identify different filname in different group

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?

Answer 1

[Alexandre Williot]

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

Answer 2

[pfb]

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.

Answer 3

[Image Analyst]

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

Answer 4

[Image Analyst]

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.

[Alexandre Williot]

Of course ! I am a beginner, thank you very much.

Now I have this message for the code below, have you an idea please? It's probably easy but ... not for me.

??? Error using ==> strcmp
Not enough input arguments.

Error in ==> exercice2__a_williot at
20
if strcmp(contenu(file).name,'S')==1

the code

contenu = dir('C:\Users\williot\Dropbox\UQTR 2013 - 2015\Matlab\MatLab EPK6064\Fichiers_Travail2');

for file = 1:length(contenu)
    if strcmp(contenu(file).name,'.')==0 && strcmp(contenu(file).name,'..')==0
        file_name = contenu(file).name;
        load(file_name)
    end
end

%--étape2--identifiez chaque groupe de personne-------

if strcmp(contenu(file).name,'S')==1 
                code_Groupe = 1;
    elseif strcmp(contenu(file).name,'C')==1
                code_Groupe = 2;
    elseif strcmp(contenu(file).name,'L')==1
                code_Groupe = 3;
end

[Image Analyst]

file is no longer defined because the for loop over "file" has ended by the time you call your next if block in etape2. I guess you probably want it INSIDE your for loop, not outside.

[Alexandre Williot]

Yes, thank you!

[Image Analyst]

By the way, you were supposed to accept my answer, not post your code and accept it.