topic Re: ifn susbtrn for multiple lengths in SAS Procedures

ifn susbtrn for multiple lengths

lillymaginta — Sat, 22 Sep 2018 12:43:47 GMT

I have the following code:

data a (drop=i);

set b;

array x{*} dx:;

hf=0;

do i=1 to dim(x);

hf=ifn(substrn(x{i}, 1, 3) in('428', 'I43'),1,hf);

end;

run;

data a (drop=i);

set b;

array x{*} dx:;

hf=0;

do i=1 to dim(x);

hf=ifn(substrn(x{i}, 1, 4) in('d282', 'd432'),1,hf);

end;

run;

Is there is a way to combine the two codes into one code given that I am requiring different lengths in the substrn statement, (For example 3 in the first and 4 in the second).

I am looking to get a code that would be something similar to:

hf=ifn(substrn(x{i}, 1, 3) in('428', 'I43'),1,hf) or hf=ifn(substrn(x{i}, 1, 4) in('d282', 'd432'),1,hf);

Is it possible to have such statement?

Thanks

Re: ifn susbtrn for multiple lengths

VDD — Sat, 22 Sep 2018 12:52:18 GMT

have you tried:

if hf=ifn(substrn(x{i}, 1, 3) in('428', 'I43'),1,hf);

else if hf=ifn(substrn(x{i}, 1, 4) in('d282', 'd432'),1,hf);

Re: ifn susbtrn for multiple lengths

lillymaginta — Sat, 22 Sep 2018 13:00:23 GMT

Thank you VDD, the statement is not correct and would generate an error (no matching if-then clause)

Re: ifn susbtrn for multiple lengths

VDD — Sat, 22 Sep 2018 13:07:18 GMT

does this link help:

http://support.sas.com/documentation/cdl/en/lrdict/64316/HTML/default/viewer.htm#a002604573.htm

Re: ifn susbtrn for multiple lengths

lillymaginta — Sat, 22 Sep 2018 13:28:57 GMT

Thank you for sharing the link. If I understand correctly, they did not use the else statement with "ifn" so I am not sure it is applicable here. Maybe one of the advisors can help (@RW9)

Re: ifn susbtrn for multiple lengths

art297 — Sat, 22 Sep 2018 14:01:41 GMT

Not sure what you are trying to do but, if i do understand correctly, the following should work:

data b;
  input (dx1-dx5) ($);
  datalines;
d234 d463 d213 d282 d435
d213 428 d360 d145 d269
d409 d231 I43 d690 d432
234 463 213 678 435
213 428 360 145 269
409 231 I43 690 609
;
data a (drop=i);
  set b;
  array x{*} dx:;
  hf=0;
  do i=1 to dim(x);
    hf=ifn(substrn(x{i}, 1, 3) in('428', 'I43')
      or
           substrn(x{i}, 1, 4) in('d282', 'd432'),
    1,hf);
  end;
run;

Art, CEO, AnalystFinder.com

Re: ifn susbtrn for multiple lengths

data_null__ — Sat, 22 Sep 2018 14:41:23 GMT

Seems like you're making this too hard. It would be helpful to see your input data. I borrowed from @art297 please indicate if that assumption is correct.

data b;
   id + 1;
   array dx[5] $8; 
   input dx[*];
   array _hf[4] $8 _temporary_ ('428' 'I43' 'd282' 'd432');
   array _st[2] $8 _temporary_ ('d269' '428');
   hf=0; st=0; 
   do i = 1 to dim(dx);;
      hf + dx[i] in _hf);
      st + dx[i] in _st);
      end;
   hf = hf gt 0;
   st = st gt 0;
   drop i;
   label hf='Heart Failure' st='Stroke';
   datalines;
d234 d463 d213 d282 d435
d213 428 d360 d145 d269
d409 d231 I43 d690 d432
234 463 213 678 435
213 428 360 145 269
409 231 I43 690 609
;;;;
   run;
proc print;
   run;

Re: ifn susbtrn for multiple lengths

lillymaginta — Sat, 22 Sep 2018 15:06:46 GMT

Thank you!