[python] 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 foo.py and I'd like to do something like this in order to get the script name:

print Scriptname

As of Python 3.5 you can simply do:

from pathlib import Path
Path(__file__).stem

See more here: https://docs.python.org/3.5/library/pathlib.html#pathlib.PurePath.stem

For example, I have a file under my user directory named test.py with this inside:

from pathlib import Path

print(Path(__file__).stem)
print(__file__)

running this outputs:

>>> python3.6 test.py
test
test.py

all that answers are great, but have some problems You might not see at the first glance.

lets define what we want - we want the name of the script that was executed, not the name of the current module - so __file__ will only work if it is used in the executed script, not in an imported module. sys.argv is also questionable - what if your program was called by pytest ? or pydoc runner ? or if it was called by uwsgi ?

and - there is a third method of getting the script name, I havent seen in the answers - You can inspect the stack.

Another problem is, that You (or some other program) can tamper around with sys.argv and __main__.__file__ - it might be present, it might be not. It might be valid, or not. At least You can check if the script (the desired result) exists !

the library lib_programname does exactly that :

• check if __main__ is present
• check if __main__.__file__ is present
• does give __main__.__file__ a valid result (does that script exist ?)
• if not: check sys.argv:
• is there pytest, docrunner, etc in the sys.argv ? --> if yes, ignore that
• can we get a valid result here ?
• if not: inspect the stack and get the result from there possibly
• if also the stack does not give a valid result, then throw an Exception.

by that way, my solution is working so far with setup.py test, uwsgi, pytest, pycharm pytest , pycharm docrunner (doctest), dreampie, eclipse

there is also a nice blog article about that problem from Dough Hellman, "Determining the Name of a Process from Python"

BTW, it will change again in python 3.9 : the file attribute of the main module became an absolute path, rather than a relative path. These paths now remain valid after the current directory is changed by os.chdir()

So I rather want to take care of one small module, instead of skimming my codebase if it should be changed somewere ...

^{Disclaimer: I'm the author of the lib_programname library.}

Note: If you are using Python 3+, then you should use the print() function instead

Assuming that the filename is foo.py, the below snippet

import sys
print sys.argv[0][:-3]

or

import sys
print sys.argv[0][::-1][3:][::-1]

As for other extentions with more characters, for example the filename foo.pypy

import sys
print sys.argv[0].split('.')[0]

If you want to extract from an absolute path

import sys
print sys.argv[0].split('/')[-1].split('.')[0]

will output foo

My fast dirty solution:

__file__.split('/')[-1:][0]

You can do this without importing os or other libs.

If you want to get the path of current python script, use: __file__

If you want to get only the filename without .py extension, use this:

__file__.rsplit("/", 1)[1].split('.')[0]

If you're doing an unusual import (e.g., it's an options file), try:

import inspect
print (inspect.getfile(inspect.currentframe()))

Note that this will return the absolute path to the file.

Try this:

print __file__

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.

Handling symbolic links

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.

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.

os.path.abspath(__file__) will give you an absolute path (relpath() available as well).

sys.argv[-1] will give you a relative path.

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])

The first argument in sys will be the current file name so this will work

import sys
print sys.argv[0] # will print the file name

 def basename():
   x=__file__
   y=x.split('\\')
   y1=y[-1]
   y2=y1.split('.')
   y3=y2[0]
 return(y2[0])
  

Since the OP asked for the name of the current script file I would prefer

import os
os.path.split(sys.argv[0])[1]

we can try this to get current script name without extension.

import os

script_name = os.path.splitext(os.path.basename(__file__))[0]

import sys
print(sys.argv[0])

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

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
print(main.__file__)

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.