How do I get the path and name of the file that is currently executing?

p1.py:

execfile("p2.py")

p2.py:

import inspect, os
print (inspect.getfile(inspect.currentframe())) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory

Update 2018-11-28:

Here is a summary of experiments with Python 2 and 3. With

main.py - runs foo.py
foo.py - runs lib/bar.py
lib/bar.py - prints filepath expressions

| Python | Run statement       | Filepath expression                    |
|--------+---------------------+----------------------------------------|
|      2 | execfile            | os.path.abspath(inspect.stack()[0][1]) |
|      2 | from lib import bar | __file__                               |
|      3 | exec                | (wasn't able to obtain it)             |
|      3 | import lib.bar      | __file__                               |

For Python 2, it might be clearer to switch to packages so can use from lib import bar - just add empty __init__.py files to the two folders.

For Python 3, execfile doesn't exist - the nearest alternative is exec(open(<filename>).read()), though this affects the stack frames. It's simplest to just use import foo and import lib.bar - no __init__.py files needed.

See also Difference between import and execfile

---

Original Answer:

Here is an experiment based on the answers in this thread - with Python 2.7.10 on Windows.

The stack-based ones are the only ones that seem to give reliable results. The last two have the shortest syntax, i.e. -

print os.path.abspath(inspect.stack()[0][1])                   # C:\filepaths\lib\bar.py
print os.path.dirname(os.path.abspath(inspect.stack()[0][1]))  # C:\filepaths\lib

Here's to these being added to sys as functions! Credit to @Usagi and @pablog

Based on the following three files, and running main.py from its folder with python main.py (also tried execfiles with absolute paths and calling from a separate folder).

C:\filepaths\main.py: execfile('foo.py')
C:\filepaths\foo.py: execfile('lib/bar.py')
C:\filepaths\lib\bar.py:

import sys
import os
import inspect

print "Python " + sys.version
print

print __file__                                        # main.py
print sys.argv[0]                                     # main.py
print inspect.stack()[0][1]                           # lib/bar.py
print sys.path[0]                                     # C:\filepaths
print

print os.path.realpath(__file__)                      # C:\filepaths\main.py
print os.path.abspath(__file__)                       # C:\filepaths\main.py
print os.path.basename(__file__)                      # main.py
print os.path.basename(os.path.realpath(sys.argv[0])) # main.py
print

print sys.path[0]                                     # C:\filepaths
print os.path.abspath(os.path.split(sys.argv[0])[0])  # C:\filepaths
print os.path.dirname(os.path.abspath(__file__))      # C:\filepaths
print os.path.dirname(os.path.realpath(sys.argv[0]))  # C:\filepaths
print os.path.dirname(__file__)                       # (empty string)
print

print inspect.getfile(inspect.currentframe())         # lib/bar.py

print os.path.abspath(inspect.getfile(inspect.currentframe())) # C:\filepaths\lib\bar.py
print os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))) # C:\filepaths\lib
print

print os.path.abspath(inspect.stack()[0][1])          # C:\filepaths\lib\bar.py
print os.path.dirname(os.path.abspath(inspect.stack()[0][1]))  # C:\filepaths\lib
print

__file__

as others have said. You may also want to use os.path.realpath to eliminate symlinks:

import os

os.path.realpath(__file__)