How do I make an array of elapsed times?

I have array, stopdates, that I need the number of seconds elapsed from t1 for each date. This is what I have been working with:

format shortg
str = 'April 18, 1995 12:00:00';
t1 = datevec(str,'mmmm dd, yyyy HH:MM:SS');
str2 = stopdate(1,runs);
t2 = datevec(str2,'mmmm dd, yyyy HH:MM:SS');
nt = etime(t2,t1);

[njj1]

What is stopdate?

Here is an easy solution that does not require etime. If you get your t# into datenum instead of datevec, then you can just subtract and cumsum everything.

%given t is a vector of datenum
t = [0;cumsum(diff(t))];

[Madison Lowe]

stopdate is a 1 by 33137 array of dates in a form like str.

Answer 1

[dpb]

str = 'April 18, 1995 12:00:00';
t1 = datetime(str,'inputformat','MMM dd, yyyy HH:mm:ss');
stoptime=datetime(stopdate,'inputformat','MMM dd, yyyy HH:mm:ss');
dt=stoptime-t1;
dtSecs=seconds(dt);

Read up on datetime and duration

Answer 2

[njj1]

OK, so this sounds like it's inside a for loop, in which case you could try something like this:

for runs=number_of_runs:-1:1
  %I run this backwards to ensure that my vector t is pre-allocated
  t(runs) = datenum(stopdate(runs,:),'mmmm dd, yyyy HH:MM:SS');
end
%after this for loop, we have a vector t where each entry is a datenum
t = [0;cumsum(diff(t))];

Hope this helps.

[njj1]

Actually, you don't need to use the for loop if all the date strings in your vector stopdate are in the same format.

t = datenum(stopdate,'mmmm dd, yyyy HH:MM:SS');
t = [0;cumsum(diff(t))];

Answer 3

[Peter Perkins]

Unless you are using a pretty old version of MATLAB (< R2014b), use datetimes:

>> t1 = datetime('April 18, 1995 12:00:00','InputFormat','MMMM dd, yyyy HH:mm:ss')
t1 = 
  datetime
   18-Apr-1995 12:00:00
>> t2 = datetime({'April 20, 1995 22:26:45' 'April 24, 1995 15:25:42' 'April 27, 1995 19:01:23'},'InputFormat','MMMM dd, yyyy HH:mm:ss')
t2 = 
  1×3 datetime array
   20-Apr-1995 22:26:45   24-Apr-1995 15:25:42   27-Apr-1995 19:01:23
>> t2 - t1
ans = 
  1×3 duration array
    58:26:45   147:25:42   223:01:23
>> between(t1,t2)
ans = 
  1×3 calendarDuration array
   2d 10h 26m 45s    6d 3h 25m 42s     9d 7h 1m 23s