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

You can use the inspect module to achieve this:

frame = inspect.stack()[1]
module = inspect.getmodule(frame[0])
filename = module.__file__

Inspired by ThiefMaster's answer but works also if inspect.getmodule() returns None:

frame = inspect.stack()[1]
filename = frame[0].f_code.co_filename

Python 3.5+

One-liner

To get the full filename (with path and file extension), use in the callee:

import inspect
filename = inspect.stack()[1].filename

Full filename vs filename only

To retrieve the caller's filename use inspect.stack(). Additionally, the following code also trims the path at the beginning and the file extension at the end of the full filename:

# Callee.py
import inspect
import os.path

def get_caller_info():
  # first get the full filename (including path and file extension)
  caller_frame = inspect.stack()[1]
  caller_filename_full = caller_frame.filename

  # now get rid of the directory (via basename)
  # then split filename and extension (via splitext)
  caller_filename_only = os.path.splitext(os.path.basename(caller_filename_full))[0]

  # return both filename versions as tuple
  return caller_filename_full, caller_filename_only

It can then be used like so:

# Caller.py
import callee

filename_full, filename_only = callee.get_caller_info()
print(f"> Filename full: {filename_full}")
print(f"> Filename only: {filename_only}")

# Output
# > Filename full: /workspaces/python/caller_filename/caller.py
# > Filename only: caller

Official docs

  • os.path.basename(): to remove the path from the filename (still includes the extension)
  • os.path.splitext(): to split the the filename and the file extension

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Python

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