Time arithmetic (no dates)
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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.
댓글 수: 3
James Tursa
2022년 1월 13일
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.
채택된 답변
Walter Roberson
2022년 1월 13일
tdiff = (hours(4)+minutes(30)) - (hours(2) + minutes(45))
[h, m, s] = hms(tdiff)
댓글 수: 1
Steven Lord
2022년 1월 14일
To make that first line look a little closer to the representation in the original question:
t1 = duration(4, 30, 00)
t2 = duration(2, 45, 00)
tdiff = t1-t2
추가 답변 (2개)
Voss
2022년 1월 13일
datetime(0,0,0,4,30,0) - datetime(0,0,0,2,45,0)
댓글 수: 1
John D'Errico
2022년 1월 13일
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
2022년 1월 13일
편집: John D'Errico
2022년 1월 13일
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]
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
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.
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