Date:         Thu, 8 Jun 2006 09:19:54 -0400
Reply-To:     "Howard Schreier <hs AT dc-sug DOT org>" <nospam@HOWLES.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         "Howard Schreier <hs AT dc-sug DOT org>" <nospam@HOWLES.COM>
Subject:      Re: How SAS Executes this Code?

On Thu, 8 Jun 2006 00:45:10 -0700, Jaggie <jagadishkr@GMAIL.COM> wrote:

>"Howard Schreier <hs AT dc-sug DOT org>" wrote:
>> On Tue, 6 Jun 2006 06:56:27 -0400, Richard A. DeVenezia
>> <rdevenezia@WILDBLUE.NET> wrote:
>>
>> >Jaggie wrote:
>> >> Hi All,
>> >>
>> >> I have a small code which is haunting me. Code is very simple but
>> >> unable to understand how SAS executes this code.
>> >>
>> >> data x;
>> >> input Price;
>> >> Cards;
>> >> 1.08
>> >> 1.10
>> >> 1.14
>> >> 1.23
>> >> 1.29
>> >> ;
>> >> run;
>> >>
>> >> data y;
>> >> do i=1 to 3;
>> >>  set x;
>> >> end;
>> >> run;
>> >>
>> >> Data set X has 5 obs. I tried with Put _ALL_ statement and found
below
>> >> observations,
>> >>
>> >> If I use do loop with ,
>> >> i =1 to 1 then it outputs 5 obs to dataset Y.
>> >> i=1 to 2 then 2 obs
>> >> i=1 to 3 then 1 obs
>> >> i=1 to 4 then 1 obs
>> >> i=1 to 5 then also 1 obs
>> >> i=1 to 6 or above then 0 obs.
>> >>
>> >> Please let me know how SAS executes this code. Especially when we use
>> >> set statement  within do loops.
>> >>
>> >> Thanks,
>> >> Jagadish
>> >
>> >
>> >- Each time the SET statement is reached a row is read from Table x
>> >- Each time the flow of control reaches the bottom of the Step, a row
is
>> >written via an implied "OUTPUT".  (If you use an explicit OUTPUT
statement
>> >anywhere in the STEP the implied output does not occur)
>>
>> and
>>
>> - If the SET statement is reached but there are no more unread rows,
>> execution ceases.
>>
>> >
>> >Let NOBS be 5 and MAX_I be the top of your loop -- thus in your code,
the
>> >bottom of the Step is reached every MAX_I'th row and you get
>> >floor(NOBS/MAX_I) rows output.
>> >
>> >There are far better discussions (conference papers, online help,
books by
>> >users, etc...) concerning the operation of the DATA Step than this
short
>> >answer.  Be sure to check them out.
>> >
>> >--
>> >Richard A. DeVenezia
>> >http://www.devenezia.com/
>
>Then what is the difference between below codes.
>
>data y;
>do i=1 to 5;
> set x;
>end;
>run;
>
>In above dataset if x has 5 obs, then out put dataset has 1 obs.
>
>data y;
> set x;
> set x;
> set x;
> set x;
> set x;
>run;
>
>I think in the above code, 5 copies of dataset x are created and
>whenever set is encountered observation in its corresponding dataset
>moves to next obs. But all the obs from last set x statement are out
>put. Am I right?

That's pretty close. I think it's a bit more accurate to say that the
mechanism for reading X (buffers, pointers, etc.) is set up 5 times and
that the 5 instances function independently.

There is also

   set x x x x x ;

which is quite different.

>
>Thanks,
>Jagadish