Date:         Tue, 15 Apr 2008 10:26:07 -0400
Reply-To:     Paul Dorfman <sashole@BELLSOUTH.NET>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Paul Dorfman <sashole@BELLSOUTH.NET>
Subject:      Re: need help in coding this...
Comments: To: SUBSCRIBE SAS-L Cherish <hawks_cherish@YAHOO.CO.IN>

Mark,

In

>data _null_;
>dsid=open('test',"i");
>n_obs=attrn(dsid,"nobs");
>call symput("n_obs",n_obs);
>run;

you left the file open, which is tantamount to troubling a trouble. Also,
would it not be simpler to just

data _null_ ;
   call symputx ('n', n) ;
   stop ;
   set test n = nobs ;
run ;

Note the use of symputX avoiding the type conversion note.

Kind regards
------------
Paul Dorfman
Jax, FL
------------
Don't shoot my pizza.

On Tue, 15 Apr 2008 12:09:11 +0100, cherish k <hawks_cherish@YAHOO.CO.IN>
wrote:

>Please find the code.
>
>drop the min and max values from the final dataset separately.
>
>data test;
>input f1 f2;
>datalines;
>1 4
>1 3
>2 2
>4 4
>4 5
>;
>run;
>
>data _null_;
>dsid=open('test',"i");
>n_obs=attrn(dsid,"nobs");
>call symput("n_obs",n_obs);
>run;
>
>
>%macro chk;
>
>%do i = 1 %to &n_obs;
>    data test;
>    set test;
>    if _n_ eq %eval(&i) then do;
>        min_val = min(f1,f2);
>        max_val = max(f1,f2);
>    end;
>    if min_val ne . and max_val ne . and f1 = max_val then f1 = min_val;
>    if min_val ne . and max_val ne . and f2 = max_val then f2 = min_val;
>    retain min_val max_val;
>    run;
>%end;
>
>%mend;
>
>%chk;
>
>Cherish
>
>Mark Marquez <sickpurple@GMAIL.COM> wrote: hi, i have a problem in coding
the following scenario...
>
>Given the following data set w/ 5 obs (the # of iterations will be equal
to
>the # of obs):
>f1, f2
>1, 4
>1, 3
>2, 2
>4, 4
>4, 5
>
>1st iteration--point to obs 1 (f1=1, f2=4); get the minimum between f1 &
f2
>(min=1) for the current obs; update starting from the current obs (obs 1)
>until the last obs for all occurrences of the higher value between f1 & f2
>(4) set equal to the minimum (1)
>f1, f2
>1, 4-->1
>1, 3
>2, 2
>4-->1, 4-->1
>4-->1, 5
>
>2nd iteration--point to obs 2 (f1=1, f2=3); get the minimum between f1 &
f2
>(min=1) for the current obs; update starting from the current obs (obs 2)
>until the last obs for all occurrences of the higher value between f1 & f2
>(3) set equal to the minimum (1)
>f1, f2
>1, 1
>1, 3-->1
>2, 2
>1, 1
>1, 5
>
>3rd iteration--point to obs 3 (f1=2, f2=2); skip; no update needed since
>both are equal
>f1, f2
>1, 1
>1, 1
>2, 2
>1, 1
>1, 5
>
>4th iteration--point to obs 4 (f1=1, f2=1); skip; no update needed since
>both are equal
>f1, f2
>1, 1
>1, 1
>2, 2
>1, 1
>1, 5
>
>5th & last iteration--point to obs 5 (f1=1, f2=5); get the minimum between
>f1 & f2 (min=1) for the current obs; update the current obs (obs 5) for
the
>occurrence of the higher value between f1 & f2 (5) set equal to the
minimum (1)
>f1, f2
>1, 1
>1, 1
>2, 2
>1, 1
>1, 5-->1
>
>Final data set--yes, both f1 & f2 will end up having the same values:
>f1, f2
>1, 1
>1, 1
>2, 2
>1, 1
>1, 1
>
>
> between 0000-00-00 and 9999-99-99       <!--11--><hr size=1></hr> Best
Jokes, Best Friends, Best Food. Get all this and more on <a
href="http://in.rd.yahoo.com/tagline_groups_11/*http://in.promos.yahoo.com/
groups/bestofyahoo/"> Best of Yahoo! Groups.</a>