzo_rc = 0;End;Run;

="No".

I have tried the following code but it is not working. Please, I need help.

Re: SAS programming

Reeza — Fri, 03 Jul 2020 20:59:22 GMT

Look up WHICHC() instead.

Benzo_RC = whichc('BENZO', of drugname(*))>1;
Opioid_RC = whichc('OPIOID', of drugname(*))>1;

Re: SAS programming

UcheOkoro — Fri, 03 Jul 2020 21:05:08 GMT

Please, could you send a link? I could not find it online.

Thank you.

Re: SAS programming

Reeza — Fri, 03 Jul 2020 21:23:08 GMT

Google: WHICHC SAS function site:documentation.sas.com

Re: SAS programming

Kurt_Bremser — Fri, 03 Jul 2020 21:32:09 GMT

Your "else" branch causes you to only pick up the value for the 48th variable.

Such loops need to be coded differently:

data Merged_data4; /* don't overwrite your input */
set Merged_data3;
array drugname $DrugFlag1-DrugFlag48;
benzo_rc = 0;
do i = 1 to 48;
  if drugname[i] = "BENZO"
  then do;
    benzo_rc = 1;
    leave;
  end;
end;
run;

The LEAVE statement will end the loop quickly if a match is found.