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In several SO's question there is these lines to access the parent directory of the code, e.g. os.path.join(os.path.dirname(__file__)) returns nothing and os.path.join(os.path.dirname(__file__)) returns nothing

import os, sys
parentddir = os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
sys.path.append(parentddir)

I understand that os.path.abspath() returns absolute path of something and sys.path.append() adds the path for the code to access. but what is this cryptic line below, what does it really mean?

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

Is there another way to achieve the same purpose of appending the parent directory of the where the code?

This problem happens because I am calling functions across directories and sometimes they share the same file name, e.g. script1/utils.py and script2/utils.py. I am calling a function from script1/test.py which calls script2/something.py contains a function that calls script2/utils.py and the following code

script1/
        utils.py
        src/
            test.py

script2/
        utils.py
        code/
            something.py

test.py

from script2.code import something
import sys
sys.path.append('../')
import utils

something.foobar()

something.py

import os, sys
parentddir = os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
sys.path.append(parentddir)
import utils

def foobar():
  utils.somefunc()
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alvas
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  • Just a question, are you saying that you need to do that in order to `import utils`? I don't get the last part right. – Paulo Bu Jan 08 '14 at 20:51

2 Answers2

55

That is a clever way to refer to paths regardless of the script location. The cryptic line you're referring is:

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

There are 3 methods and a 2 constants present:

  1. abspath returns absolute path of a path
  2. join join to path strings
  3. dirname returns the directory of a file
  4. __file__ refers to the script's file name
  5. pardir returns the representation of a parent directory in the OS (usually ..)

Thus, the expression returns the full path name of the executing script in a multiplatform-safe way. No need to hardwire any directions, that's why it is so useful.

There might be other approaches to get a parent directory of where a file is located, for example, programs have the concept of current working directory, os.getcwd(). So doing os.getcwd()+'/..' might work. But this is very dangerous, because working directories can be changed.

Also, if the file is intended to be imported, the working directory will point to the importing file, not the importee, but __file__ always points to the actual module's file so it is safer.

Hope this helps!

Edit: P.S. - Python 3 greatly simplifies this situation by letting us treat paths in an object-oriented manner, so the above line becomes:

from pathlib import Path
Path(__file__).resolve().parent.parent
Louis Maddox
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Paulo Bu
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  • You can't take de path `dirname` of `__file__` because __file__ is only the name of the script. You need to use `os.getcwd()` to obtain that path. – juanfal Nov 21 '19 at 11:42
  • In this case, what would be the difference between `os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))` and `os.path.abspath(os.path.dirname(os.path.dirname(__file__)))`? won't they be the same? – rredondo Jul 12 '22 at 08:01
21

__file__ represents the file the code is executing from

os.path.dirname(__file__) gives you the directory the file is in

os.path.pardir stands for ".." which means one directory above the current one

os.path.join(os.path.dirname(__file__), os.path.pardir) joins the directory name and ".."

os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)) resolves the above path and gives you an absolute path for the parent directory of the directory your file is in

praveen
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