Make kwargs directly accessible

Question

I am refactoring a piece of code, and I have run into the following problem. I have a huge parameter list, which now I want to pass as kwargs. The code is like this:

def f(a, b, c, ...):
  print a
  ...

f(a, b, c, ...)

I am refactoring it to:

data = dict(a='aaa', b='bbb', c='ccc', ...)
f(**data)

Which means I have to do:

def f(**kwargs):
  print kwargs['a']
  ...

But this is a pita. I would like to keep:

def f(**kwargs):
  # Do some magic here to make the kwargs directly accessible
  print a
  ...

Is there any straightforward way of making the arguments in the kwargs dict directly accessible, maybe by using some helper class / library?

Answer 1

There are some ways - but you can also wrap your function like this:

def f(**kwargs):
    arg_order = ['a', 'b', 'c', ...]
    args = [kwargs.get(arg, None) for arg in arg_order]

    def _f(a, b, c, ...):
        # The main code of your function

    return _f(*args)

Sample:

def f(**kwargs):
    arg_order = ['a', 'b', 'c']
    args = [kwargs.get(arg, None) for arg in arg_order]

    def _f(a, b, c):
        print a, b, c

    return _f(*args)

data = dict(a='aaa', b='bbb', c='ccc')
f(**data)

Output:

>>> 
aaa bbb ccc