Date:         Fri, 28 Oct 2005 17:32:47 -0400
Reply-To:     Tony Yang <tonyyangsxz@GMAIL.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Tony Yang <tonyyangsxz@GMAIL.COM>
Subject:      Re: A complicated data management problem
Comments: To: Peter Crawford <peter.crawford@blueyonder.co.uk>
In-Reply-To:  <200510281049.j9SAk0va002617@malibu.cc.uga.edu>
Content-Type: text/plain; charset=ISO-8859-1

Thanks much, Peter, and sorry for the confusing question.
 Let me re-state the question and add one more subject for my question:
 1. I want to calculate different time interval based on the IOP value,
firstly we fixed the baseline value of IOP for each subject, here
corresponding to *1 9/23/1997 0.40* for subject 2.
 2. Then we make a difference between the subsequent value and baseline
value, if the difference, >=0.2, is first detected, then we calculate the
interval between the date between founded IOP and the date for the baseline
IOP value, we can set it as INTERVAL1, and the number of visit time can be
set as VISIT1;
 3. Since we only need the first detection, difference >=0.2, hence for the
remaining part of the date variable, we can calculate the interval between
last date for the visit(here it is 12/5/2003 for subject 2) and the first
date after detection(here it is 5/12/2000 for subject 2), and we can set it
as INTERVAL2, and the number of visit for this period is VISIT2.
 4. For these subjects that we can not detect the condition >=0.2, then we
calculate the total number of visits, and the time interval, and put the
corresponding value to INTERVAL1 and INTERVAL2, VISIT1 and VISIST2
simutaneously.
 Therefore, the final output can take this form(I put the concrete date for
length calculation, actually it should be a number, please do not be misled)
 id n1 length1 n2 length2
1 5 4/26/1999-9/23/1997 5 2/5/2003-5/12/2000
2 9 1/11/2005-12/8/1993 9 1/11/2005-12/8/1993

data er;
input id date mmddyy10. IOP;
format date mmddyy10.;
cards;
1 12/8/1993 0.50
1 1/19/1999 0.45
1 2/16/1999 0.60
1 8/24/2000 0.40
1 2/12/2002 0.45
1 1/22/2003 0.45
1 7/23/2003 0.50
1 1/21/2004 0.40
1 1/11/2005 0.45
2 9/23/1997 0.40
2 12/17/1997 0.45
2 6/4/1998 0.50
2 10/12/1998 0.30
2 4/26/1999 0.65
2 5/12/2000 0.60
2 6/15/2001 0.60
2 2/5/2002 0.30
2 6/18/2002 0.30
2 2/5/2003 0.35
;
proc print data=er;
run;
 BTW, this is a Glaucoma medical record data. Thanks much for your
suggestions. Have a nice weekend to you all!
 Best regards,
Tony