How to get path of the pyd file aka equivalent of __file__

问题

I have a file package.py that i am trying to package into package.pyd. I have the following statement in package.py

    CURR = os.path.dirname(os.path.realpath(__file__))

which works fine when I run package.py but when I import package.pyd into another file wrapper.py I get the following error message

Traceback (most recent call last):
  File "C:\Projects\Wrapper.py", line 1, in <module>
    import package
  File "package.py", line 40, in init package (package.c:4411)
NameError: name '__file__' is not defined

How can I get the location of the .pyd file. Also is there a way to check if it is being run as a .pyd or .py.

Thank you!

回答1:

It seems that __file__ variable not available in module init. But you can get __file__ after module was loaded:

def get_file():
    return __file__

You can check the __file__ variable to know what file was loaded. Also keep in mind python's search order: pyd (so), py, pyw(for windows), pyc.

More information about it is in this this question

Found two working methods.

Involving inspect module:

import inspect
import sys
import os

if hasattr(sys.modules[__name__], '__file__'):
    _file_name = __file__
else:
    _file_name = inspect.getfile(inspect.currentframe())

CURR = os.path.dirname(os.path.realpath(_file_name))

import some file from the same level and using its __file__ attribute:
```
import os
from . import __file__ as _initpy_file
CURR = os.path.dirname(os.path.realpath(_initpy_file))
```
Actually, it doesn't have to be __init__.py module, you can add and import any [empty] file to make it work.

回答2:

__file__ now works in recent more versions of Cython (0.27 ish) when on run on a version of Python that supports multi-phase module initialization (Python >=3.5). See https://github.com/cython/cython/issues/1715 for the point at which it was added.

来源：https://stackoverflow.com/questions/40857480/how-to-get-path-of-the-pyd-file-aka-equivalent-of-file

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python

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cythonize