Very simple Python program help...

[footloose61]

I am extremely new to python and I need to write a program that simply outputs all the filenames within a specific directory to a text file. Thanks!

[bartonc]

I am extremely new to python and I need to write a program that simply outputs all the filenames within a specific directory to a text file. Thanks!

The glob module may be what you are looking for:

6.24 glob -- Unix style pathname pattern expansion

The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell. No tilde expansion is done, but *, ?, and character ranges expressed with [] will be correctly matched. This is done by using the os.listdir() and fnmatch.fnmatch() functions in concert, and not by actually invoking a subshell. (For tilde and shell variable expansion, use os.path.expanduser() and os.path.expandvars().)

For example, consider a directory containing only the following files: 1.gif, 2.txt, and card.gif. glob() will produce the following results. Notice how any leading components of the path are preserved.

>>> import glob
>>> glob.glob('./[0-9].*')
['./1.gif', './2.txt']
>>> glob.glob('*.gif')
['1.gif', 'card.gif']
>>> glob.glob('?.gif')
['1.gif']

[dshimer]

Depending on what you want to do, once you get all the filenames into a list using glob you may also find some help in parsing filenames into individual parts (drive, path, base name, extension) using the os.path module. I find these two modules to be a huge help in working with file names.

[bvdet]

Depending on what you want to do, once you get all the filenames into a list using glob you may also find some help in parsing filenames into individual parts (drive, path, base name, extension) using the os.path module. I find these two modules to be a huge help in working with file names.

There are a several threads on the Python forum on compiling a list of file names from a specific directory. In short, the file list can be created like this:

>>> dn = 'directory_path'
>>> fileStr = '\n'.join([f for f in os.listdir(dn) if os.path.isfile(os.path.join(dn, f))])

Directory listings are skipped in the above list comprehension. To write the resulting string to a disk file:

f = open('file_name', 'w')
f.write(fileStr)
f.close()