CODEWITHSUNDEEP
pythonfunctiondictionaryparameterskeyword-argument
Elliptica
Elliptica

Reputation: 4322

Can I Skip Parameters in Python When Using Kwargs

I have a function that takes in kwargs. I want to use the kwargs parameters by name without having to declare each explicitly as either a parameter or variable. Is there a way to use my_var_key by passing in kwargs without having to specifically define it in the function call, or is the only way to use kwargs[ "my_var_key" ]?

E.g. I want something like

def func(**kwargs):
    print(my_var_key)

as opposed to

def func(my_var_key, **kwargs):
    print(my_var_key)

or 

def func(**kwargs):
    print(kwargs[ "my_var_key" ])

I'm okay with it breaking if the key doesn't exist.

Upvotes: 4

Views: 2474

Answers (2)

adrtam
adrtam

Reputation: 7211

Don't do this. It makes your code harder to read.

But if you insist, try modify your globals() scope:

def func(**kwargs):
    for key, val in kwargs.items():
        globals()[key] = val
    print(globals())
    print(my_var_key)

func(my_var_key='foobar')

and the output is:

{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <_frozen_importlib_external.SourceFileLoader object at 0x10f058f28>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>, '__file__': 'test.py', '__cached__': None, 'func': <function func at 0x10ef52268>, 'my_var_key': 'foobar'}
foobar

This will surely pollute your global namespace (and you cannot use locals() because the interpreter confuses). Again, don't do this.

Upvotes: 0

Blckknght
Blckknght

Reputation: 104712

No, there's no practical way to get the syntax you want.

One part of Python's design philosophy is that "explicit is better than implicit", so there are not many situations where names will get added to your namespace without your input. If you want my_var_key to be put into your function's namespace when it's passed as an argument, you should just name it in the def statement. If you want it to only be passable as a keyword argument (not a positional one), you can put a * in the argument list:

def func(*, my_var_key): # arg can only be passed as a keyword: func(my_var_key=1)
   ...

Upvotes: 2

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