change filename from different folder

Hi everybody,

I've lots of files, in different folder, which I'd liked to change the name. So I wanted to get the name of the folders using 'dir' :

>> c=dir 

    c = 

    125x1 struct array with fields:
        name
        date
        bytes
        isdir
        datenum

    >> c(1)

    ans = 

           name: '.'
           date: '14-Nov-2012 15:27:02'
          bytes: 0
          isdir: 1
        datenum: 7.3519e+05

So as you see, it doesn't return me any name. Normally, it should be '364D' (first folder)>

If I get a variable with my folder name, I'd like to do something like that:

foldername=dir 

for i=1:length(foldername);
    cd foldername(i)
    filename=dir *.sac.inv;

      for i=1:length(filename);
          loadsac(waveform,filename(1))
          cmp=get(h,'component');
          station=get(h,'station');
          time=get(h,'time');
          name=cmp_station_foldername(i)_time.sac.inv;
          save(name)
      end

      cd ..
  end

Because my origin file form is 'a3114001.sac.inv' and I'd like 'EW_BACA_343A_031306.sac.inv' with the information contains in the header of the file. Do you think it's possible?

Thank you in advance,

Virginie

[Marcelo Costa]

for return the name of file use c(1).name

[Virginie]

It does't work... I still get the same thing.

[MatlabPro]

why not try loop/

[Virginie]

I've tried:

for i=1:125
    c=dir;
    c=c(i).name;
    c(i)=[c(i)]
end

It gives me name but c=c(i).name; doesn't keep all the name so I tried to create a matrix c(i)=[c(i)] to keep them. But I've some difficulties... Could you help me please.

Thanks in advance,

Virginie

Answer 1

[John Petersen]

The first filename in a directory is the "." file which is just specifying the path to this directory... not really a file at all. The second "file" in a DOS directory is ".." which specifies the path just one folder above the current one. All the rest of the files or folders within this folder are after this. The first real file or directory other than these two would be at c(3).name.

[Walter Roberson]

Correction: some names that dir() outputs will be directory names. This includes the names '.' and '..' on all operating systems that MATLAB currently runs on (I can't swear it was the case back when VMS was supported.) '.' and '..' are not necessarily the first two entries, though. The order defined for dir() is "whatever the operating system returns". The order defined for Unix-based operating systems is "whatever the file system returns". There have been file systems that did not return '.' and '..' first: in particular in some Unix filesystems, it is possible to create directories that will sort before '..' in the filesystem.

To eliminate directories, do not expect them to be at any particular location: instead use

c([c.isdir]) = []

to remove the entries.

[Virginie]

I'm not sure to have well understood.'dir' gives you the path and not the name? Because I'd liked to get the string name of the folder in order to enter in every of them automatically using "cd foldername" and use the files that they are containing. I re-read the help of 'dir' and I think I'm using it wrongly. 'dir' gives the files inside the folder than I just want to get the folder names. I tryed 'ls', which list it but after I don't know how to keep this list in a variable. Do you know which command I could use.

Thank you very much for your help and your patience!

Virginie

[Walter Roberson]

Apply the dir() at the directory that contains the directories you want to process.

If you want to keep only the folders, then

c(~[c.isdir]) = [];

and then you will need to check whether c(i).name matches '.' or '..' to eliminate those two.

The names '.' and '..' are considered to be subdirectories of every directory.

When you

cd(c(i).name)

remember that you need to cd back again before you can do the next cd.

[Virginie]

Sorry but it doesn't work... When I do "cd(c(1).name)", nothing happend. I think, it's because "c(1).name" = '.'. So when I do 'cd .' it just stayed where is it. I don't understand how it works! Thank you, Virignie

[Image Analyst]

You're best off not using cd anyway. Use fullfile() instead. See the FAQ.

[Walter Roberson]

As I wrote, "and then you will need to check whether c(i).name matches '.' or '..' to eliminate those two."

[Virginie]

Sorry, but I don't understand what you mean by "matches". With what?

[Walter Roberson]

if strcmp(c(i).name, '.') || strcmp(c(i).name, '..'); continue; end

[Virginie]

Yes, they are matching. What does it mean?

[Virginie]

It's working now. Thanks a lot!

[Walter Roberson]

Every folder has a subfolder named '.', which refers to the exact same directory. This allows filenames such as ./startup.m to be processed more easily; in this eaxample it means "stay in the current directory and look for startup.m here". If you were to give a command such as

edit startup.m

then normally MATLAB would search along the MATLAB path for a startup.m file along that path and open the first one it found, but if you

edit ./startup.m

then it would mean "only look for startup.m here in the current directory".

Each folder also contains a subfolder named '..' which refers to the parent folder. For example if you were in /User/people/Virginie/matlab then '..' relative to that folder would refer to /User/people/Virginie/ . This is very useful for creating and using bundles of related directories without needing to figure out where you started from.

Usually when you encounter those two directories, you just want to skip them. That is what the "continue" command is for in the code sample I showed: "continue" inside a "for" or "while" loop means to continue on with the next loop iteration, skipping over everything else until the matching "end" statement.

Answer 2

[Image Analyst]

If all your files have a .inv extension, why not simply say

folder = pwd; % not foldername=dir  like you had!!!
filePattern = fullfile(folder, '*.inv');
filenames = dir(filePattern );

?????? Of course that could all be done in one line if you wanted to be less explicit. Then you wouldn't have to worry about dot, dot dot, etc.

[Virginie]

Thanks for your help.