# Time arithmetic (no dates)

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N on 13 Jan 2022
Commented: Steven Lord on 14 Jan 2022
I want to subtract two times only (no dates):
e.g. 4:30 - 2:45
Is there a function to do that? I do not want any dates involved.
James Tursa on 13 Jan 2022
Well, the method for subtracting them does depend on how they are stored. E.g., you could store them as decimal seconds in double variables and just subtract them.

Walter Roberson on 13 Jan 2022
tdiff = (hours(4)+minutes(30)) - (hours(2) + minutes(45))
tdiff = duration
1.75 hr
[h, m, s] = hms(tdiff)
h = 1
m = 45
s = 0
Steven Lord on 14 Jan 2022
To make that first line look a little closer to the representation in the original question:
t1 = duration(4, 30, 00)
t1 = duration
04:30:00
t2 = duration(2, 45, 00)
t2 = duration
02:45:00
tdiff = t1-t2
tdiff = duration
01:45:00

_ on 13 Jan 2022
datetime(0,0,0,4,30,0) - datetime(0,0,0,2,45,0)
ans = duration
01:45:00
John D'Errico on 13 Jan 2022
LOL. That is really a better answer than mine. But it does use dates, so I tried to avoid using time functionality in MATLAB. Even so, +1.

John D'Errico on 13 Jan 2022
Edited: John D'Errico on 13 Jan 2022
4:30 is not a number. In order for you to work with something like that, you need to first convert it to a vector of TWO numbers. Thus something like [4,30], representing hours and minutes. In order to subtract them, do this:
timediff = [4 30] - [2 45]
timediff = 1×2
2 -15
The problem is, now you need to work in base 60. That is, if the second element in that time vector is greater than 60, or less than 0, you need to resolve the issue with a carry. Something like...
while timediff(2) < 0
timediff = timediff + [-1 60];
end
while timediff(2) > 60
timediff = timediff + [1 -60];
end
timediff
timediff = 1×2
1 45
You can make a simple function of this. If the first element exceeds 24, or is less than zero, you will need to think about how this would work for time differentials that are more than days, or even weeks, etc. The same idea would apply then. Just remember there are 24 hours in a day.