how to calculate the differences start and stop time

[HitmonTran]

hello,

I need to calculate the # mins from start and stop time; however, i have one subject with a special situation. 

Most subjects have start/stop time within the same day, but one subject (subject dd) has a stop time after midnight, which causes the calculation to be a negative number eg. 00:57 - 19:10 = -1093 mins.  How would I solve this issue?

data:

subject	start time (time5. format)	stop time (time5. format)
aa	14:35	14:02
bb	10:46
cc	13:05	11:37
dd	00:57	19:10

want:

subject	mins
aa	33
bb	.
cc	88
dd	347

[PeterClemmensen]

Shouldn't start and stop time be reversed? If starttime and stoptime is within the same day for subject = 'aa', then it can not start at 14:35 and stop at 14:02?

[PeterClemmensen · Accepted solution]

See if this gives you the desired result. Should be easy to follow

data have;
input subject $ (starttime stoptime)(:time5.);
infile datalines missover;
format starttime stoptime time5.;
datalines;
aa 14:02 14:35 
bb 10:46       
cc 11:37 13:05 
dd 19:10 00:57 
;

data want(drop = starttime stoptime);
   set have;
   
   if stoptime < starttime then stoptime = stoptime + 3600*24;

   mins = intck('minute', starttime, stoptime);
run;

Result:

subject  mins
aa       33
bb       .
cc       88
dd       347