在Python中提取一部分文件路径（目录）

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?

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.

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

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.

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:

>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'

which shows retrieving the directory just above the file, and the directory just above that.

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.

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.

You have to put the entire path as a parameter to os.path.split. See The docs. It doesn’t work like string split.