Date:         Wed, 5 Mar 2008 08:03:13 -0500
Reply-To:     Arthur Tabachneck <art297@NETSCAPE.NET>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Arthur Tabachneck <art297@NETSCAPE.NET>
Subject:      Re: Question abount counting strings
Comments: To: Guylhem Aznar <sasllist@GUYLHEM.NET>

Guylhem,

I, too, probably don't understand what you are trying to do.  Does the
following provide the result you are seeking to accomplish?

data want (drop=want: i j transitions);
  array want(4);
  input patient_id $ states $;
  transitions='NH HD HP PH';
  call missing (of want:);
  do i=1 to length(states)-1;
    do j=1 to 4;
      if substr(states,i,2) eq scan(transitions,j) then want(j)+1;
    end;
  end;
  nh=want(1);
  hd=want(2);
  hp=want(3);
  ph=want(4);
datalines;
124883 NHD
124884 N
124885 NHPHD
124886 NHPHP
;

Art
---------
On Wed, 5 Mar 2008 11:53:34 +0100, Guylhem Aznar <sasllist@GUYLHEM.NET>
wrote:

>Hello
>
>On 3/4/08, Mike Rhoads <RHOADSM1@westat.com> wrote:
>>  So, a "state" is equivalent to a single letter, and a "transition" is
>>  the combination of 2 adjacent states/letters, right?
>
>True.
>
>>  If my understanding is correct, I wouldn't bother with regular
>>  expressions.  I would just use a DATA step with a DO-loop to create a
>>  restructured data set with one record for each patient and transition.
>
>Sounds like a much better idea.
>
>I did a minor adaptation to your code, because I did end the
>transition (called "parcours in french" with a dot as a terminator
>char.
>
>DATA Transitions;
>/* parcours == transition */
>SET Work.Parcours (KEEP=patient_id parcours);
>
>DO I = 1 TO LENGTH(parcours) - 2;
> transition = SUBSTR(parcours,I,2);
> OUTPUT;
>END;
>RUN;
>
>This gives something like
>
>patient_id parcours
>80465      NHD
>
>patient_id parcours I transition
>80465       NHD      1 NH
>80465       NHD      2 HD
>
>However the result isn't exactly what I expected.
>
>Ideally, it would create for each patient a list of the transition
>found, ie of the 2 consecutive letters in the string, matched by a
>count:
>
>transition
>NHD
>N
>NHPHD
>
>should give
>
>transition nh hd hp ph
>NHD        1   1   .   .
>N             .    .   .   .
>NHPHD    1   1  1   1
>NHPHP    1   .   2   1
>
>I guess I would need a transpose somewhere. Yet I wonder how I will be
>able to count identical transition in the proc transpose (NHPHP
>example : 2 HP transitions)
>
>>  Note that this does not create any records for patients like 124886,
who
>>  have only one state and therefore no transitions.  You could modify the
>>  code to create a record for these patients if necessary.
>
>No - that's the right way to do. If they are in a continuous state,
>they have no transition.
>
>Guylhem