How do I get the parent directory in Python?

Question

Could someone tell me how to get the parent directory of a path in Python in a cross platform way. E.g.

C:\Program Files ---> C:\

and

C:\ ---> C:\

If the directory doesn't have a parent directory, it returns the directory itself. The question might seem simple but I couldn't dig it up through Google.

score 756 · Accepted Answer · edited Apr 30 '21 at 18:01

756

Python 3.4

Use the pathlib module.

from pathlib import Path
path = Path("/here/your/path/file.txt")
print(path.parent.absolute())

Old answer

Try this:

import os
print os.path.abspath(os.path.join(yourpath, os.pardir))

where yourpath is the path you want the parent for.

edited Apr 30 '21 at 18:01

Neuron

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answered May 18 '10 at 19:00

kender

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Your answer is correct but convoluted; `os.path.dirname` is the function for this, like `a+=5-4` is more convoluted than `a+=1`. The question requested only the parent directory, not whether is exists or the *true* parent directory assuming symbolic links get in the way. – tzot May 24 '10 at 12:03
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I tried to change the code snippet to `os.pardir`, but unfortunately Stack Overflow "Edits must be at least 6 non-space characters". It seems smart enough to detect whitespace padding; maybe the OP can correct this someday... – monsur Aug 09 '12 at 15:27
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@tzot: unfortunately `os.path.dirname` gives different results depending on whether a trailing slash is included in the path. If you want reliable results you need to use the `os.path.join` method in answer above. – Artfunkel Jun 28 '13 at 10:32
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@Artfunkel Thanks. Your comment should be appended in the answer for its completeness' sake. – tzot Jun 29 '13 at 08:18
@tzot: I guess the OP wanted to know how to move from one subdirectory to its parent. But `dirname` only returns the directory part of the full pathname to a *file* by removing the last component auf the path string (here: `..`/`pardir`) via `split`. `abspath` on the other hand correctly interprets and resolves `..` as a reference to the top-level directory, which, I guess, is what the OP really asked for. – Frank May 27 '15 at 08:45
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Since this is apparently complicated enough to warrant a StackOverflow question, I feel that this should be added to the os.path library as a built-in function. – antred Mar 04 '16 at 12:44
Don't miss @benjarobin's solution below if you are interested to keep relative path relative. – Maxim Oct 24 '17 at 19:10
1

This doesn't work if you are trying to work with paths on a different file system – alta Aug 24 '18 at 01:56
1

Try `from pathlib import Path` and next `print((Path().parent.absolute()).parent)` – Artem S. Zhelonkin Nov 16 '20 at 17:04
`Path("/here/your/path/file.txt")` does not work. Error message: `'WindowsPath' object is not callable` – mihca Aug 31 '21 at 13:18
I prefer ```os.path.dirname``` since it gave me the path in linux format when converting back to string. May not matter for most people but the Pathlib way gave me it back in stinky windows format on my pc – brando f Mar 03 '23 at 14:35
Works fine. print(os.path.abspath(os.path.join(yourpath, os.pardir))) No need 'pathlib', enough os packages. – muthukumar Apr 11 '23 at 07:14

score 395 · Answer 2 · edited Apr 09 '17 at 10:55

395

Using os.path.dirname:

>>> os.path.dirname(r'C:\Program Files')
'C:\\'
>>> os.path.dirname('C:\\')
'C:\\'
>>>

Caveat: os.path.dirname() gives different results depending on whether a trailing slash is included in the path. This may or may not be the semantics you want. Cf. @kender's answer using os.path.join(yourpath, os.pardir).

edited Apr 09 '17 at 10:55

LarsH

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answered May 18 '10 at 19:14

Wai Yip Tung

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`os.path.dirname(r'C:\Program Files')` what? Python's just giving you the directory where the file 'Program Files' would be. What's more, it doesn't even have to exist, behold: `os.path.dirname(r'c:\i\like\to\eat\pie')` outputs `'c:\\i\\like\\to\\eat'` – Nick T May 18 '10 at 19:28
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The original poster does not state that the directory have to exist. There are a lot of pathname methods that does nothing but string manipulation. To verify if the pathname actually exist requires a disk access. Depends on the application this may or may not be desirable. – Wai Yip Tung May 18 '10 at 19:45
1

Seems that this would be a good solution if you knew that the input directory was valid (which can be checked). – new name May 18 '10 at 19:49
1

Most of the other provided solutions don't check if the parent exists either (as most os.path functions are indeed string manipulations _only_), so no big loss here. And all the other solutions here seem unnecessary complex to me, so I'm upvoting this one :) Btw, imho the os.path is one of the worst classes for python (especially implementation-wise -- platform-independence screwed up is where it hurted me most :( ) – riviera Aug 11 '12 at 23:43
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this solution is sensitive to trailing os.sep. Say os.sep=='/'. dirname(foo/bar) -> foo, but dirname(foo/bar/) -> foo/bar – marcin Sep 10 '12 at 13:28
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That's by design. It comes down to the interpretation of a path with a trailing /. Do you consider "path1" equals to "path1/"? The library use the most general interpretation that they are distinct. In some context people may want to treat them as equivalent. In this case you can do a rstrip('/') first. Had the library pick the other interpretation you will lost fidelity. – Wai Yip Tung Sep 11 '12 at 16:57
1

@WaiYipTung "Do you consider "path1" equals to "path1/"" --> Yes, because they **are**. In what context would you ever _not_ treat them as equivalent? – Ryan Jul 24 '14 at 06:52
4

@Ryan, I don't know about that. There is an entire RFC 1808 written to address the issue of relative path in URI and all the subtlety of the presence and absence of a trailing /. If you know of any documentation that says they should be treated equivalent in general please point it out. – Wai Yip Tung Jul 24 '14 at 15:12
2

@WaiYipTung Hmm, okay, seems silly but I thought this was just a library decision. Makes more sense then. – Ryan Jul 24 '14 at 16:41

hostingutilities.com · Answer 3 · 2020-02-28T20:52:51.523

The Pathlib method (Python 3.4+)

from pathlib import Path
Path('C:\Program Files').parent
# Returns a Pathlib object

The traditional method

import os.path
os.path.dirname('C:\Program Files')
# Returns a string

Which method should I use?

Use the traditional method if:

You are worried about existing code generating errors if it were to use a Pathlib object. (Since Pathlib objects cannot be concatenated with strings.)
Your Python version is less than 3.4.
You need a string, and you received a string. Say for example you have a string representing a filepath, and you want to get the parent directory so you can put it in a JSON string. It would be kind of silly to convert to a Pathlib object and back again for that.

If none of the above apply, use Pathlib.

What is Pathlib?

If you don't know what Pathlib is, the Pathlib module is a terrific module that makes working with files even easier for you. Most if not all of the built in Python modules that work with files will accept both Pathlib objects and strings. I've highlighted below a couple of examples from the Pathlib documentation that showcase some of the neat things you can do with Pathlib.

Navigating inside a directory tree:

>>> p = Path('/etc')
>>> q = p / 'init.d' / 'reboot'
>>> q
PosixPath('/etc/init.d/reboot')
>>> q.resolve()
PosixPath('/etc/rc.d/init.d/halt')

Querying path properties:

>>> q.exists()
True
>>> q.is_dir()
False

This is the only sane answer. If you're forced to use Python 2, just `pip install pathlib2` and use the backport. — Navin, Nov 02 '17 at 05:51
You don't really have to "worry about existing code generating errors if it were to use a Pathlib object" because you can just wrap the pathlib object: `path_as_string = str(Path())`. — John, Nov 22 '20 at 21:33

score 47 · Answer 4 · edited Dec 31 '16 at 23:07

47

import os
p = os.path.abspath('..')

C:\Program Files ---> C:\\\

C:\ ---> C:\\\

edited Dec 31 '16 at 23:07

Kenly

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answered May 18 '10 at 19:00

ivo

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This only gets the parent of the CWD, not the parent of an arbitrary path as the OP asked. – Sergio Jun 04 '13 at 14:05
Add the double dots at the end of your URL and it will work E.g `os.path.abspath(r'E:\O3M_Tests_Embedded\branches\sw_test_level_gp\test_scripts\..\..')` Result: `E:\\O3M_Tests_Embedded\\branches` – Arindam Roychowdhury Dec 02 '15 at 11:31
This means: `/`. – loretoparisi Jul 26 '19 at 09:28

benjarobin · Answer 5 · 2017-10-27T09:07:38.557

An alternate solution of @kender

import os
os.path.dirname(os.path.normpath(yourpath))

where yourpath is the path you want the parent for.

But this solution is not perfect, since it will not handle the case where yourpath is an empty string, or a dot.

This other solution will handle more nicely this corner case:

import os
os.path.normpath(os.path.join(yourpath, os.pardir))

Here the outputs for every case that can find (Input path is relative):

os.path.dirname(os.path.normpath('a/b/'))          => 'a'
os.path.normpath(os.path.join('a/b/', os.pardir))  => 'a'

os.path.dirname(os.path.normpath('a/b'))           => 'a'
os.path.normpath(os.path.join('a/b', os.pardir))   => 'a'

os.path.dirname(os.path.normpath('a/'))            => ''
os.path.normpath(os.path.join('a/', os.pardir))    => '.'

os.path.dirname(os.path.normpath('a'))             => ''
os.path.normpath(os.path.join('a', os.pardir))     => '.'

os.path.dirname(os.path.normpath('.'))             => ''
os.path.normpath(os.path.join('.', os.pardir))     => '..'

os.path.dirname(os.path.normpath(''))              => ''
os.path.normpath(os.path.join('', os.pardir))      => '..'

os.path.dirname(os.path.normpath('..'))            => ''
os.path.normpath(os.path.join('..', os.pardir))    => '../..'

Input path is absolute (Linux path):

os.path.dirname(os.path.normpath('/a/b'))          => '/a'
os.path.normpath(os.path.join('/a/b', os.pardir))  => '/a'

os.path.dirname(os.path.normpath('/a'))            => '/'
os.path.normpath(os.path.join('/a', os.pardir))    => '/'

os.path.dirname(os.path.normpath('/'))             => '/'
os.path.normpath(os.path.join('/', os.pardir))     => '/'

Normalizing the path is always a good practice, especially when doing cross-platform work. — DevPlayer, Jan 28 '16 at 13:50
@Maxim This solution was not perfect, I did improved it since the orginal solution does not handle one case — benjarobin, Oct 27 '17 at 09:10

Dan Menes · Answer 6 · 2016-01-31T09:44:44.833

20

os.path.split(os.path.abspath(mydir))[0]

edited Jan 31 '16 at 09:44

answered May 18 '10 at 18:58

Dan Menes

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This won't work for paths which are to a directory, it'll just return the directory again. – Anthony Briggs Feb 20 '13 at 02:37
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@AnthonyBriggs, I just tried this using Python 2.7.3 on Ubuntu 12.04 and it seems to work fine. `os.path.split(os.path.abspath("this/is/a/dir/"))[0]` returns `'/home/daniel/this/is/a'` as expected. I don't at the moment have a running Windows box to check there. On what setup have you observed the behavior that you report? – Dan Menes Feb 22 '13 at 00:59
You could do `parentdir = os.path.split(os.path.apspath(dir[:-1]))[0]`. This - I am certain - works because if there is a slash on the end, then it is removed; if there is no slash, this will still work (even if the last part of the path is only one char long) because of the preceding slash. This of course assumes that the path is proper and not say something like `/a//b/c///d////` (in unix this is valid still), which in most cases they are (proper) especially when you do something like `os.path.abspath` or any other `os.path` function. – dylnmc Oct 04 '14 at 16:51
Also, to counteract a lot of slashes on the end, you could just write a small for loop that removes those. I'm sure there could even be a clever one-liner to do it, or maybe do that and os.path.split in one line. – dylnmc Oct 04 '14 at 16:57
@Den Menes I just saw you comment. It doesn't work if you have something like `os.path.split("a/b//c/d///")` and, for example, `cd //////dev////// is equivalent to `cd /dev/` or `cd /dev`; all of these are valid in linux. I just came up with this and it may be useful, though: `os.path.split(path[:tuple(ind for ind, char in enumerate(path) if char != "/" and char != "\\")[-1]])[0]`. (This essentially searches for the last non-slash, and gets the substring of the path up to that char.) I used `path = "/a//b///c///d////"` and then ran the aforementioned statement and got `'/a//b///c'`. – dylnmc Oct 04 '14 at 17:13
With the understanding that the line above is for demonstration, I'd still recommend to not use "dir" as a variable/reference. It is also a built-in function. – DevPlayer Jan 28 '16 at 13:48

score 16 · Answer 7 · answered May 18 '10 at 18:58

16

os.path.abspath(os.path.join(somepath, '..'))

Observe:

import posixpath
import ntpath

print ntpath.abspath(ntpath.join('C:\\', '..'))
print ntpath.abspath(ntpath.join('C:\\foo', '..'))
print posixpath.abspath(posixpath.join('/', '..'))
print posixpath.abspath(posixpath.join('/home', '..'))

answered May 18 '10 at 18:58

Ignacio Vazquez-Abrams

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Does not work with Symbolic links – CodingTil Aug 13 '22 at 17:02

score 10 · Answer 8 · answered Feb 22 '13 at 23:06

import os
print"------------------------------------------------------------"
SITE_ROOT = os.path.dirname(os.path.realpath(__file__))
print("example 1: "+SITE_ROOT)
PARENT_ROOT=os.path.abspath(os.path.join(SITE_ROOT, os.pardir))
print("example 2: "+PARENT_ROOT)
GRANDPAPA_ROOT=os.path.abspath(os.path.join(PARENT_ROOT, os.pardir))
print("example 3: "+GRANDPAPA_ROOT)
print "------------------------------------------------------------"

Soumendra · Answer 9 · 2018-04-03T12:50:16.977

10

>>> import os
>>> os.path.basename(os.path.dirname(<your_path>))

For example in Ubuntu:

>>> my_path = '/home/user/documents'
>>> os.path.basename(os.path.dirname(my_path))
# Output: 'user'

For example in Windows:

>>> my_path = 'C:\WINDOWS\system32'
>>> os.path.basename(os.path.dirname(my_path))
# Output: 'WINDOWS'

Both examples tried in Python 2.7

edited Apr 03 '18 at 12:50

answered Apr 03 '18 at 12:44

Soumendra

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score 9 · Answer 10 · answered Aug 10 '19 at 13:31

Suppose we have directory structure like

1]

/home/User/P/Q/R

We want to access the path of "P" from the directory R then we can access using

ROOT = os.path.abspath(os.path.join("..", os.pardir));

2]

/home/User/P/Q/R

We want to access the path of "Q" directory from the directory R then we can access using

ROOT = os.path.abspath(os.path.join(".", os.pardir));

score 6 · Answer 11 · answered Aug 29 '15 at 10:02

6

If you want only the name of the folder that is the immediate parent of the file provided as an argument and not the absolute path to that file:

os.path.split(os.path.dirname(currentDir))[1]

i.e. with a currentDir value of /home/user/path/to/myfile/file.ext

The above command will return:

myfile

answered Aug 29 '15 at 10:02

8bitjunkie

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os.path.basename(os.path.dirname(current_dir)) also works here. – DevPlayer Jan 28 '16 at 13:58

score 4 · Answer 12 · answered Nov 05 '16 at 05:01

4

import os

dir_path = os.path.dirname(os.path.realpath(__file__))
parent_path = os.path.abspath(os.path.join(dir_path, os.pardir))

answered Nov 05 '16 at 05:01

Miguel Mota

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score 3 · Answer 13 · answered Jul 30 '14 at 14:08

3

import os.path

os.path.abspath(os.pardir)

answered Jul 30 '14 at 14:08

Washington Botelho

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This presumes you want the parent directory of "the current working directory" and not the parent directory any path in general. – DevPlayer Jan 28 '16 at 14:00

samsamara · Answer 14 · 2022-11-14T03:53:20.520

3

Just adding something to the Tung's answer (you need to use rstrip('/') to be more of the safer side if you're on a unix box).

>>> input1 = "../data/replies/"
>>> os.path.dirname(input1.rstrip('/'))
'../data'
>>> input1 = "../data/replies"
>>> os.path.dirname(input1.rstrip('/'))
'../data'

But, if you don't use rstrip('/'), given your input is

>>> input1 = "../data/replies/"

would output,

>>> os.path.dirname(input1)
'../data/replies'

which is probably not what you're looking at as you want both "../data/replies/" and "../data/replies" to behave the same way.

edited Nov 14 '22 at 03:53

answered Nov 16 '15 at 05:19

samsamara

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I would recommend to not use "input" as a variable/reference. It is a built-in function. – DevPlayer Jan 28 '16 at 14:01

score 2 · Answer 15 · edited Mar 27 '12 at 16:24

2

print os.path.abspath(os.path.join(os.getcwd(), os.path.pardir))

You can use this to get the parent directory of the current location of your py file.

edited Mar 27 '12 at 16:24

jofel

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answered May 03 '11 at 15:24

Eros Nikolli

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That suggestion often leads to bugs. os.getcwd() is often NOT where "your py file" is. Think packages. If I "import some_package_with_subpackages" many modules will not be in that package's top-most directory. os.getcwd() returns where you execute top-most script. And that also presumes you are doing it from a command line. – DevPlayer Jan 28 '16 at 13:56
Like DevPlayer noted os.getcwd() is not necessarily the location of your python file. sys.argv[0] is what you're looking for. – Anonymous1847 Jul 21 '20 at 01:08

score 1 · Answer 16 · answered Jun 13 '22 at 12:26

import os 

def parent_directory():
  # Create a relative path to the parent of the current working directory 
  relative_parent = os.path.join(os.getcwd(), "..") # .. means parent directory

  # Return the absolute path of the parent directory
  return os.path.abspath(relative_parent)

print(parent_directory())

score 0 · Answer 17 · answered Apr 21 '14 at 12:02

GET Parent Directory Path and make New directory (name new_dir)

Get Parent Directory Path

os.path.abspath('..')
os.pardir

Example 1

import os
print os.makedirs(os.path.join(os.path.dirname(__file__), os.pardir, 'new_dir'))

Example 2

import os
print os.makedirs(os.path.join(os.path.dirname(__file__), os.path.abspath('..'), 'new_dir'))

score 0 · Answer 18 · edited Jul 02 '14 at 16:27

0

os.path.abspath('D:\Dir1\Dir2\..')

>>> 'D:\Dir1'

So a .. helps

edited Jul 02 '14 at 16:27

Patrick Hofman

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answered Jul 02 '14 at 16:02

Arindam Roychowdhury

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score 0 · Answer 19 · answered Sep 08 '16 at 01:35

import os

def parent_filedir(n):
    return parent_filedir_iter(n, os.path.dirname(__file__))

def parent_filedir_iter(n, path):
    n = int(n)
    if n <= 1:
        return path
    return parent_filedir_iter(n - 1, os.path.dirname(path))

test_dir = os.path.abspath(parent_filedir(2))

score 0 · Answer 20 · answered Aug 01 '18 at 08:21

The answers given above are all perfectly fine for going up one or two directory levels, but they may get a bit cumbersome if one needs to traverse the directory tree by many levels (say, 5 or 10). This can be done concisely by joining a list of N os.pardirs in os.path.join. Example:

import os
# Create list of ".." times 5
upup = [os.pardir]*5
# Extract list as arguments of join()
go_upup = os.path.join(*upup)
# Get abspath for current file
up_dir = os.path.abspath(os.path.join(__file__, go_upup))

score 0 · Answer 21 · answered Sep 20 '21 at 19:21

0

To find the parent of the current working directory:

import pathlib
pathlib.Path().resolve().parent

answered Sep 20 '21 at 19:21

Ondrej Sotolar

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