=20

This assumes no missing characters in the ID structure you described and
no names less than 5 characters.

=20

data _null_;

length id2 $12;

id =3D 'LOLOBRIDGIDA195703F';

id2 =3D substr(id,1,5)||reverse((substr(reverse(id),1,7)));

put _all_;

run;

=20

=20

Jack Clark

Research Analyst

Center for Health Program Development and Management

University of Maryland, Baltimore County

=20

________________________________

From: SAS(r) Discussion [mailto:SA...@LISTSERV.UGA.EDU] On Behalf Of
Irin later
Sent: Wednesday, April 23, 2008 8:36 AM
To: SA...@LISTSERV.UGA.EDU
Subject: How to construct the string ID ?

=20

 I have ID that includes a name, year of birth, month of birch an gender

LOLOBRIDGIDA195703F

=20

I need to construct id  of just 5 letters of that first name and the
rest of the ID above:

LOLOB195703F

=20

How can I implement it with a code (I mean all names has different
length)?

=20

Thank you in advance,

=20

Irin

=20

________________________________

Try GINA1 :). Now seriously (untested),

NewID = cats (substr (ID, 1, 5), substr (ID, anydigit (ID)) ;

Kind regards
------------
Paul Dorfman
Jax, FL
------------

  -------------- Original message from Irin later <irinfig...@YAHOO.COM>: --------------

 I have ID that includes a name, year of birth, month of birch an gender
LOLOBRIDGIDA195703F

I need to construct id  of just 5 letters of that first name and the rest of the ID above:
LOLOB195703F

How can I implement it with a code (I mean all names has different length)?

Thank you in advance,

You could write this out in one statement but I find it easier to
understand and follow if I see the parts.

Data newid;

string ='LOLOBRIDGIDA195703F';

name   = substr( string , 1 , 5 );           * extract five letters of the
name;
string = reverse( string );                  * reverse the string so that
you may read from the right;
stats  = reverse( substr( string , 1 , 7 )); * read the needed info and
reverse;
newid  = compress( name || stats) ;          * create the new variable;
Drop stats name;
run;

Nat Wooding
Environmental Specialist III
Dominion, Environmental Biology
4111 Castlewood Rd
Richmond, VA 23234
Phone:804-271-5313, Fax: 804-271-2977

             Irin later
             <irinfigvam@YAHOO
             .COM>                                                      To
             Sent by: "SAS(r)          SA...@LISTSERV.UGA.EDU
             Discussion"                                                cc
             <SA...@LISTSERV.U
             GA.EDU>                                               Subject
                                       How to construct the string ID ?

             04/23/2008 08:35
             AM

             Please respond to
             irinfigvam@yahoo.
                    com

  I have ID that includes a name, year of birth, month of birch an gender
 LOLOBRIDGIDA195703F

 I need to construct id  of just 5 letters of that first name and the rest
 of the ID above:
 LOLOB195703F

 How can I implement it with a code (I mean all names has different
 length)?

 Thank you in advance,

 Irin

Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it
now.

This assumes no missing characters in the ID structure you described and
no names less than 5 characters.

=20

data _null_;

length id2 $12;

id =3D 'LOLOBRIDGIDA195703F';

id2 =3D substr(id,1,5)||reverse((substr(reverse(id),1,7)));

put _all_;

run;

=20

=20

Jack Clark

Research Analyst

Center for Health Program Development and Management

University of Maryland, Baltimore County

=20

________________________________

From: SAS(r) Discussion [mailto:SA...@LISTSERV.UGA.EDU] On Behalf Of
Irin later
Sent: Wednesday, April 23, 2008 8:36 AM
To: SA...@LISTSERV.UGA.EDU
Subject: How to construct the string ID ?

=20

 I have ID that includes a name, year of birth, month of birch an gender

LOLOBRIDGIDA195703F

=20

I need to construct id  of just 5 letters of that first name and the
rest of the ID above:

LOLOB195703F

=20

How can I implement it with a code (I mean all names has different
length)?

=20

Thank you in advance,

=20

Irin

=20

________________________________

   data _null_;
   do ID = 'Goodnight123456A', 'Sall234567B';
      NewID = cats (substr (ID, 1, min(anydigit (ID) - 1, 5) )
                    , substr (ID, anydigit (ID) ) );
      put id= newid=;
      end;
   run;

Data newid;

id ='L LO195703F';

NewID = cats( compress( substr (ID, 1, 5) , '0123456789' ), substr (ID,
anydigit (ID) )) ;
put newid;
run;

Nat Wooding
Environmental Specialist III
Dominion, Environmental Biology
4111 Castlewood Rd
Richmond, VA 23234
Phone:804-271-5313, Fax: 804-271-2977

             Irin later
             <irinfigvam@YAHOO
             .COM>                                                      To
             Sent by: "SAS(r)          SA...@LISTSERV.UGA.EDU
             Discussion"                                                cc
             <SA...@LISTSERV.U
             GA.EDU>                                               Subject
                                       Re: How to construct the string ID
                                       ?
             04/23/2008 09:14
             AM

             Please respond to
             irinfigvam@yahoo.
                    com

 Jack, I do not have spaces, but scenarion with less than 5 letters did not
 even  come to my mind .It actually can….

 Is it possible to do it in such a case?

 --- On Wed, 4/23/08, Jack Clark <JCl...@CHPDM.UMBC.EDU> wrote:
  From: Jack Clark <JCl...@CHPDM.UMBC.EDU>
  Subject: Re: How to construct the string ID ?
  To: SA...@LISTSERV.UGA.EDU
  Date: Wednesday, April 23, 2008, 8:49 AM

  Irin,

  This assumes no missing characters in the ID structure you described and
  no names less than 5 characters.

  data _null_;

  length id2 $12;

  id = 'LOLOBRIDGIDA195703F';

  id2 = substr(id,1,5)||reverse((substr(reverse(id),1,7)));

  put _all_;

  run;

  Jack Clark

  Research Analyst

  Center for Health Program Development and Management

  University of Maryland, Baltimore County

  ________________________________

  From: SAS(r) Discussion [mailto:SA...@LISTSERV.UGA.EDU] On Behalf Of
  Irin later
  Sent: Wednesday, April 23, 2008 8:36 AM
  To: SA...@LISTSERV.UGA.EDU
  Subject: How to construct the string ID ?

   I have ID that includes a name, year of birth, month of birch an gender

  LOLOBRIDGIDA195703F

  I need to construct id  of just 5 letters of that first name and the
  rest of the ID above:

  LOLOB195703F

  How can I implement it with a code (I mean all names has different
  length)?

  Thank you in advance,

  Irin

  ________________________________

  Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try
  it now.
  <http://us.rd.yahoo.com/evt=51733/*http:/mobile.yahoo.com/;_ylt=Ahu06i62
  sR8HDtDypao8Wcj9tAcJ%20>

If you don't, then you have to go to the more ugly approach of explicitly extracting the last 7 characters (the "YYYYMMX" at the end).  This function will deal with that.

    newid2 = cats(   substr(id,1,min(length(id)-7,4))
                    ,substr(id,length(id)-6)
                 );

I'd also suggest that you might want all the NEWID's to have the same length, which won't happen with short names.  In that case, let's say you want to put an underscore as a filler whenever the name part of the id is too short.  Which would introduce the complications in this version:

  newid3 =  cats(  substr(cats(substr(id,1,min(length(id)-7,5)),'____'),1,5)
                  ,substr(id,length(id)-6)
                );

459  data _null_;
460     length id ref $50;
461     input id @1 ref;
462     if length(id) gt 12 then substr(id,6)=substr(id,length(id)-6);
463     else id =
subpad(substr(id,1,length(id)-7),1,5)||substr(id,length(id)-6);
464     put (_all_)(:);
465     cards;

LOLOB195703F LOLOBRIDGIDA195703F
DudeB200804M DudeBornYesterday200804M
Dude 200804M Dude200804M
Du   200804M Du200804M

On Wed, Apr 23, 2008 at 8:26 AM, Howard Schreier <hs AT dc-sug DOT

Data newid;

id =3D'L LO195703F';

NewID =3D cats( compress( substr (ID, 1, 5) , '0123456789' ), substr
(ID,
anydigit (ID) )) ;
put newid;
run;

Nat Wooding
Environmental Specialist III
Dominion, Environmental Biology
4111 Castlewood Rd
Richmond, VA 23234
Phone:804-271-5313, Fax: 804-271-2977

                                                                          =
=20
             Irin later                                                   =
=20
             <irinfigvam@YAHOO                                          =
  =20
             .COM>                                                      =
To=20
             Sent by: "SAS(r)          SA...@LISTSERV.UGA.EDU            =
=20

             Discussion"                                                cc

             <SA...@LISTSERV.U                                          =
  =20
             GA.EDU>                                               Subje=
ct=20
                                       Re: How to construct the string ID =
=20
                                       ?                                  =
=20
             04/23/2008 09:14                                             =
=20
             AM                                                           =
=20
                                                                          =
=20
                                                                          =
=20
             Please respond to                                            =
=20
             irinfigvam@yahoo.                                            =
=20
                    com                                                   =
=20
                                                                          =
=20
                                                                          =
=20

I am using Windows XP and SAS 9.1...I think this might get what you want.

Data Have;
  Input @1 Old $14.;

Datalines;
Adam19480204m
Stlo19620408m
;

Data Need (Drop = Temp1 Temp2 New);
  Length New $14.;
  Set Have;
  Temp1 = Left(Reverse(Old));
  Sex   = Substr(Temp1,1,1);
  Temp2 = Substr(Temp1,4,10);
  New   = Reverse(Temp2);
  Fin   = Cats(New,Sex);
Run;

Proc Print;
Var Old Fin Sex;
Run;

On Thu, 1 May 2008 10:47:13 -0700, Irin later <irinfig...@YAHOO.COM>
wrote:

  Temp1 = Left(Reverse(Old));

I am using Windows XP and SAS 9.1...I think this might get what you want.

Data Have;
  Input @1 Old $14.;

Datalines;
Adam19480204m
Stlo19620408m
;

Data Need (Drop = Temp1 Temp2 New);
  Length New $14.;
  Set Have;
  Temp1 = Left(Reverse(Old));
  Sex   = Substr(Temp1,1,1);
  Temp2 = Substr(Temp1,4,10);
  New   = Reverse(Temp2);
  Fin   = Cats(New,Sex);
Run;

Proc Print;
Var Old Fin Sex;
Run;

 Temp1 = Left(Reverse(Old));

Next, isolate M or F.

 Sex   = Substr(Temp1,1,1);

Now skip 4 characters over and read the rest

 Temp2 = Substr(Temp1,4,10);

Now place it back in the original order (could have said temp2 = reverse
(temp2), but wanted to break out the process so you could see it)

 New   = Reverse(Temp2);

Use Cats to remove leading and trailing blanks and || together the two
variables into one

 Fin   = Cats(New,Sex);

Clear as mud?

Fin has just Gender letter! I am not sure if I properly explained the issue

This is my old id

ADAM19480204M
This should be my new id
ADAM194802M (Short one without "04" - day):

SESIL20071210F
SESIL200712F (short one without "10" - day):