Batch process text files

I am trying to save series of cells in a loop. I need to batch process text files having a 10 lines of both numbers and characters ( 12 12 34 54 rr). I am trying to save them in cell array so that i can access them later. The code I have written is giving error :

error : Error using textscan Invalid file identifier. Use fopen to generate a valid file identifier.

This code is working perfectly without the loop. Please help I am new to matlab. Is there any other better method of saving the files.

My code :

input_directory = 'd2/';
filelabels = dir([input_directory '*.txt']);
wav_label1 = cell(12,12);
wav_label2 = cell(12,12);
for i = 1: numel(filelabels)
  fileName =  filelabels(i).name;
  fid = fopen(fileName);
  results{i}= textscan(fid,'%f %f %f ........
          %s','HeaderLines',2,'Delimiter',',','CollectOutput',1);
  fclose(fid); 
  %testscan is giving output as a cell{{1,1}{1,2}} as I have numbers 
  %and characters both in text file
  wav_label1{i} = results{1,1}; 
  wav_label2{i} = results{1,2};
end

Thanks

[dpb]

The error indicates the file handle is invalid which means the fopen didn't succeed. Use the optional second output argument to see where the problem arises...

[fid,msg] = fopen(fileName);
if fid<0
   error(['FOPEN: ' msg ' while opening: ' fileName])
end

[Guillaume]

Or better, use the formatting facility of error:

if fid<0
    error('FOPEN: %s while opening ''%s''', msg, fileName);
end

With any code that deals with entities external to your program (user input, file system, database, etc.), always be prepared to deal with unexpected failure.

Answer 1

[Udit Gupta]

You can use -

fileName =  [input_directory filelabels(i).name];

That should solve the issue.

[Guillaume]

That would probably solve the problem, but it is always good practice to check that a fopen succeeds.

It is also good practice to use path manipulation functions instead of string manipulation functions when dealing with path. In this case:

fileName = fullfile(input_directory, filelabels(i).name);

would be safer. You don't have to worry whether or not your path ends by a path separator, nor what that separator exactly is (so your code works on any platform).

[alia]

The code works now. Thanks a lot. You correctly pointed my mistake. Thanks a ton :)

Answer 2

[Image Analyst]

I would say the most likely reason is that you do not have a subdirectory of the current working directory called "d2". You can use mkdir() to create it.