CODEWITHSUNDEEP
pythonregex
MHS
MHS

Reputation: 2350

How to get rid of extensions from file basename using python

I have got the complete path of files in a list like this:

a = ['home/robert/Documents/Workspace/datafile.xlsx', 'home/robert/Documents/Workspace/datafile2.xls', 'home/robert/Documents/Workspace/datafile3.xlsx']

what I want is to get just the file NAMES without their extensions, like:

b = ['datafile', 'datafile2', 'datafile3']

What I have tried is:

xfn = re.compile(r'(\.xls)+')
for name in a:
    fp, fb = os.path.split(fp)
    ofn = xfn.sub('', name)
    b.append(ofn)

But it results in:

b = ['datafilex', 'datafile2', 'datafile3x']

Upvotes: 12

Views: 19566

Answers (4)

jjisnow
jjisnow

Reputation: 1496

This is a repeat of: How to get the filename without the extension from a path in Python?

https://docs.python.org/3/library/os.path.html

In python 3 pathlib "The pathlib module offers high-level path objects." so,

>>> from pathlib import Path
>>> p = Path("/a/b/c.txt")
>>> print(p.with_suffix(''))
\a\b\c
>>> print(p.stem)
c

Upvotes: 1

kennytm
kennytm

Reputation: 523184

  1. The regex you've used is wrong. (\.xls)+ matches strings of the form .xls, .xls.xls, etc. This is why there is a remaining x in the .xlsx items. What you want is \.xls.*, i.e. a .xls followed by zero or more of any characters.

  2. You don't really need to use regex. There are specialized methods in os.path that deals with this: basename and splitext.

    >>> import os.path
>>> os.path.basename('home/robert/Documents/Workspace/datafile.xlsx')
'datafile.xlsx'
>>> os.path.splitext(os.path.basename('home/robert/Documents/Workspace/datafile.xlsx'))[0]
'datafile'

    so, assuming you don't really care about the .xls/.xlsx suffix, your code can be as simple as:

    >>> a = ['home/robert/Documents/Workspace/datafile.xlsx', 'home/robert/Documents/Workspace/datafile2.xls', 'home/robert/Documents/Workspace/datafile3.xlsx']
>>> [os.path.splitext(os.path.basename(fn))[0] for fn in a]
['datafile', 'datafile2', 'datafile3']

    (also note the list comprehension.)

Upvotes: 27

Amyth
Amyth

Reputation: 32949

Why not just use the split method?

def get_filename(path):
    """ Gets a filename (without extension) from a provided path """

    filename = path.split('/')[-1].split('.')[0]
    return filename


>>> path = '/home/robert/Documents/Workspace/datafile.xlsx'
>>> filename = get_filename(path)
>>> filename
'datafile'

Upvotes: 0

Paulo Scardine
Paulo Scardine

Reputation: 77251

Oneliner:

>>> filename = 'file.ext'
>>> '.'.join(filename.split('.')[:-1]) if '.' in filename else filename
'file'

Upvotes: 4

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