Time stamp formatting question
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Sergio Mendoza
am 17 Jun. 2019
Bearbeitet: Adam Danz
am 17 Jun. 2019
Hi everyone -
I have a time stamp in the following form : '0d 00:00:5.03000020980835'
I am trying to convert it into a vector in hh:mm:ss, any ideas?
2 Kommentare
Walter Roberson
am 17 Jun. 2019
Do you need the seconds to come out as 5.03 with no 000020980835 stored?
Adam's suggestion is not bad, but it does end up storing as 00:00:05.030000208 (I might have expected 00:00:05.030000210) . If the 0000208 is not desired, then Adam's code would need to be modified.
Question: is the number of days ever non-zero ?
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Adam Danz
am 17 Jun. 2019
Bearbeitet: Adam Danz
am 17 Jun. 2019
dtstr = '0d 00:00:5.03000020980835';
dtstr = regexprep(dtstr,'\d+d',''); %remove days
dt = datetime(dtstr,'InputFormat','HH:mm:ss.SSS','Format','HH:mm:ss')
[update]
To preserve the number of days within the hour-count and to account for missing decimals in the seconds,
dtstr = {'0d 23:59:58.439998626709'
'0d 23:59:59'
'1d 00:00:0'
'1d 00:00:1.44000005722046'
'1d 00:00:2.44000005722046'
'1d 00:00:3.44000005722046'};
% Add .0 to time stamps that are missing the decimal
noDecIdx = ~cellfun(@(x)contains(x,'.'),dtstr);
dtstr(noDecIdx) = cellfun(@(x)[x,'.0'],dtstr(noDecIdx),'UniformOutput',false);
% replace "d " with ":"
dtstr = strrep(dtstr,'d ',':');
% Convert to your desired format as durations
D = duration(dtstr,'InputFormat', 'dd:hh:mm:ss.S','Format','hh:mm:ss');
Result:
6×1 duration array
23:59:58
23:59:59
24:00:00
24:00:01
24:00:02
24:00:03
7 Kommentare
Adam Danz
am 17 Jun. 2019
Hi Sergio, I updated my answer to address this. Sorry it took some dialog for me to understand what your goal was. My updated answer uses durations rather than datetime.
The updated solution uses durations rather than datetime but it's easy to convert those durations to datetime if that's what you need to do.
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