Date:         Fri, 15 Oct 1999 11:13:02 -0400
Reply-To:     "Brucken, Nancy" <Nancy.Brucken@WL.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         "Brucken, Nancy" <Nancy.Brucken@WL.COM>
Subject:      Re: What's wrong with the lag function?
Content-Type: text/plain; charset="iso-8859-1"

Hi Ya,
   This description of the LAG function comes from the SAS 6.12 online help:

The LAG functions, LAG1, LAG2, . . .  , LAG100 return values from a queue.
LAG1 can also be written as LAG.  A LAGn  function stores a value in a queue
and returns a value stored previously in that queue.  Each occurrence of a
LAGn  function in a program generates its own queue of values. The queue for
a LAGn function is initialized with n  missing values, where n  is the
length of the queue (for example, a LAG2 queue is initialized with two
missing values).  When the LAGn  function is executed, the value at the top
of the queue is removed and returned, the remaining values are shifted
upwards, and the new value of the argument is placed at the bottom of the
queue.  Hence, missing values are returned for the first n  executions of a
LAGn  function, after which the lagged values of the argument begin to
appear. Storing values at the bottom of the queue and returning values from
the top of the queue occurs only when the function is executed.  A LAGn
function that is executed conditionally will store and return only values
from the observations for which the condition is satisfied. If the argument
of a LAGn  function is an array name, a separate queue is maintained for
each variable in the array.

Copyright (c) 1995, SAS Institute Inc., Cary, NC 27513-2414 USA.  All rights
reserved.

So, what I think it's saying is that the first time the LAG function is
invoked, it returns a missing value.  In your first example, it's only
invoked when C=0.  That condition is met for the first time on the third
record in your dataset, so the LAG function comes back with a missing value.
In your second example, it's called for every record.  So, by the time your
third record is read, it has already been called twice before, and thus
returns the value that you expect.  Does this make any sense?

Hope it helps,
    Nancy

Nancy Brucken
Parke-Davis, QIP
(734) 622-5767
E-mail address:  Nancy.Brucken@wl.com


-----Original Message-----
From: Huang, Ya [mailto:Y.Huang@ORGANONINC.COM]
Sent: Friday, October 15, 1999 10:27 AM
To: SAS-L@LISTSERV.UGA.EDU
Subject: What's wrong with the lag function?


Hello All,

This question seems to have been asked by somebody long time ago, I don't
remember the answer, so forgive me to ask it again.

Can somebody explain to me why I have to use a intermediate variable
to keep the lag value and then I can use it to assign to other variable?

DATA XX;
INPUT A B C;
CARDS;
1 2 1
3 4 1
5 6 0
7 8 1
;
data yy; set xx;
if c=0 then a=lag(b); /** not working ***/

PROC PRINT;

data zz; set xx;
d=lag(b);
if c=0 then a=d; /** this way works **/

proc print;
run;

Syntaxly no error in both of the data steps, logically I really don't see
the difference between these two data steps, so what is really going on?

Thanks

Ya Huang



Nancy Brucken
Parke-Davis, QIP
(734) 622-5767
E-mail address:  Nancy.Brucken@wl.com