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Pass keyword arguments as required arguments in Python

I have, for example, 3 functions with the required arguments (some arguments are separated by functions in different order):

def function_one(a,b,c,d,e,f):
   value = a*b/c ...
   return value

def function_two(b,c,e):
   value = b/e ..
   return value

def function_three(f,a,c,d):
   value = a*f ...
   return value

If I have the following dictionary:

argument_dict = {'a':3,'b':3,'c':23,'d':6,'e':1,'f':8}

Can functions be called this way ?:

value_one = function_one(**argument_dict)
value_two = function_two (**argument_dict)
value_three = function_three (**argument_dict)
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2 answers


Not the way you wrote these functions, no: they don't expect additional arguments, so will raise the TypeError.



If you define all the pending functions as well **kwargs

, everything works the way you want.

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I am assuming you are trying to create a function with undefined number of arguments. You can do this using args (arguments) or kwargs (keyword arguments like foo = 'bar'), for example, for example:

for arguments

def f(*args): print(args)

f(1, 2, 3)
(1, 2, 3)`

then for kwargs

def f2(**kwargs): print(kwargs)

f2(a=1, b=3)
{'a': 1, 'b': 3}

Try a couple more things.



def f(my_dict): print (my_dict['a'])

f(dict(a=1, b=3, c=4))
1

It works!!! so you can do it that way and pad it with kwargs if you don't know what else the function might get.

Of course, you could do:

argument_dict = {'a':1, 'b':3, 'c':4}
f(argument_dict)
1

So you don't have to use kwargs and args all the time. It all depends on the level of abstraction of the object you are passing to the function. In your case, you pass the dictionary so you can handle this guy only.

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