获取Python中调用者的相对路径

Question

I've got this function:

def relative_path(*paths):
    return os.path.join(os.path.dirname(__file__), *paths)

How would I change it to return the path relative to the caller?

For example, if I called relative_path('index.html') from another script, is it possible to get the path relative the script from where it was called implicitly or would I need to modify relative_path to pass __file__ across as well like this?

def relative_path(__file__, *paths):
    return os.path.join(os.path.dirname(__file__), *paths)

Answer 1

adapting the solution in Get __name__ of calling function's module in Python

file1.py

import os
import inspect

def relative_path(*paths):
    return os.path.join(os.path.dirname(__file__), *paths)

def relative_to_caller(*paths):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    return os.path.join(os.path.dirname(mod.__file__), *paths)

if __name__ == '__main__':
    print(relative_path('index.html'))

sub/sub_file.py

import sys
sys.path.append(r'/Users/xx/PythonScripts/!scratch')

import file1

if __name__ == '__main__':
    print(file1.relative_path('index.html'))
    print(file1.relative_to_caller('index.html'))

Running sub_file.py gives the following output:

/Users/xx/PythonScripts/!scratch/index.html
/Users/xx/PythonScripts/!scratch/sub/index.html

There are some caveats in the comments to the question in the link above...

Answer 2

Note that tracing stack would be a possibility here, but it could cause some serious trouble ( like confusing 'garbage collector', or may not even work in eggs )

I believe the cleanest way would be to pass the caller to the rel_path function.

However, as you may know, there is usually an ugly way of doing things in python. You can do for example something like this:

consider following two scripts:

# relpath.py

import os


def rel_path(path):
    if os.path.isfile(__name__):
        return os.path.relpath(path, start=__name__)

    print("Warning: %s is not a file: returning path relative to the current working dir" % __name__, file=sys.stderr)
    return os.path.relpath(path)

# caller.py

import importlib.util


spec = importlib.util.spec_from_file_location(name=__file__, location="/workspace/relpath.py")

rel =  importlib.util.module_from_spec(spec)

spec.loader.exec_module(rel)
print(rel.rel_path("/tmp"))

What we did here: when loading the module using importlib.util, we passed the name=__file__ , which gave our module the name consisting of the caller script path . Hence we needn't pass it as an argument to the relpath.py .

Note that this is not clean solution and might not be readable for future developers reading your code. I just wanted to demonstrate the possibilities of python.