How do I get the path and name of the file that is currently executing?


I have scripts calling other script files but I need to get the filepath of the file that is currently running within the process.

For example, let’s say I have three files. Using execfile:

  • calls
  • In turn, calls

How can I get the file name and path of, from code within, without having to pass that information as arguments from

(Executing os.getcwd() returns the original starting script’s filepath not the current file’s.)

Asked By: Ray



It’s not entirely clear what you mean by “the filepath of the file that is currently running within the process”.
sys.argv[0] usually contains the location of the script that was invoked by the Python interpreter.
Check the sys documentation for more details.

As @Tim and @Pat Notz have pointed out, the __file__ attribute provides access to

the file from which the module was
loaded, if it was loaded from a file

Answered By: Blair Conrad

I think it’s just __file__ Sounds like you may also want to checkout the inspect module.

Answered By: Pat Notz

The __file__ attribute works for both the file containing the main execution code as well as imported modules.


Answered By: readonly


import inspect, os
print (inspect.getfile(inspect.currentframe())) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory
Answered By: Pat Notz

You can use inspect.stack()

import inspect,os
inspect.stack()[0]  => (<frame object at 0x00AC2AC0>, 'g:\Python\Test\', 15, '<module>', ['print inspect.stack()[0]n'], 0)
os.path.abspath (inspect.stack()[0][1]) => 'g:\Python\Test\'
Answered By: PabloG

as others have said. You may also want to use os.path.realpath to eliminate symlinks:

import os

Answered By: user13993
import sys

print sys.path[0]

this would print the path of the currently executing script

Answered By: appusajeev

The suggestions marked as best are all true if your script consists of only one file.

If you want to find out the name of the executable (i.e. the root file passed to the python interpreter for the current program) from a file that may be imported as a module, you need to do this (let’s assume this is in a file named

import inspect

print inspect.stack()[-1][1]

Because the last thing ([-1]) on the stack is the first thing that went into it (stacks are LIFO/FILO data structures).

Then in file if you import foo it’ll print, rather than, which would be the value of all of these:

  • __file__
  • inspect.getfile(inspect.currentframe())
  • inspect.stack()[0][1]
Answered By: user282406
import sys
print sys.argv[0]
Answered By: WBAR

I used the approach with __file__
but there is a little trick, it returns the .py file
when the code is run the first time,
next runs give the name of *.pyc file
so I stayed with:

Answered By: mik80

I think this is cleaner:

import inspect
print inspect.stack()[0][1]

and gets the same information as:

print inspect.getfile(inspect.currentframe())

Where [0] is the current frame in the stack (top of stack) and [1] is for the file name, increase to go backwards in the stack i.e.

print inspect.stack()[1][1]

would be the file name of the script that called the current frame. Also, using [-1] will get you to the bottom of the stack, the original calling script.

Answered By: Usagi

I have a script that must work under windows environment.
This code snipped is what I’ve finished with:

import os,sys
PROJECT_PATH = os.path.abspath(os.path.split(sys.argv[0])[0])

it’s quite a hacky decision. But it requires no external libraries and it’s the most important thing in my case.

Answered By: garmoncheg
import os
print os.path.basename(__file__)

this will give us the filename only. i.e. if abspath of file is then 2nd line will print

Answered By: vishal ekhe
import os

No need for inspect or any other library.

This worked for me when I had to import a script (from a different directory then the executed script), that used a configuration file residing in the same folder as the imported script.

Answered By: Kwuite
import os
os.path.dirname(__file__) # relative directory path
os.path.abspath(__file__) # absolute file path
os.path.basename(__file__) # the file name only
Answered By: Neal Xiong

This should work:

import os,sys
Answered By: Jahid

Update 2018-11-28:

Here is a summary of experiments with Python 2 and 3. With – runs – runs lib/
lib/ – prints filepath expressions

| Python | Run statement       | Filepath expression                    |
|      2 | execfile            | os.path.abspath(inspect.stack()[0][1]) |
|      2 | from lib import bar | __file__                               |
|      3 | exec                | (wasn't able to obtain it)             |
|      3 | import      | __file__                               |

For Python 2, it might be clearer to switch to packages so can use from lib import bar – just add empty files to the two folders.

For Python 3, execfile doesn’t exist – the nearest alternative is exec(open(<filename>).read()), though this affects the stack frames. It’s simplest to just use import foo and import – no files needed.

See also Difference between import and execfile

Original Answer:

Here is an experiment based on the answers in this thread – with Python 2.7.10 on Windows.

The stack-based ones are the only ones that seem to give reliable results. The last two have the shortest syntax, i.e. –

print os.path.abspath(inspect.stack()[0][1])                   #
print os.path.dirname(os.path.abspath(inspect.stack()[0][1]))  # C:filepathslib

Here’s to these being added to sys as functions! Credit to @Usagi and @pablog

Based on the following three files, and running from its folder with python (also tried execfiles with absolute paths and calling from a separate folder). execfile('') execfile('lib/')

import sys
import os
import inspect

print "Python " + sys.version

print __file__                                        #
print sys.argv[0]                                     #
print inspect.stack()[0][1]                           # lib/
print sys.path[0]                                     # C:filepaths

print os.path.realpath(__file__)                      #
print os.path.abspath(__file__)                       #
print os.path.basename(__file__)                      #
print os.path.basename(os.path.realpath(sys.argv[0])) #

print sys.path[0]                                     # C:filepaths
print os.path.abspath(os.path.split(sys.argv[0])[0])  # C:filepaths
print os.path.dirname(os.path.abspath(__file__))      # C:filepaths
print os.path.dirname(os.path.realpath(sys.argv[0]))  # C:filepaths
print os.path.dirname(__file__)                       # (empty string)

print inspect.getfile(inspect.currentframe())         # lib/

print os.path.abspath(inspect.getfile(inspect.currentframe())) #
print os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))) # C:filepathslib

print os.path.abspath(inspect.stack()[0][1])          #
print os.path.dirname(os.path.abspath(inspect.stack()[0][1]))  # C:filepathslib
Answered By: Brian Burns

I wrote a function which take into account eclipse debugger and unittest.
It return the folder of the first script you launch. You can optionally specify the __file__ var, but the main thing is that you don’t have to share this variable across all your calling hierarchy.

Maybe you can handle others stack particular cases I didn’t see, but for me it’s ok.

import inspect, os
def getRootDirectory(_file_=None):
    Get the directory of the root execution file
    Can help:
    For eclipse user with unittest or debugger, the function search for the correct folder in the stack
    You can pass __file__ (with 4 underscores) if you want the caller directory
    # If we don't have the __file__ :
    if _file_ is None:
        # We get the last :
        rootFile = inspect.stack()[-1][1]
        folder = os.path.abspath(rootFile)
        # If we use unittest :
        if ("/pysrc" in folder) & ("org.python.pydev" in folder):
            previous = None
            # We search from left to right the :
            for el in inspect.stack():
                currentFile = os.path.abspath(el[1])
                if ("unittest/" in currentFile) | ("org.python.pydev" in currentFile):
                previous = currentFile
            folder = previous
        # We return the folder :
        return os.path.dirname(folder)
        # We return the folder according to specified __file__ :
        return os.path.dirname(os.path.realpath(_file_))
Answered By: hayj

To get directory of executing script

 print os.path.dirname( inspect.getfile(inspect.currentframe()))
Answered By: pbaranski

Simplest way is:


import subprocess['python3',<>])



P.S.: I’ve tried execfile, but since it reads as a string, sys.argv[0] returned <string>.

Answered By: Lucas Azevedo

Try this,

import os
Answered By: Soumyajit
Answered By: BaiJiFeiLong

Here is what I use so I can throw my code anywhere without issue. __name__ is always defined, but __file__ is only defined when the code is run as a file (e.g. not in IDLE/iPython).

if '__file__' in globals():
    self_name = globals()['__file__']
elif '__file__' in locals():
    self_name = locals()['__file__']
    self_name = __name__

Alternatively, this can be written as:

self_name = globals().get('__file__', locals().get('__file__', __name__))
Answered By: craymichael

Since Python 3 is fairly mainstream, I wanted to include a pathlib answer, as I believe that it is probably now a better tool for accessing file and path information.

from pathlib import Path

current_file: Path = Path(__file__).resolve()

If you are seeking the directory of the current file, it is as easy as adding .parent to the Path() statement:

current_path: Path = Path(__file__).parent.resolve()
Answered By: Doug

The following returns the path where your current main script is located at. I tested this with Linux, Win10, IPython and Jupyter Lab. I needed a solution that works for local Jupyter notebooks as well.

import builtins
import os
import sys

def current_dir():
    if "get_ipython" in globals() or "get_ipython" in dir(builtins):
        # os.getcwd() is PROBABLY the dir that hosts the active notebook script.
        # See also
        return os.getcwd()
        return os.path.abspath(os.path.dirname(sys.argv[0]))
Answered By: nichoio

Finding the home directory of the path in which your Python script resides

As an addendum to the other answers already here (and not answering the OP’s question, since other answers already do that), if the path to your script is /home/gabriel/GS/dev/eRCaGuy_dotfiles/useful_scripts/, and you wish to obtain the home directory part of that path, which is /home/gabriel, you can do this:

import os

# Obtain the home dir of the user in whose home directory this script resides
script_path_list = os.path.normpath(__file__).split(os.sep)
home_dir = os.path.join("/", script_path_list[1], script_path_list[2])

To help make sense of this, here are the paths for __file__, script_path_list, and home_dir. Notice that script_path_list is a list of the path components, with the first element being an empty string since it originally contained the / root dir path separator for this Linux path:

__file__         = /home/gabriel/GS/dev/eRCaGuy_dotfiles/useful_scripts/
script_path_list = ['', 'home', 'gabriel', 'GS', 'dev', 'eRCaGuy_dotfiles', 'useful_scripts', '']
home_dir         = /home/gabriel


  1. Python: obtain the path to the home directory of the user in whose directory the script being run is located [duplicate]
Answered By: Gabriel Staples
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