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Python Python Collections (2016, retired 2019) Dictionaries String Formatting with Dictionaries

Ong Jia Rui
Ong Jia Rui
1,731 Points
Posted by Ong Jia Rui
Ong Jia Rui
1,731 Points

string_factory() takes 0 positional arguments but 1 was given?

What does this mean? string_factory() takes 0 positional arguments but 1 was given

string_factory.py 
# Example:
# values = [{"name": "Michelangelo", "food": "PIZZA"}, {"name": "Garfield", "food": "lasagna"}]
# string_factory(values)
# ["Hi, I'm Michelangelo and I love to eat PIZZA!", "Hi, I'm Garfield and I love to eat lasagna!"]
def string_factory(**values):
    if name and food:
        template = "Hi, I'm {} and I love to eat {}!".format(name,food)
    else:
        print("Hi no name!")
    return template

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

You haven't specified the argument as the challenge expects.

The challenge says your function should accept a "list of dictionaries as an argument", but you defined your function to take a variable-sized keyworded list using the packing operator (**).

Here's a few more hints that may help:

  • you accept an argument named values, but it is never used inside the function
  • to handle a list of items you will need some kind of loop
  • you will need to construct a dictionary instead of a string as the return value
  • your function does not need to print anything