Is there a way to convert named function arguments to dict

Question

I am trying to find out if there is a way to convert named arguments to dict.

I understand using **kwargs in place of individual named arguments would be pretty straight forward.

def func(arg1=None, arg2=None, arg3=None):
    # How can I convert these arguments to {'arg1': None, 'arg2': None, 'arg3': None}`

Because the answer is just `{'arg1': None, 'arg2': None, 'arg3': None}` — Mad Physicist, Feb 12 '19 at 21:26
I can't imagine a use case for this. Would you be able to post the use-case as well explaining why you need this kind of structure? — mad_, Feb 12 '19 at 21:29
`func(..., **kwargs)` is for when you're converting named args on the **calling**-side to a dict on the **function-definition** side. But you want to convert one into the other, **both on the function-definition side**. That would be weird. Please explain why you want to do this (i.e. also show the calling-side syntax, and state what actual problem you're trying to solve). — smci, Feb 12 '19 at 21:29
Possible duplicate of [Getting list of parameter names inside python function](https://stackoverflow.com/questions/582056/getting-list-of-parameter-names-inside-python-function) — Iluvatar, Feb 12 '19 at 21:39

score 5 · Accepted Answer · answered Feb 12 '19 at 21:33

5

You can use locals() to get the local arguments:

def func(arg1=None, arg2=None, arg3=None):
    print(locals())

func()  # {'arg3': None, 'arg2': None, 'arg1': None}

answered Feb 12 '19 at 21:33

Darius

1 Answers1