How to fill in optional function arguments automatically with corresponding dictionary values?

Question

Let's say I have a function in Python:

def foo(x,a=".",b=2):
    ...

and I have a dictionary:

param_dict = {'a':'!', 'b':5}

How would I be able to communicate the values in param_dict to foo, matching the corresponding dictionary key to the argument name, and do so in an automatic way? (i.e., not manually writing out foo(x,a=param_dict['a'],b=param_dict['b']))

Also would like it so that if the corresponding key isn't found in param_dict, the default value of the argument is kept.

What about passing the entire dictionary to `foo`! – Elmex80s Feb 13 '19 at 23:26 — Elmex80s, Feb 13 '19 at 23:26

score 1 · Answer 1 · answered Feb 13 '19 at 23:27

1

foo(x, **param_dict) Should work

answered Feb 13 '19 at 23:27

Benoît P

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If the dictionary had keys that didn't match any of the arguments in the function, would the function just ignore them or would there be an error? – NeutronStar Feb 14 '19 at 03:10
An error, to avoid it: `def foo(x,a=".",b=2, **rest):` – Benoît P Feb 14 '19 at 15:20

How to fill in optional function arguments automatically with corresponding dictionary values?

1 Answers1