Looping with datetime greater and less than 24 hour

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eko supriyadi
eko supriyadi 2022-6-3
回答： Stephen23 2022-6-4
Hi community,
i'm stuck with looping and datetime
suppose, i have datetime array (actualiy i have large array):
a=[duration(0,22,0);duration(0,52,0);duration(24,06,0);duration(0,1,0);duration(-24,-5,0)];
so, it will produce
a =
5×1 duration array
00:22:00
00:52:00
24:06:00
00:01:00
-24:05:00
Now, i want to subtract those array that are greater than 24 hours by 24 hours and add times that are less than 24 hours by 24 hours..
the result what i want is like this:
a =
5×1 duration array
00:22:00
00:52:00
00:06:00
00:01:00
-00:05:00
this is the construct loop i have done but it still fails:
a=[duration(0,22,0);duration(0,52,0);duration(24,06,0);duration(0,1,0);duration(-24,-5,0)];
a1=[];
for i=1:length(a)
if a(i) >= hours(24);
a1 = a(i)-hours(24);
elseif a(i) < hours(-24);
a1 = a(i)+hours(-24);
else a1 == a;
end;end
appreciated much help!

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采纳的回答

Steven Lord
Steven Lord 2022-6-3
Ran in:
If the duration is less than -24 hours you want to add 24 hours to it not add -24 hours, right? Also FYI: you can pass vectors into the duration function to create a vector of durations.
a= duration([0; 0; 24; 0; -24], [22; 52; 6; 1; -5], [0; 0; 0; 0; 0])
a = 5×1 duration array
00:22:00 00:52:00 24:06:00 00:01:00 -24:05:00
a(a > hours( 24)) = a(a > hours( 24)) - hours(24)
a = 5×1 duration array
00:22:00 00:52:00 00:06:00 00:01:00 -24:05:00
a(a < hours(-24)) = a(a < hours(-24)) + hours(24)
a = 5×1 duration array
00:22:00 00:52:00 00:06:00 00:01:00 -00:05:00

更多回答（4 个）

Walter Roberson
Walter Roberson 2022-6-3
编辑：Walter Roberson 2022-6-3
h24 = hours(24);
a = mod(a, h24);
mod() works for duration!
  2 个评论
dpb
dpb 2022-6-3
Clever; hadn't thought of it.
I was going to post the more-or-less brute force w/o explicit loop of
ix=abs(hours(a))>=24;
a(ix)-hours(24*sign(hours(a(ix))))
ans =
2×1 duration array
00:06:00
-00:05:00
Steven Lord
Steven Lord 2022-6-3
Ran in:
Instead of mod you want rem.
a= duration([0; 0; 24; 0; -24], [22; 52; 6; 1; -5], [0; 0; 0; 0; 0])
a = 5×1 duration array
00:22:00 00:52:00 24:06:00 00:01:00 -24:05:00
aMod = mod(a, hours(24))
aMod = 5×1 duration array
00:22:00 00:52:00 00:06:00 00:01:00 23:55:00
aRem = rem(a, hours(24))
aRem = 5×1 duration array
00:22:00 00:52:00 00:06:00 00:01:00 -00:05:00

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eko supriyadi
eko supriyadi 2022-6-3
编辑：eko supriyadi 2022-6-3
and when we interest with <24 hours, we can do indexing one more again:
h24 = hours(24);
mask = a>h24;
a(mask) = a(mask) - h24;
nh24 = hours(-24);
nmask = a<nh24;
a(nmask) = a(nmask) + nh24
and thank you, but I'm still curious about the looping step
Edit mod:
no, mod function still give error result for -24 hour but work for 24 hour
  2 个评论
Walter Roberson
Walter Roberson 2022-6-3
mod(-hours(24),hours(24)) works for me and gives me duration of 0 hours
dpb
dpb 2022-6-3
What about the looping step?
Other than
a1 = a(i)+hours(-24);
is turning -24 into -48 instead of 0 because of sign, it works.
One doesn't need the else if would just assign a1=a first, then operate on a1 inside the loop.
I didn't test Walters mod() but he rarely makes a goof -- far less frequent than I do, anyways -- I posted a way that works w/o the looping construct.

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dpb
dpb 2022-6-3
编辑：dpb 2022-6-3
Since indeed mod() doesn't do what is wanted here for negative hours, I'll go ahead and post the alternative way that's akin to @Steven Lord's --
ix=abs(hours(a))>=24;
a(ix)-hours(24*sign(hours(a(ix))))
ans =
2×1 duration array
00:06:00
-00:05:00
that uses sign to perform the logical test and make correction in proper direction.
Stephen23
Stephen23 2022-6-4
Ran in:
The simple and efficient approach is to use REM:
a = duration([0; 0; 24; 0; -24], [22; 52; 6; 1; -5], [0; 0; 0; 0; 0])
a = 5×1 duration array
00:22:00 00:52:00 24:06:00 00:01:00 -24:05:00
b = rem(a,hours(24))
b = 5×1 duration array
00:22:00 00:52:00 00:06:00 00:01:00 -00:05:00

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