Python's inspect module implements a method to obtain the load module path

This article mainly describes how to get the path of the module. It is necessary to declare that the module mentioned here can be a functional implementation of the module, or other modules.

The.getsourcefile for the inspect module is used (the module name you need to get)

Create test.py as follows:

import os
import inspect
 
class pathManager(object):
 
	def __init__(self):
		pass
 
	def _abPath(self):
		modulePath = inspect.getsourcefile(os)
		abPath = os.path.split(modulePath)
		return abPath[0]
 
if __name__ == "__main__":
    getPath = pathManager()
    getPath._abPath()

Execute python ES13en. py to see the results as follows:

clay@aclgcl-ubnt:~/Desktop/python$ python test.py 
/usr/local/lib/python2.7/os.py
('/usr/local/lib/python2.7', 'os.py')
clay@aclgcl-ubnt:~/Desktop/python$

Can see we have direct access to: / usr local lib/python2. 7 / os. py, through os. path. split can intercept a simple path.