Get name of current script in Python
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__) .
Similar question
Since the OP asked for the name of the current script file I would prefer
import os os.path.split(sys.argv[0])[1] 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.
bitranox 1368
Some more answer related to the same question
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:
we can try this to get current script name without extension.
import os script_name = os.path.splitext(os.path.basename(__file__))[0] 
Krishn Singh 61
Related question
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__) >>> python3.6 test.py test test.py elad silver 8529
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.
Gajendra D Ambi 3418
ceth 42790
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.
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]) 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.
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.
Extracting the full file path
__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.
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.
import inspect source_file_path = inspect.getfile(inspect.currentframe()) 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.
Yoel 8639
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 .)
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