Time format conversion command
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This post is related to this thread:
If I want to convert
36:40.0
to
time in seconds, how do I do this?
2 comentarios
Matt Fig
el 16 de Oct. de 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?
Respuesta aceptada
Azzi Abdelmalek
el 16 de Oct. de 2012
Editada: Azzi Abdelmalek
el 16 de Oct. de 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 comentarios
Azzi Abdelmalek
el 18 de Oct. de 2012
Editada: Azzi Abdelmalek
el 18 de Oct. de 2012
Anthony, I suggest that you reformulate and repost your question, let it brief and very clear.
Más respuestas (1)
Matt Fig
el 16 de Oct. de 2012
Editada: Matt Fig
el 16 de Oct. de 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 comentarios
Matt Fig
el 16 de Oct. de 2012
Editada: Matt Fig
el 16 de Oct. de 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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