Listing file names w/o the extentions

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November 16th, 2004, 08:24 PM

Boceifus

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Listing file names w/o the extensions

Greets,

I have a small program I use to keep track of some of my customers info.All customer info is written into a text file(each file named after the customer),and I have a option to list the files in the customers directory.I am currently using os.listdir() to display the files,but all the ".txt" extensions makes reading a long list a little hard on the eye.I could slice the extensions off,but I suspect there is a better way to do it.

Can anyone recommend a way to list the files without showing the extensions?

*edit*
Sorry for this,but since I can't delete the thread,I guess I'll just post what I have done using slices,and fix spelling errors

Code:

list=os.listdir('path/to/dir/')
            for i in list:
                print i[:-4]

This works just fine for me,and allowes me to display the file name w/o the extension

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November 18th, 2004, 05:25 AM

Konstandinos

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Ignoring file extensions on dir.list

Hi there.

I had a similar problem in the past, and solved it using the exact same method as you.

I believe that the functionality/method you're looking for doesn't actually exist.

Why not send the guys over at python.org an e-mail requesting an extension of the glob module?

The glob.glob('*') which is similar/identical to your os.listdir() method could be extended to return a list without file extensions by calling it with glob.glob('*', NO_EXTENSION) for example.

Just a thought.

Cheers,
K

November 18th, 2004, 06:58 AM

Grim Archon

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Just a thought - what if the directory does not just contain files with the right extension?

I think it would be safer to try this:

Code:

>>> import glob
>>> textfiles = [x[:-4] for x in glob.glob("*.txt")]
>>> textfiles
['MOTD', 'LICENSE', 'README', 'NEWS', 'text', 'log1', 'log2', 'diff', 'Document1', 'Document2']

grim

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November 18th, 2004, 10:06 AM

sfb

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What if some of the files have longer extensions?

Code:

>>> import os, sys
>>> os.path.extsep
'.'
>>> os.path.splitext("c:\\a.txt")
('c:\\a', '.txt')

>>> import glob
>>> f = glob.glob("e:\\*.txt")
>>> f[0]
'e:\\contacts.txt'

>>> os.path.basename(f[0])
'contacts.txt'

>>> [os.path.splitext(os.path.basename(x))[0] for x in glob("d:\\*.txt")

November 18th, 2004, 11:03 AM

Grim Archon

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Yep, that was a more robust solution

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