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:
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
import sys print(sys.argv)
This will print
python dir/foo.py, etc. It’s the first argument to
python. (Note that after py2exe it would be
__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__)
__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)
For completeness’ sake, I thought it would be worthwhile summarizing the various possible outcomes and supplying references for the exact behaviour of each.
The answer is composed of four sections:
A list of different approaches that return the full path to the currently executing script.
A caveat regarding handling of relative paths.
A recommendation regarding handling of symbolic links.
An account of a few methods that could be used to extract the actual file name, with or without its suffix, from the full file path.
__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.
sys.argv (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:
argvis the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the
-ccommand line option to the interpreter,
argvis set to the string
'-c'. If no script name was passed to the Python interpreter,
argvis 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 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, as suggested in another answer to this question:
import inspect source_file_path = inspect.getfile(inspect.currentframe())
However, inspect.currentframe() would raise an exception when running in an implementation without Python stack frame.
Note that inspect.getfile(…) is preferred over inspect.getsourcefile(…) because the latter raises a TypeError exception when it can determine only a binary file, not the corresponding source file (see also this answer to another question).
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
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.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.
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
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.
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.
For modern Python versions (3.4+),
Path(__file__).name should be more idiomatic. Also,
Path(__file__).stem gives you the script name without the
Since the OP asked for the name of the current script file I would prefer
import os os.path.split(sys.argv)
we can try this to get current script name without extension.
import os script_name = os.path.splitext(os.path.basename(__file__))
As of Python 3.5 you can simply do:
from pathlib import Path Path(__file__).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
__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 :
__main__.__file__a valid result (does that script exist ?)
by that way, my solution is working so far with
pycharm pytest ,
pycharm docrunner (doctest),
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.
You can do this without importing os or other libs.
If you want to get the path of current python script, use:
If you want to get only the filename without .py extension, use this: