How do I get the path of the current executed file in Python?

You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.

However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.

some_path/module_locator.py:

def we_are_frozen():    # All of the modules are built-in to the interpreter, e.g., by py2exe    return hasattr(sys, "frozen")def module_path():    encoding = sys.getfilesystemencoding()    if we_are_frozen():        return os.path.dirname(unicode(sys.executable, encoding))    return os.path.dirname(unicode(__file__, encoding))

some_path/main.py:

import module_locatormy_path = module_locator.module_path()

If you have several main scripts in different directories, you may need more than one copy of module_locator.

Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.

First, you need to import from inspect and os

from inspect import getsourcefilefrom os.path import abspath

Next, wherever you want to find the source file from you just use

abspath(getsourcefile(lambda:0))

This solution is robust even in executables:

import inspect, os.pathfilename = inspect.getframeinfo(inspect.currentframe()).filenamepath     = os.path.dirname(os.path.abspath(filename))