python parent directory

Question

I want to get the parent directory of a file. This code doesn't work:

projectRoot = os.path.dirname(os.pardir(os.path.abspath(__file__)))

What's wrong?

(I am developing a django application, and use this variable in the settings file of my application)

Thanks.

Romain

Check out the UniPath module, this helps with this problem excellently. — Henrik Andersson, Apr 10 '13 at 11:53
`os.pardir` is not what you think it is: `>>> os.pardir` `'..'` — jamylak, Apr 10 '13 at 11:59

karthikr · Accepted Answer · 2013-04-10T12:14:45.893

5

You can access the parent directory by using ..

import os
os.path.join(os.path.dirname(__file__), os.pardir)

OR:

import os
split_limit = 1 if os.path.isdir(__file__) else 0
parent_dir = os.path.abspath(__file__).rsplit(os.path.sep, split_limit)[0]

edited Apr 10 '13 at 12:14

answered Apr 10 '13 at 11:54

karthikr

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1

I used your first solution, works well! Thanks! – rom Apr 10 '13 at 13:15

score 1 · Answer 2 · answered Apr 10 '13 at 11:55

1

As simple as this:

import os
def get_parent_dir(file_path):
    return os.path.dirname(file_path)

To get the file_path of the script being run, you could do this:

import sys
file_path = sys.argv[0]

answered Apr 10 '13 at 11:55

Inbar Rose

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`os.path.dirname` appears to return the absolute *path* to the file, not the *name* of the file's immediate parent directory – 8bitjunkie Aug 29 '15 at 09:58

score 0 · Answer 3 · answered Apr 10 '13 at 12:06

in our django project we use next structure: we have prodaction_settings.py where placed settings for prodaction server and all variables. on local machine in file settings.py

from os.path import dirname, abspath, join
TEST_DIR = dirname(abspath(__file__))
from production_settings import *

to add your local folder with templates:

TEMPLATE_DIRS = (
    join(TEST_DIR, 'tmpl'), 
)

python parent directory

3 Answers3