Date:         Sun, 26 Dec 2010 22:35:02 -0500
Reply-To:     Nat Wooding <nathani@VERIZON.NET>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Nat Wooding <nathani@VERIZON.NET>
Subject:      Re: Is there an easier way to solve this?
In-Reply-To:  <000301cba573$d1fea620$75fbf260$@com>
Content-Type: text/plain; charset="us-ascii"

Max

That is neat and I don't have to take Excedrin in order to look at the code.


Nat
-----Original Message-----
From: SAS(r) Discussion [mailto:SAS-L@LISTSERV.UGA.EDU] On Behalf Of bbser
2009
Sent: Sunday, December 26, 2010 10:12 PM
To: SAS-L@LISTSERV.UGA.EDU
Subject: Re: Is there an easier way to solve this?

Wow, so many people like me are SASing while Christmasing.

Art

Alternatively, I tried using tranwrd().

data b;
        x="New York NY 85044";
        Length zip $ 5 state $ 2 city $ 30;
        Zip=scan(x, -1);
        State=scan(x, -2);
        City=tranwrd(x, state||" "||zip, " ");
run;

proc print;
run;

Cheers, Max

-----Original Message-----
From: SAS(r) Discussion [mailto:SAS-L@LISTSERV.UGA.EDU] On Behalf Of Arthur
Tabachneck
Sent: December-26-10 11:07 AM
To: SAS-L@LISTSERV.UGA.EDU
Subject: [SAS-L] Is there an easier way to solve this?

The following was a question that was raised on the SAS discussion forum.
You are confronted with data that has 3 lines per subject, but the third
line has variables that may contain embedded spaces, but there is only one
space between variables.

The only suggestion I could think of was the one shown below.  Is there an
easier way?

data work.Address (drop=_:);
  infile cards;
  input FirstName $ LastName $ /
        Address $ 1 - 20 /
        _Third_Line & $80.;
  format City $10.;
  Zip=scan(_Third_Line,-1);
  State=scan(_Third_Line,-2);
  call scan(_Third_Line, -2, _position, _length);
  City=substr(_Third_Line,1,_position-1);
  cards;
Lee Athnos
1215 Raintree Circle
New York NY 85044
Heidie Baker
1751 Diehl Road
Vienna VA 22124
;

Thanks in advance,
Art