How to edit own kwargs from inside a function?

Question

This code:

#!/usr/bin/env python

def f(a):
    print(f"{locals()}")

if __name__ == '__main__':
    f(1)

Output: {'a': 1}

Want to do something like this:

#!/usr/bin/env python

def f(a):
    kwargs["b"] = 2
    print(f"{locals()}")

if __name__ == '__main__':
    f(1)

to get the output {'a': 1, 'b': 2}

What's the right syntax for that?

Perhaps you can use the function signature `f(**kwargs)` and then you can access and modify kwargs. — jkr, Jan 04 '22 at 15:02
Related to keyword arguments: https://stackoverflow.com/questions/36901/what-does-double-star-asterisk-and-star-asterisk-do-for-parameters — Tzane, Jan 04 '22 at 15:03
Why are you trying to do this? For example, are you trying to add keyword arguments to pass on to another function? — Solomon Ucko, Jan 04 '22 at 15:04
@SolomonUcko some code gets a variable from the context, I'm trying to detangle what part of it gets that variable. I thought it does this by calling `locals()`, but if just `b = 2` solves that then it's some other part. — Velkan, Jan 04 '22 at 15:08

score 0 · Answer 1 · answered Jan 04 '22 at 15:07

0

You have to create your kwargs dict before trying to index it:

def f(a):
    kwargs = locals()
    kwargs["b"] = 2
    print(kwargs)

if __name__ == '__main__':
    f(1)

answered Jan 04 '22 at 15:07

Tzane

score 0 · Answer 2 · answered Jan 04 '22 at 15:14

0

Ok, it's just b = 2.

My actual problem was on the outside with one of the decorators that wanted the function to have one specific argument. Can't do much about that.

answered Jan 04 '22 at 15:14

Velkan

2 Answers2