python - os.path.dirname(__file__) returns empty

173
votes

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?

6
python
It works in Pycharm - kmiklas

6 Answers

267
votes

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__))
9
votes

can be used also like that:

dirname(dirname(abspath(__file__)))
7
votes
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

7
votes
import os.path

dirname = os.path.dirname(__file__) or '.'
5
votes
print(os.path.join(os.path.dirname(__file__)))

You can also use this way

0
votes

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

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