Extract a part of the filepath (a directory) in Python


Question

I need to extract the name of the parent directory of a certain path. This is what it looks like:

c:\stuff\directory_i_need\subdir\file

I am modifying the content of the "file" with something that uses the directory_i_need name in it (not the path). I have created a function that will give me a list of all the files, and then...

for path in file_list:
   #directory_name = os.path.dirname(path)   # this is not what I need, that's why it is commented
   directories, files = path.split('\\')

   line_replace_add_directory = line_replace + directories  
   # this is what I want to add in the text, with the directory name at the end 
   # of the line.

How can I do that?

1
121
5/31/2019 10:47:48 AM

Accepted Answer

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...

And you can continue doing this as many times as necessary...

Edit: from os.path, you can use either os.path.split or os.path.basename:

dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir) 
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.
187
4/16/2012 4:29:14 PM

In Python 3.4 you can use the pathlib module:

>>> from pathlib import Path
>>> p = Path('C:\Program Files\Internet Explorer\iexplore.exe')
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.root
'\\'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True

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