How to obtain a dictionary of the named arguments in a Python function

Question

I know it's a common case for people to give a function an arbitrary number of kwargs with **kwargs , and then access them as a dictionary; however, I want to explicitly specify my functions kwargs, but still be able to access them as a dictionary.

This is because I want my function to only receive specific kwargs , but I need to perform an identical operation with all of them, which I can put into a for loop.

def my_func(kwarg1=None, kwarg2=None, kwarg3=None):
    kwargs = {} # After somehow getting all my kwargs into a dictionary

    for k in kwargs:
        # Do my operation

I do not want my function to receive an arbitrary number of kwargs , but I do want to access my kwargs in a dictionary.

Answer 1

Assuming you have no positional arguments, you could get access to your kwargs via locals if you put it at the top of your function:

def my_func(kwarg1=None, kwarg2=None, kwarg3=None):
    # Don't add any more variables before the next line!
    kwargs = dict(locals())

    for k in kwargs:
        # Do my operation

This is hacky (at best) and it's probably better to just spell it out:

kwargs = {'kwarg1': kwarg1, ...}

Answer 2

This is Python3.3+ code that creates the list of keyword argument names automatically. Just for completness. I would prefer any of the simpler solutions.

import inspect

def my_func(*, kwarg1=None, kwarg2=None, kwarg3=None):
    local_vars = locals()
    kwargs = {k: local_vars[k] for k in KWARGS_my_func}
    print(kwargs)

KWARGS_my_func = [p.name for p in inspect.signature(my_func).parameters.values()
                  if p.kind == inspect.Parameter.KEYWORD_ONLY]

my_func(kwarg2=2)

Answer 3

Simply create a dictionary as normal, retrieving the values of each argument.

def my_func(kwarg1=None, kwarg2=None, kwarg3=None):
    kwargs = {'kwarg1':kwarg1, 'kwarg2':kwarg2, 'kwarg3':kwarg3}

    for k in kwargs:
        # Do my operation