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identify different filname in different group

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Alexandre Williot
Alexandre Williot 2015-4-17
评论： Image Analyst 2015-4-22
Hello,
I just realized a program that allows to load data files from a folder. These data files are all different names but I would group them into three groups in matlab. A group 1 whose file names begin with 'X', a group 2 whose file names begin with 'Y' and a group 3 whose file names begin with 'Z' but I do not know how. Do you have any idea? with strcmp?

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采纳的回答

Alexandre Williot
Alexandre Williot 2015-4-22
this is what I needed,
contenu = dir('C:\Users\alexandre\Dropbox\UQTR 2013 - 2015\Matlab\MatLab EPK6064\Fichiers_Travail2');
for file = 1:length(contenu)
Fichiers_Travail2 = contenu(file).name;
if strcmp(Fichiers_Travail2,'.')==0 && strcmp(Fichiers_Travail2,'..')==0 && strcmp(Fichiers_Travail2(1,1),'e')==0
load(Fichiers_Travail2)
if strcmp(Fichiers_Travail2(1,1),'S')==1
code_Groupe = 1;
elseif strcmp(Fichiers_Travail2(1,1),'C')==1
code_Groupe = 2;
elseif strcmp(Fichiers_Travail2(1,1),'L')==1
code_Groupe = 3;
end
end
end

更多回答（3 个）

pfb
pfb 2015-4-17
It's not very clear to me what you mean by "group the files in matlab".
Anyway, one possibility is perhaps to discriminate them before loading them. I suppose you use dir to make a list of the files to be loaded. Why don't you use something like
dirX = dir('X*');
dirY = dir('Y*');
dirZ = dir('Z*');
or something to that effect? I mean, you might have to refine the string you feed to dir(), possibly adding a path or using some more information about the file names.
Anyway, now dirX, dirY and dirZ are structures containing the names of the files starting with the desired letter.
Image Analyst
Image Analyst 2015-4-17
It is all there in the FAQ. Code samples and everything. Adapt as needed: http://matlab.wikia.com/wiki/FAQ#How_can_I_process_a_sequence_of_files.3F
Image Analyst
Image Analyst 2015-4-21
Of course, because you're passing the whole structure:
file_name = contenu(file).name;
load(contenu) % Tries to load the entire structure of all filenames.
Instead, try to load only one filename:
file_name = contenu(file).name;
load(file_name) % Tries to load only one single filename.
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Image Analyst
Image Analyst 2015-4-22
By the way, you were supposed to accept my answer, not post your code and accept it.

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