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Batch process text files

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alia on 6 Feb 2015
Commented: alia on 8 Feb 2015
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 ........
%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};


dpb on 6 Feb 2015
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])
Guillaume on 6 Feb 2015
Or better, use the formatting facility of error:
if fid<0
error('FOPEN: %s while opening ''%s''', msg, fileName);
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.

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Accepted Answer

Udit Gupta
Udit Gupta on 6 Feb 2015
You can use -
fileName = [input_directory filelabels(i).name];
That should solve the issue.


Guillaume on 6 Feb 2015
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 on 8 Feb 2015
The code works now. Thanks a lot. You correctly pointed my mistake. Thanks a ton :)

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More Answers (1)

Image Analyst
Image Analyst on 6 Feb 2015
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.


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