determine from which file a function is defined in python

Question

I am programmatically printing out a list of function in python. I can get the name from __name__

for i, func in enumerate(list_of_functions): print(str(i) + func.__name__)

How to get the source filename where the function is defined as well?

and in case the function it is attribute of a object, how to get the type of parent object?

portability python2/3 is a must

The title says "from which file a function is called", but you are actually asking in which file a function is defined, which is different... — zvone
– zvone, Commented May 31, 2018 at 10:31

Smart Manoj · Accepted Answer · 2023-09-17 12:58:56Z

func.__module__

Will return the module in which it is defined

func.__globals__['__file__']

will return the whole path of the file where it is defined. Only for user-defined functions

JobJob · Accepted Answer · 2020-11-04 23:43:25Z

How to get the source filename where the function is defined ... ?

import inspect inspect.getfile(func)

and in case the function it is attribute of a object, how to get the type of parent object?

To get the class of a method (i.e. when the function is an attribute of a object):

if inspect.ismethod(func): the_class = func.__self__.__class__

Moberg · Accepted Answer · 2018-05-31 08:43:09Z

As stated here https://docs.python.org/3/library/inspect.html func.__globals__ returns the global namespace where the function was defined.

Ran Sasportas · Accepted Answer · 2018-05-31 08:42:56Z

1

For getting the filename just use -

print(os.path.basename(__file__))

if you want the full file path you can just use

print __file__

answered May 31, 2018 at 8:42

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1 Comment

not quite. this will give me the filename of the current script. I am editing to clarify

AlexShein · Accepted Answer · 2018-05-31 08:48:11Z

0

You can use __file__ variable.

print(__file__)

answered May 31, 2018 at 8:42

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