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

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Brad 2020-4-9

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评论： Brad 2020-4-13

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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Answer 1

Walter Roberson 2020-4-9

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编辑：Walter Roberson 2020-4-9

在 MATLAB Online 中打开

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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Walter Roberson 2020-4-9

编辑：Walter Roberson 2020-4-9

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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.

Brad 2020-4-13

I forgot all about datetime. Thanks again Walter!

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Answer 2

Kelly Kearney 2020-4-9

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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.