get variables by **kwargs but throws error about positional arguments

Question

Try to pass the dictionary into the function to print them out, but it throws error: most_courses() takes 0 positional arguments but 1 was given

def most_courses(**diction):
    for key, value in diction.items():
        print("{} {}".format(key,value))

most_courses({'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],'Kenneth Love': ['Python Basics', 'Python Collections']})

I have used **kwargs but why cant python unpack the dictionary?

Answer 1

arguments denoted with a ** in the definition of a function need to be passed with a keyword:

example:

def test(**diction):
    print(diction)

Argument passed without keyword:

test(8)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-9-5092d794a50d> in <module>
      2     print(diction)
      3 
----> 4 test(8)
      5 test(test_arg=9)

TypeError: test() takes 0 positional arguments but 1 was given

With arbitrary keyword:

test(test_arg=8)

output:

{'test_arg': 8}

edit:

helpful links:

Use of *args and **kwargs

What does ** (double star/asterisk) and * (star/asterisk) do for parameters?

Answer 2

When you pass your dict as a param, you can either do it as you wrote:

most_courses({'Andrew Chalkley':  ... 

in this case most_cources should accept a "positional" param. That's why it raises: most_courses() takes 0 positional arguments but 1 was given.

You gave it 1 positional param, while most_cources (which looks like: most_courses(**d)) isn't expecting any..

You should either do:

most_courses(**{'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],'Kenneth Love': ['Python Basics', 'Python Collections']})

OR change the signiture of your method:

def most_courses(diction):
    for key, value in diction.items():
        print("{} {}".format(key,value))

Answer 3

There is no reason to use ** here. You want to pass a dict and have it processed as a dict. Just use a standard argument.

def most_courses(diction):