How to build arguments for a python function in a variable

Question

I have a Python (3) function

def f(a, b, c=1, d=2):
    pass

I would like to build up a list of arguments in code. The actual use-case is to do this based on some command line argument parsing, but it's a general question (I'm fairly new to python) e.g.

myargs = {}
myargs['positionalParams'] = ['foo', 'bar']
myargs['c']=2
myargs['d']=3
f(myargs)

I know that I can do f(*somelist) to expand a list into separate args, but how to do that with a mix of positional and optional params? Or do I do it twice, once for positionals and once for a dict of others?

You'd use both *args and **kwargs

Cory Kramer
– Cory Kramer

2015-03-11 12:37:15 +00:00
Commented Mar 11, 2015 at 12:37 — Cory Kramer
– Cory Kramer, Commented Mar 11, 2015 at 12:37

deceze · Accepted Answer · 2015-03-11 12:40:12Z

5

args = ['foo', 'bar']
kwargs = {'c': 2, 'd': 3}
f(*args, **kwargs)

Note that this will also do just fine:

kwargs = {'a': 'foo', 'b': 'bar', 'c': 2, 'd': 3}
f(**kwargs)

answered Mar 11, 2015 at 12:40

deceze♦

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1 Comment

Wayne Over a year ago

Ned Deily discusses a style that preserves the keyword=value style here in a similar topic.

thefourtheye · Accepted Answer · 2015-03-11 12:43:46Z

2

how to do that with a mix of positional and optional params?

You can pass values to positional arguments also with a dictionary by unpacking it like this

>>> def func(a, b, c=1, d=2):
...     print(a, b, c, d)
...     
... 
>>> myargs = {"a": 4, "b": 3, "c": 2, "d": 1}
>>> func(**myargs)
4 3 2 1

Or do I do it twice, once for positionals and once for a dict of others?

Yes. You can unpack the values for the positional arguments also, like this

>>> pos_args = 5, 6
>>> keyword_args = {"c": 7, "d": 8}
>>> func(*pos_args, **keyword_args)
5 6 7 8

Or you might choose to pass the values explicitly, like this

>>> func(9, 10, **keyword_args)
9 10 7 8

edited Mar 11, 2015 at 12:43

answered Mar 11, 2015 at 12:38

thefourtheye

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