Time format conversion command
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If I want to convert
36:40.0
to
time in seconds, how do I do this?
2 Comments
Matt Fig
on 16 Oct 2012
What form is that? Is it a string and many in a character array or a cell array or what?
A = ['36:40.0';'36:41.0';'34:40.3']; % Like this?
A = {'36:40.0';'36:41.0';'34:40.3'}; % Like this?
Accepted Answer
Azzi Abdelmalek
on 16 Oct 2012
Edited: Azzi Abdelmalek
on 16 Oct 2012
t={'36:40.0' ,'37:40.0' ;'39:40.0' ,'31:40.0'}
out=cellfun(@(x) sum(cellfun(@str2double, regexp(x,'[:.]','split')).*[3600 60 1]),t)
7 Comments
Azzi Abdelmalek
on 18 Oct 2012
Edited: Azzi Abdelmalek
on 18 Oct 2012
Anthony, I suggest that you reformulate and repost your question, let it brief and very clear.
More Answers (1)
Matt Fig
on 16 Oct 2012
Edited: Matt Fig
on 16 Oct 2012
If you have a cell array, I would do this:
A = {'36:40.0';'36:40.1';'34:40.3'}; % A cell array
B = '${num2str(str2num($1)*60+str2num($2))}';
B = regexprep(A,'(\d+):(\d+\.\d*)',B)
If you have a character array, then:
A = ['36:40.0';'36:41.0';'34:40.3']; % A character array.
B = '${num2str(str2num($1)*60+str2num($2))}';
B = char(regexprep(cellstr(A),'(\d+):(\d+\.\d*)',B))
3 Comments
Matt Fig
on 16 Oct 2012
Edited: Matt Fig
on 16 Oct 2012
You seem to show where you converted to datenumbers using the DATENUM command. So why would it be surprising that you get datenumbers?
Show what this shows:
data{2}(1:3) % Or, what is in data{2}... strings?
If you don't see the strings in there, take the time to explore the data cell array before you run all these conversions on it. What is in data{1}? How about data{3}, etc...
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