how to get the caller's filename, method name in python

Question

for example, a.boo method calls b.foo method. In b.foo method, how can I get a\'s file name (I don\'t want to pass __file__

Answer 1

Python 3.5+

Caller's Filename

To retrieve just the caller's filename with no path at the beginning or extension at the end, use:

import inspect
import os

# First get the full filename (including path and file extension)
caller_frame = inspect.stack()[1]
caller_filename = caller_frame.filename

# Now get rid of the directory and extension
filename = os.path.splitext(os.path.basename(caller_filename))[0]

# Output
print(f"Caller Filename: {caller_filename}")
print(f"Filename: {filename}")

Sources:

• inspect.stack()
• os.path.basename()
• os.path.splitext()