Using both explicit arguments and **kwargs in python

user_11970084 0
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I have this piece of code:

kwargs_input = {"name": "lala", "age": 25, "postcode": 17867}

def print_kwargs(**kwargs):
    for k,v in kwargs.items():
        print(k,v)

def func_print(name="lolo", age=56, **kwargs):
    print_kwargs(**kwargs)

print(func_print(**kwargs_input))

It prints:

postcode 17867
None

I have two questions:

  1. Why name and age are not printed? I was expecting to override the default value of name (lolo) and age (56) with the arguments passed as **kwargs_input. Is that possible?
  2. Where does None come from?
python
3 Answers
  1. Neither name nor age are printed, because they have been removed from kwargs during the function call. Inside the function call you have access to both name and age through those variables. kwargs will contain the rest of the 'variables'.
  2. The None is returned from func_print() by default and your print(func_print()) prints it out.
quamrana_48340

name and age aren't packed into the first kwargs in func_print() therefore aren't passed further into other functions. Or in other words, it's not collecting all arguments, it's only collecting arguments that are unspecified, "dangling", so that it doesn't throw an error. Similar to Varargs in C.

None comes from the func_print()'s return as it's the default value of any function.

For example:

def func():
    ...  # check help(...) and type(...)
    # implicit "return None"

def func():
    pass
    # implicit "return None"

def func():
    return
    # implicit "return None"

def func():
    # explicit return None
    return None

What you can however do is something that's common in JavaScript nowadays, send an object as an argument. Fortunately, you don't need to create and pass a custom object instance or anything like that because just **kwargs itself is a syntax that creates a dictionary (alternatively *args creates a tuple), so an object's instance preserving the arguments. (Similarly, you can use *args for positional arguments, more on that here).

From that you can do this:

def other(**kwargs):
    print(kwargs)

def func(**kwargs):
    print(kwargs.get("name", "lolo"))
    other(**kwargs)

print(func(name=123, hello="world"))
# 123
# {'name': 123, 'hello': 'world'}
# None
print(func(hello="world"))
# lolo
# {'hello': 'world'}
# None
user_59940410

Key here is that **dict in python unrolls the dict to key value pairs in the function call.

kwargs_input = {"name": "lala", "age": 25, "postcode": 17867}

def print_kwargs(**kwargs):
    for k,v in kwargs.items():
        print(k,v)

def func_print(name="lolo", age=56, **kwargs):
    print_kwargs(**kwargs)

# Equivalent to passing **kwargs_input
print(func_print(name='lala', age=25, postcode=17867))

**kwargs then captures all those 'extra' key word arguments not specified in the function definiton.

As you can see name and age are.

user_159817830
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