A one data step approach will be

data have; input id cycle $ date :ddmmyy.; format date ddmmyy10.; cards; 1 c1 11/01/2008 1 c1 12/01/2008 1 c1 13/01/2008 1 c2 15/02/2008 1 c2 16/02/2008 1 c2 17/02/2008 1 c2 19/02/2008 1 c3 11/03/2008 1 c3 20/03/2008 1 c4 15/04/2008 1 c4 20/04/2008 ; run;

data want; retain prev .; do until(last.cycle); set have end = eof; by cycle; if first.cycle then do; if prev = . then prev = date; else do; gap = date - prev; output; prev = date; end; end; if eof then do; gap = date - prev; output; end; end; drop prev; run;

Kind regards,

Muthia Kachirayan

On Thu, Oct 22, 2009 at 5:20 PM, gkk <gayakrup@gmail.com> wrote:

> Hi , > > I have a condition where the subjects are treated in cycles some can > have just one cycle whereas others can have upto 10 cycles. > > I have to calculate the duration of the cycle which will be first date > of first cycle - first date of second cycle which will be the duration > for first cycle .similarly for second its first day of second cycle - > first date of third .If its the last cycle then its first date of last > cycle - last date of last cycle. > > eg: > > id cycle date > 1 c1 11/01/2008 > 1 c1 12/01/2008 > 1 c1 13/01/2008 > 1 c2 15/02/2008 > 1 c2 16/02/2008 > 1 c2 17/02/2008 > 1 c2 19/02/2008 > 1 c3 11/03/2008 > 1 c3 20/03/2008 > 1 c4 15/04/2008 > 1 c4 20/04/2008 > > > > > Thanks in advance. >