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

User · Answer

This is what I did to extract the piece of the directory:

for path in file_list:
  directories = path.rsplit('\\')
  directories.reverse()
  line_replace_add_directory = line_replace+directories[2]

Thank you for your help.

User · Answer

In Python 3 4 you can use the pathlib module    gt  gt  gt  from pathlib import Path  gt  gt  gt  p   Path  C  Program Files Internet Explorer iexplore exe    gt  gt  gt  p name  iexplore exe   gt  gt  gt  p suffix   exe   gt  gt  gt  p root       gt  gt  gt  p parts   C       Program Files    Internet Explorer    iexplore exe    gt  gt  gt  p relative to  C  Program Files   WindowsPath  Internet Explorer iexplore exe    gt  gt  gt  p exists   True

User · Answer

First  see if you have splitunc   as an available function within os path  The first item returned should be what you want    but I am on Linux and I do not have this function when I import os and try to use it   Otherwise  one semi-ugly way that gets the job done is to use    gt  gt  gt  pathname      C   mystuff  project  file py   gt  gt  gt  pathname    C   mystuff  project  file py   gt  gt  gt  print pathname  C  mystuff project file py  gt  gt  gt       join pathname split        -2      C   mystuff   gt  gt  gt       join pathname split        -1      C   mystuff  project    which shows retrieving the directory just above the file  and the directory just above that

User · Answer

All you need is parent part if you use pathlib   from pathlib import Path p   Path r C  Program Files Internet Explorer iexplore exe   print p parent     Will output   C  Program Files Internet Explorer       Case you need all parts  already covered in other answers  use parts   p   Path r C  Program Files Internet Explorer iexplore exe   print p parts     Then you will get a list     C       Program Files    Internet Explorer    iexplore exe     Saves tone of time

User · Answer

import os  directory   os path abspath         root directory print directory    e g   C     directory   os path abspath        current directory print directory    e g   C  Users User Desktop   parent directory  directory name   os path split directory  print directory name    e g   Desktop  parent parent directory  parent directory name   os path split parent directory  print parent directory name    e g   User    This should also do the trick

User · 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

User · Answer

You have to put the entire path as a parameter to os path split  See The docs  It doesn t work like string split

[python] Extract a part of the filepath (a directory) in Python

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