os.path.dirname(file) returns empty

Question

I want to get the path of the current directory under which a .py file is executed.

For example a simple file D:test.py with code:

import os

print os.getcwd()
print os.path.basename(__file__)
print os.path.abspath(__file__)
print os.path.dirname(__file__)

It is weird that the output is:

D:
test.py
D:test.py
EMPTY

I am expecting the same results from the getcwd() and path.dirname().

Given os.path.abspath = os.path.dirname + os.path.basename, why

os.path.dirname(__file__)

returns empty?

Asked By: Flake

||

Source

Answer 1

Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have

os.path.dirname(filename) + os.path.basename(filename) == filename

Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.

To get the dirname of the absolute path, use

os.path.dirname(os.path.abspath(__file__))

Answered By: Sven Marnach

Answer 2

print(os.path.join(os.path.dirname(__file__)))

You can also use this way

Answered By: Mikhail

Answer 3

import os.path

dirname = os.path.dirname(__file__) or '.'

Answered By: Deve

Answer 4

can be used also like that:

dirname(dirname(abspath(__file__)))

Answered By: adnan dogar

Answer 5

os.path.split(os.path.realpath(__file__))[0]

os.path.realpath(__file__)return the abspath of the current script; os.path.split(abspath)[0] return the current dir

Answered By: RY_ Zheng

Answer 6

I guess this is a straight forward code without the os module..

__file__.split(__file__.split("/")[-1])[0]

Answered By: MohammadArik

Answer 7

Since Python 3.4, you can use pathlib to get the current directory:

from pathlib import Path

# get parent directory
curr_dir = Path(__file__).parent

file_path = curr_dir.joinpath('otherfile.txt')

Answered By: Melle

Answer 8

None of the above answers is correct. OP wants to get the path of the current directory under which a .py file is executed, not stored.

Thus, if the path of this file is /opt/script.py…

#! /usr/bin/env python3
from pathlib import Path

# -- file's directory -- where the file is stored
fd = Path(__file__).parent

# -- current directory -- where the file is executed
# (i.e. the directory of the process)
cwd = Path.cwd()

print(f'{fd=} {cwd=}')

Only if we run this script from /opt, fd and cwd will be the same.

$ cd /
$ /opt/script.py
cwd=PosixPath('/') fd=PosixPath('/opt')

$ cd opt
$ ./script.py
cwd=PosixPath('/opt') fd=PosixPath('/opt')

$ cd child
$ ../script.py
cwd=PosixPath('/opt/child') fd=PosixPath('/opt/child/..')

Answered By: Nuno André

os.path.dirname(file) returns empty

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