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How can I subtract the times without considering the date in question?

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Newbie
Newbie 2021 年 8 月 11 日
コメント済み: Scott MacKenzie 2021 年 8 月 12 日
I have a vector of time which are linked to dates but I only want to subtract the times from each other without considering the dates too. For example, I want to do 4:30 - 3:30 is equal to 1 hour and then 3:30 - 2:50 is equal to 40 minutes (0.6666667) and etc. I dont want Matlab to consider the date period basically.
I have this code for the time from 25-Apr to 01-May but it does take into account the date too:
out1m=24-(seconds(diff(datetime([T_Smooth{2,4};T_Smooth{3,4}])))/3600);
Can anyone help please? Thank you.
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Newbie
Newbie 2021 年 8 月 11 日
1 hour

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採用された回答

Cris LaPierre
Cris LaPierre 2021 年 8 月 11 日
編集済み: Cris LaPierre 2021 年 8 月 11 日
Use the timeofday function.
% create vector of datetimes
date = datetime(2019,[4;4;5;5],[24;25;1;2],[4;3;2;4],[30;30;50;30],0)
date = 4×1 datetime array
24-Apr-2019 04:30:00 25-Apr-2019 03:30:00 01-May-2019 02:50:00 02-May-2019 04:30:00
% Take difference of the times.
dt = diff(timeofday(date))
dt = 3×1 duration array
-01:00:00 -00:40:00 01:40:00
Use the hours function to turn hh:mm:ss into decimal hours if you want.
hours(dt)
ans = 3×1
-1.0000 -0.6667 1.6667
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Newbie
Newbie 2021 年 8 月 11 日
Thanks, this works!

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その他の回答 (1 件)

Scott MacKenzie
Scott MacKenzie 2021 年 8 月 11 日
編集済み: Scott MacKenzie 2021 年 8 月 11 日
Here's one way to do this:
% test data (one month in 30-min increments)
dt = datetime(2021,4,1):minutes(30):datetime(2021,5,1);
% indices of datetime elements to compute hour difference between
n1 = 154;
n2 = 255;
x = datenum(dt(n2) - dt(n1));
hrs = rem(x,1) * 24
hrs = 2.5000
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Scott MacKenzie
Scott MacKenzie 2021 年 8 月 12 日
Oops, yes, I used datenum as well. Thanks for pointing that out. The language to discourage use of this function could be a tad stronger, however. Below is what appears for xlsread, as an example.

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