Пропустить аргумент функции по имени динамически

Question

Возможно ли динамическое задание имени передаваемого аргумента функции?

так:

def func(one=None, two=None, three=None, four=None): 
    ... 

params = ("one","two","three","four",) 
for var in params: 
    tmp = func(var=value) 

Answer 1

Да, с keyword argument unpacking:

def func(one=None, two=None, three=None, four=None): 
    return (one, two, three, four) 

params = ("one", "two", "three", "four") 
for var in params: 
    tmp = func(**{var: "!!"}) 
    print(tmp) 

Выход:

('!!', None, None, None) 
(None, '!!', None, None) 
(None, None, '!!', None) 
(None, None, None, '!!')