Get name of current script in Python


I'm trying to get the name of the Python script that is currently running.

I have a script called and I'd like to do something like this in order to get the script name:

print Scriptname
8/27/2019 11:59:06 PM

Accepted Answer

You can use __file__ to get the name of the current file. When used in the main module, this is the name of the script that was originally invoked.

If you want to omit the directory part (which might be present), you can use os.path.basename(__file__).

4/30/2020 1:22:21 PM

import sys
print sys.argv[0]

This will print for python, dir/ for python dir/, etc. It's the first argument to python. (Note that after py2exe it would be foo.exe.)


For completeness' sake, I thought it would be worthwhile summarizing the various possible outcomes and supplying references for the exact behaviour of each:

  • __file__ is the currently executing file, as detailed in the official documentation:

    __file__ is the pathname of the file from which the module was loaded, if it was loaded from a file. The __file__ attribute may be missing for certain types of modules, such as C modules that are statically linked into the interpreter; for extension modules loaded dynamically from a shared library, it is the pathname of the shared library file.

    From Python3.4 onwards, per issue 18416, __file__ is always an absolute path, unless the currently executing file is a script that has been executed directly (not via the interpreter with the -m command line option) using a relative path.

  • __main__.__file__ (requires importing __main__) simply accesses the aforementioned __file__ attribute of the main module, e.g. of the script that was invoked from the command line.

    From Python3.9 onwards, per issue 20443, the __file__ attribute of the __main__ module became an absolute path, rather than a relative path.

  • sys.argv[0] (requires importing sys) is the script name that was invoked from the command line, and might be an absolute path, as detailed in the official documentation:

    argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string.

    As mentioned in another answer to this question, Python scripts that were converted into stand-alone executable programs via tools such as py2exe or PyInstaller might not display the desired result when using this approach (i.e. sys.argv[0] would hold the name of the executable rather than the name of the main Python file within that executable).

  • If none of the aforementioned options seem to work, probably due to an atypical execution process or an irregular import operation, the inspect module might prove useful. In particular, invoking inspect.stack()[-1][1] should work, although it would raise an exception when running in an implementation without Python stack frame.

  • From Python3.6 onwards, and as detailed in another answer to this question, it's possible to install an external open source library, lib_programname, which is tailored to provide a complete solution to this problem.

    This library iterates through all of the approaches listed above until a valid path is returned. If all of them fail, it raises an exception. It also tries to address various pitfalls, such as invocations via the pytest framework or the pydoc module.

    import lib_programname
    # this returns the fully resolved path to the launched python program
    path_to_program = lib_programname.get_path_executed_script()  # type: pathlib.Path

Handling relative paths

When dealing with an approach that happens to return a relative path, it might be tempting to invoke various path manipulation functions, such as os.path.abspath(...) or os.path.realpath(...) in order to extract the full or real path.

However, these methods rely on the current path in order to derive the full path. Thus, if a program first changes the current working directory, for example via os.chdir(...), and only then invokes these methods, they would return an incorrect path.

If the current script is a symbolic link, then all of the above would return the path of the symbolic link rather than the path of the real file and os.path.realpath(...) should be invoked in order to extract the latter.

Further manipulations that extract the actual file name

os.path.basename(...) may be invoked on any of the above in order to extract the actual file name and os.path.splitext(...) may be invoked on the actual file name in order to truncate its suffix, as in os.path.splitext(os.path.basename(...)).

From Python 3.4 onwards, per PEP 428, the PurePath class of the pathlib module may be used as well on any of the above. Specifically, pathlib.PurePath(...).name extracts the actual file name and pathlib.PurePath(...).stem extracts the actual file name without its suffix.


Note that __file__ will give the file where this code resides, which can be imported and different from the main file being interpreted. To get the main file, the special __main__ module can be used:

import __main__ as main

Note that __main__.__file__ works in Python 2.7 but not in 3.2, so use the import-as syntax as above to make it portable.


The Above answers are good . But I found this method more efficient using above results.
This results in actual script file name not a path.

import sys    
import os    
file_name =  os.path.basename(sys.argv[0])

For modern Python versions (3.4+), Path(__file__).name should be more idiomatic. Also, Path(__file__).stem gives you the script name without the .py extension.


Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow
Email: [email protected]