Pass keyword arguments as required arguments in Python

Question

I have, for example, 3 functions, with required arguments (some arguments are shared by the 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 next dictionary:

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

Is posible to call the functions in this way??:

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

Yes I have tried and I have TypeError as Daniel says

José Luis
– José Luis

2015-04-19 16:12:46 +00:00
Commented Apr 19, 2015 at 16:12 — José Luis
– José Luis, Commented Apr 19, 2015 at 16:12

Daniel Roseman · Accepted Answer · 2015-04-19 15:45:42Z

4

Not the way you have written those functions, no: they are not expecting the extra arguments so will raise a TypeError.

If you define all the functions as also expecting **kwargs, things will work as you want.

answered Apr 19, 2015 at 15:45

Daniel Roseman

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2 Comments

José Luis Over a year ago

Thanks Daniel, so the pythonic way of write funcions is this (e.g.): def function_two(b,c,e, **kwargs): ... def function_three(f,a,c,d, **kwargs): ... Or this??: def function_two(**kwargs): ... def function_three(**kwargs): ...

zom-pro Over a year ago

the key there is to understand what do you know you will be giving to the function. Let's say I know will give the function a and b and then a bunch of other arguments, you can use def f(a, b, *args) then if I also will use keyword arguments, def f(a, b, *args, **kwargs). (Kwargs are of the type c=1, d=2)

zom-pro · Accepted Answer · 2015-04-19 16:29:21Z

I assume what you're trying to do is to create a function with an undefined number of arguments. You can do this by using args (arguments) or kwargs (key word arguments kind of foo='bar') style so 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}

Let's 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 could do it that way and complement it with kwargs if you don't know what else the function could receive.

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 the level of abstraction of the object you're passing to the function. In your case, you're passing a dictionary so you can handle that guy without only.

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