diagnosis code

[Davidliu494]

the goal is to write a SAS program to make diagnosis on a disease. idno = id of patient, visit = visit number, test = result from test (can be A or B),

to make a positive diagnosis, the patient have to have test=A in two consecutive visits. so if test=A in visit 1 and test=A in visit 2, then patient is positive. If A in visit 1 and B in visit 2, patient is negative.

problem occurs when there's visits with missing data. For example, if patient 11 tested A in visit 3, did not get tested in visit 4, and tested again A in visit 5, he should be positive (because technically he was tested A 2 times consecutively).

note, once a patient is labeled as positive, he remains positive forever, regardless of later results

idno visit test

11     1     A

11     2     B

11     3     A

11     4    

11     5     A               = positive

idno visit test

11     1     A

11     2    

11     3     A

11     4     B

11     5     B               = positive

idno visit test

11     1     A

11     2    

11     3     B

11     4    

11     5     A               = negative

thank you for the help! here's the test code to start with:

data test;

    input idno visit test $;

    datalines;

    11    1    A

    11    2    B

    11    3    A

    11    4    .

    11    5    A

    ;

run;

[Linlin]

Hi,

Try the code below:

data test;

    input idno visit test $;

    datalines;

    11    1    A

    11    2    B

    11    3    A

    11    4    .

    11    5    A

    ;

run;

data want;

length diag $ 10;

set test (where=(not missing(test)));

by idno;

if not first.idno and test=lag(test) then diag='positive';

proc print;run;

[Astounding]

Linlin, I think you will have a problem with consecutive values of B.  Here's an alternative:

data want;

   set have;

   by idno;

   length result $ 8;

   retain result;

   if first.idno then do;

      A_count=0;

      result=' ';

   end;

   if test='A' then A_count + 1;

   else if test='B' then A_count=0;

   if A_count > 1 then result='Positive';

run;

This identifies the "Positive" subjects only, but that's easy to change if needed.  It also shows when they first were designated as "Positive".

[Linlin]

Hi Astounding,

I thought it should be "Positive" when there are consecutive values of B.

Thanks - Linlin

[Haikuo]

Oh, My, can't you be sharper? Astounding, you have been a huge asset to this forum, and I am among one of those who have been benefited the most. I have been a big fan since 2 years ago when I first finished one of your books on SAS program efficiency.

Haikuo

[Astounding]

Thanks for the more than kind words.  We all learn here.  Between you and KSharp, I'm actually learning something about hash tables.  Grrrrr.  Watch out, the old dog has a new trick. 

[art297]

: You're not the only old dog who continues to learn new tricks!  Not a play on your screen name, but I've always been astounded that more people don't take advantage of how much help and knowledge sites like this forum, and SAS-L, have to offer.  After almost 40 years working with SAS, I still learn something new every day.

[Haikuo]

LinLin's code is surely working for you, but In case you want to keep all of the rows and just add a variable, here is another way:

data test;

  input idno visit test $;

  datalines;

  11 1 A

  11 2 B

  11 3 A

  11 4 .

  11 5 A

  ;

data want;

do until (last.idno);

set test;

by idno;

length flag _r $10.;

if test=_r and _r='A' then _c+1;

if _c>=1 then flag='Positive';

_r=coalescec(test, _r);

output;

  end;

call missing (of _:);

drop _:;

run;

Haikuo