Date:         Wed, 13 Sep 2006 08:51:47 -0700
Reply-To:     "Nordlund, Dan (DSHS)" <NordlDJ@DSHS.WA.GOV>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         "Nordlund, Dan (DSHS)" <NordlDJ@DSHS.WA.GOV>
Subject:      Re: Lead function?
Comments: To: "Dunnigan, Keith" <Keith.Dunnigan@TYCOHEALTHCARE.COM>
Content-Type: text/plain; charset=utf-8

> -----Original Message-----
> From: SAS(r) Discussion [mailto:SAS-L@LISTSERV.UGA.EDU] On Behalf Of
> Dunnigan, Keith
> Sent: Wednesday, September 13, 2006 8:38 AM
> To: SAS-L@LISTSERV.UGA.EDU
> Subject: Lead function?
>
> Dear SAS-L'rs,
>
>   I can state my question best with an example, the below dataset gives
> me what I want.  The first 2 variables are the key ones in the input, I
> and t_i.  They correspond to timepoint and the time at timepoint i in
> hours.  I would prefer to calculate the second two variables which are
> t_i lagged or lead by one timepoint.  SAS has a lag function which will
> give me the second variable.  From what I have been able to find they
> don't have a similar 'lead' function to calculate the second one.  Can
> someone give my some code to do this (the simpler the better)?  I should
> say also that the number of timepoints is in general unknown (that is,
> don't hardcode a '4'  into the code).  The incoming dataset can vary,
> but will include I and t_i.
>
>
> ** Times dataset.;
>
>
>
> data times;
>
> input i t_i t_im1 t_ip1;
>
> cards;
>
> 1  0.5  .    1
>
> 2  1    0.5  2
>
> 3  2    1    3
>
> 4  3    2    .
>
> run;
>
>
>   Thanks for any help or advice you can give!
>
>     Keith Dunnigan
>     Sr. Biostatistician
>     Mallinckrodt Medical / Tyco Healthcare

Keith,

Here is a brute force method that is simple enough.  I am sure someone will
provide a much more elegant one.

data times;
  input i t_i /*t_im1 t_ip1*/;
cards;
1  0.5  .    1
2  1    0.5  2
3  2    1    3
4  3    2    .
;
run;
proc sort data=times;
  by descending i;
run;
data times2;
  set times ;
  t_ip1=lag(t_i);
run;
proc sort data=times2;
  by i;
run;
data times3;
  set times2 ;
  t_im1=lag(t_i);
run;
proc print data=times3;
run;

Hope this is helpful,

Dan

Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA  98504-5204