Time arithmetic (no dates)

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]

How are your times currently stored? As character strings, or ... ?

[N]

I haven't stored them yet. it doesn't matter. I can store them as anything.

[James Tursa]

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.

Answer 1

[Walter Roberson]

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]

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

Answer 2

[Voss]

datetime(0,0,0,4,30,0) - datetime(0,0,0,2,45,0)

ans = duration
   01:45:00

[John D'Errico]

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.

Answer 3

[John D'Errico]

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.