How to Retrieve a Module's Path?
I Want to Detect Whether Module Has Changed. Now, Using Inotify Is Simple, You Just Need to Know the Directory You Want to Get Notifications From. How Do I...
I want to detect whether module has changed. Now, using inotify is simple, you just need to know the directory you want to get notifications from.
How do I retrieve a module's path in python?
22 Answers
import a_module
print(a_module.__file__)
Will actually give you the path to the .pyc file that was loaded, at least on Mac OS X. So I guess you can do:
import os
path = os.path.abspath(a_module.__file__)
You can also try:
path = os.path.dirname(a_module.__file__)
To get the module's directory.
There is inspect module in python.
Must Read
Official documentation
The inspect module provides several useful functions to help get information about live objects such as modules, classes, methods, functions, tracebacks, frame objects, and code objects. For example, it can help you examine the contents of a class, retrieve the source code of a method, extract and format the argument list for a function, or get all the information you need to display a detailed traceback.
Example:
>>> import os
>>> import inspect
>>> inspect.getfile(os)
'/usr/lib64/python2.7/os.pyc'
>>> inspect.getfile(inspect)
'/usr/lib64/python2.7/inspect.pyc'
>>> os.path.dirname(inspect.getfile(inspect))
'/usr/lib64/python2.7'
As the other answers have said, the best way to do this is with __file__ (demonstrated again below). However, there is an important caveat, which is that __file__ does NOT exist if you are running the module on its own (i.e. as __main__).
For example, say you have two files (both of which are on your PYTHONPATH):
#/path1/foo.py
import bar
print(bar.__file__)
and
#/path2/bar.py
import os
print(os.getcwd())
print(__file__)
Running foo.py will give the output:
/path1 # "import bar" causes the line "print(os.getcwd())" to run
/path2/bar.py # then "print(__file__)" runs
/path2/bar.py # then the import statement finishes and "print(bar.__file__)" runs
HOWEVER if you try to run bar.py on its own, you will get:
/path2 # "print(os.getcwd())" still works fine
Traceback (most recent call last): # but __file__ doesn't exist if bar.py is running as main
File "/path2/bar.py", line 3, in <module>
print(__file__)
NameError: name '__file__' is not defined
Hope this helps. This caveat cost me a lot of time and confusion while testing the other solutions presented.
I will try tackling a few variations on this question as well:
- finding the path of the called script
- finding the path of the currently executing script
- finding the directory of the called script
(Some of these questions have been asked on SO, but have been closed as duplicates and redirected here.)