Given a path, how can I extract just the containing folder name?

os.path.dirname is what you are looking for –

os.path.dirname(r"C:folder1folder2filename.xml")

Make sure you prepend r to the string so that its considered as a raw string.

Demo –

In [46]: os.path.dirname(r"C:folder1folder2filename.xml")
Out[46]: 'C:\folder1\folder2'

If you just want folder2 , you can use os.path.basename with the above, Example –

os.path.basename(os.path.dirname(r"C:folder1folder2filename.xml"))

Demo –

In [48]: os.path.basename(os.path.dirname(r"C:folder1folder2filename.xml"))
Out[48]: 'folder2'

You are looking to use dirname. If you only want that one directory, you can use os.path.basename,

When put all together it looks like this:

os.path.basename(os.path.dirname('dir/sub_dir/other_sub_dir/file_name.txt'))

That should get you “other_sub_dir”

The following is not the ideal approach, but I originally proposed,using os.path.split, and simply get the last item. which would look like this:

os.path.split(os.path.dirname('dir/sub_dir/other_sub_dir/file_name.txt'))[-1]

this is pretty old, but if you are using Python 3.4 or above use PathLib.

# using OS
import os
path=os.path.dirname("C:/folder1/folder2/filename.xml")
print(path)
print(os.path.basename(path))

# using pathlib
import pathlib
path = pathlib.PurePath("C:/folder1/folder2/filename.xml")
print(path.parent)
print(path.parent.name)

You could get the full path as a string then split it into a list using your operating system’s separator character.
Then you get the program name, folder name etc by accessing the elements from the end of the list using negative indices.

Like this:

import os
strPath = os.path.realpath(__file__)
print( f"Full Path    :{strPath}" )
nmFolders = strPath.split( os.path.sep )
print( "List of Folders:", nmFolders )
print( f"Program Name :{nmFolders[-1]}" )
print( f"Folder Name  :{nmFolders[-2]}" )
print( f"Folder Parent:{nmFolders[-3]}" )

The output of the above was this:

Full Path    :C:UsersterryDocumentsappsenvironmentsdevapp_02app_02.py
List of Folders: ['C:', 'Users', 'terry', 'Documents', 'apps', 'environments', 'dev', 'app_02', 'app_02.py']
Program Name :app_02.py
Folder Name  :app_02
Folder Parent:dev

you can use pathlib

from pathlib import Path
Path(r"C:folder1folder2filename.xml").parts[-2]

The output of the above was this:

'folder2'

I’m using 2 ways to get the same response:
one of them use:

   os.path.basename(filename)

due to errors that I found in my script I changed to:

Path = filename[:(len(filename)-len(os.path.basename(filename)))]

it’s a workaround due to python’s '\'

I made an improvement on the solutions available, namely the snippet that works with all of,

1. File
2. Directory with a training slash
3. Directory without a training slash

My solution is,

from pathlib import Path

def path_lastname(s):
    Path(s).with_name("foo").parts[-2]

Explanation

• Path(s) – Creates a custom Path object out of s without resolving it.

• .with_name("foo") – Adds a fake file foo to the path

• .parts[-2] returns second last part of the string. -1 part will be foo