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How do I extract the HH:MM:SS.FFF portion of a julian day time stamp?

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Brad
Brad on 9 Apr 2020
Commented: Brad on 13 Apr 2020
I have a 2x1 cell array containing the following data;
Data_Time_Stamps_Cells =
' 086 2020 18:30:19.578'
' 086 2020 18:30:18.569'
I'm attempting to extract the HH:MM:SS.FFF portion of these time stamps using the following:
exp = '([\d:\.]+)';
times = regexp(Data_Time_Stamps_Cells, exp, 'tokens');
The result is a 2 x 1 cell array;
times =
'086' '2020' '18:30:19.578'
'086' '2020' '18:30:18.569'
Why am I getting the entire contents of the Data Time Stamps Cells in 3 individual cells?

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Accepted Answer

Walter Roberson
Walter Roberson on 9 Apr 2020
Edited: Walter Roberson on 9 Apr 2020
times = regexp(Data_Time_Stamps_Cells, '\d{1,2}:\d{2}:\d{2}\.\d{3}', 'match', 'once');
But you might as well do
cellfun(@(D) D(end-11:end), Data_Time_Stamps_Cells, 'uniform', 0)
exp = '([\d:\.]+)';
[\d:\.] means any one character that is a digit or a colon or a period. The + after that pattern means one or more occurances.
In ' 086 ', the 0 matches a digit, the 8 matches a digit, the 6 matches a digit, the space after does not match a digit or colon or period. So '086' would be matched.

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Brad
Brad on 9 Apr 2020
Walter, thanks for taking a look at this. I got a question regarding the 1st part of the expression: \d{1,2}:
\d - is any numeric digit; equivalent to [0-9]
{1,2} - matches between 1 and 2 numeric digits
Am I safe to assume this is a quantifier to account for a single or double digit hour?
Walter Roberson
Walter Roberson on 9 Apr 2020
Yes, exactly, it is accounting for the possibility of single or double digit hour. If you can be certain that the hour is double digit you can change it to \d{2}
The cellfun I posted earlier (and just corrected) assumes double-digit hour.
If you want the parts broken out, hour separated from minute and so on, then
times = regexp(Data_Time_Stamps_Cells, '(\d{1,2}):(\d{2}):(\d{2}\.\d{3})', 'tokens', 'once');
would return a size(Data_Time_Stamps_Cells) cell array, each entry of which is a 1 x 3 cell of character vectors with the parts broken out.
Somes the easiest approach is just to datetime()
[H,M,S] = hms(datetime(Data_Time_Stamps_Cells, 'InputFormat', 'DDD uuuu HH:mm:ss.SSS'));
Now H is a vector of (numeric) hours, M is numeric minutes, S of seconds.

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More Answers (1)

Kelly Kearney
Kelly Kearney on 9 Apr 2020
An alternative method that avoids regular expressions would be to let datetime to the parsing for you:
Data_Time_Stamp_Cells = {...
' 086 2020 18:30:19.578'
' 086 2020 18:30:18.569'};
% To datetime...
t = datetime(Data_Time_Stamp_Cells, 'inputformat', 'DDD uuuu HH:mm:ss.SSS');
% And back to your preferred format
datestr(t, 'HH:MM:SS.FFF') % Option A
char(t, 'HH:mm:ss.SSSS') % Option B
Using datestr (Option A) is a tiny bit faster than char (Option B) to convert back to a string, but then that adds in the annoyance of mixing datetime vs datenum/datestr formatting codes.

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