Calling a Python function with *args,**kwargs and optional / default arguments

Question

In python, I can define a function as follows:

def func(kw1=None,kw2=None,**kwargs):
   ...

In this case, i can call func as:

Answer 1

Clear and concise:

In Python 3.5 or greater:

def foo(a, b=3, *args, **kwargs):
  defaultKwargs = { 'c': 10, 'd': 12 }
  kwargs = { **defaultKwargs, **kwargs }
  
  print(a, b, args, kwargs)
  
  # Do something else

foo(1) # 1 3 () {'c': 10, 'd': 12}
foo(1, d=5) # 1 3 () {'c': 10, 'd': 5}
foo(1, 2, 4, d=5) # 1 2 (4,) {'c': 10, 'd': 5}

Note: you can use In Python 2

kwargs = merge_two_dicts(defaultKwargs, kwargs)

In Python 3.5

kwargs = { **defaultKwargs, **kwargs }

In Python 3.9

kwargs = defaultKwargs | kwargs  # NOTE: 3.9+ ONLY