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.jpg I would like to extract directory_i_need.
- 1You might want to check this answer out: stackoverflow.com/a/4580931/311220Acorn– Acorn2012-04-13 23:01:27 +00:00Commented Apr 13, 2012 at 23:01
- The above link helped me understand how to fix what I did wrong. Thank you.Thalia– Thalia2012-04-13 23:18:28 +00:00Commented Apr 13, 2012 at 23:18
- or this one: stackoverflow.com/a/31273488/1048186Josiah Yoder– Josiah Yoder2018-06-20 19:24:57 +00:00Commented Jun 20, 2018 at 19:24
Add a comment |
7 Answers
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. 
3 Comments
Thalia
It does extract parts of the path - but I don't know how to extract the actual directory name from the path.
rasul
+1: A scenario where
os.path.dirname(path) is handy compared to pathlib.Path(path).parent: You are given a string path. You want to create the directory of path if it does not exist, whether path itself is a directory or not. For example, path could be /home/me/directory_to_create/file_to_create.txt or /home/me/directory_to_create/. In the second case, pathlib.Path(path).parent returns /home/me/ which is not desired.
Raleigh L.
I wouldn't recommend copying the second snippet verbatim,
dir() is a reserved keyword in Python now: docs.python.org/3/library/functions.html#dirFor Python 3.4+, try the pathlib module:
>>> from pathlib import Path >>> p = Path('C:\\Program Files\\Internet Explorer\\iexplore.exe') >>> str(p.parent) 'C:\\Program Files\\Internet Explorer' >>> p.name 'iexplore.exe' >>> p.suffix '.exe' >>> p.parts ('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe') >>> p.relative_to('C:\\Program Files') WindowsPath('Internet Explorer/iexplore.exe') >>> p.exists() True 
4 Comments
Nadim Farhat
nice demonstration of the API
johnthagen
This has also been backported to older versions of Python: pathlib2
maugch
You example is wrong if your path contains for example \a. you should add r in front of the path' string.
horace
I'm doing something very similar to the OP. This is the right answer. The accepted answer uses os.path.split which missed some path elements in my testing. pathlib Path works perfect for me
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.
Comments
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.
3 Comments
Thalia
I edited my entry to show use of rsplit which does what you suggest - but still gives me the path not just the directory name.
Thalia
I ended up splitting the path and taking the piece I wanted, it didn't work before but after reading all these answers, I found out what I did wrong.
TazgerO
I like the way this semi-ugly way works. I change the "\\" by a simple os.sep and it works perfectly to retrieve just a fraction of a path.
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.
Comments
You have to put the entire path as a parameter to os.path.split. See The docs. It doesn't work like string split.
1 Comment
ely
This won't work on UNC type pathnames on Windows, as the Python docs for os.path stuff state.
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.