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

William Warren
William Warren
2,519 Points
Posted by William Warren
William Warren
2,519 Points

Not sure how to use ** in string_factory

I'm not sure exactly how to use ** in this exercise, or if its possible to loop through kwargs at all.

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!"]
template = "Hi, I'm {} and I love to eat {}!"

def string_factory(name=None, food=None, **kwargs):
    string_list = []
    for item in kwargs:
        if name and food:
            string_list.append(template.format(name, food))
    return string_list

2 Answers

leonardbode
leonardbode
Courses Plus Student 4,011 Points

Hello William,

Your template's placeholders need to look like this, I assume you changed them:

template = "Hi, I'm {name} and I love to eat {food}!"

The reason for this is that Python does not know which key: value to assign to which placeholder because dictionaries are unordered.

Here is my code:

template = "Hi, I'm {name} and I love to eat {food}!"

def string_factory(list_of_dicts):
    return [template.format(**dictionary) for dictionary in list_of_dicts]

If you're not comfortable with list comprehension, here is a simplification of the code:

template = "Hi, I'm {name} and I love to eat {food}!"

def string_factory(list_of_dicts):
    li = []
    for dictionary in list_of_dicts:
        li.append(template.format(**dictionary))
    return li

If you have any other questions I will update my answer, if you do not have any other questions:

Remember to upvote and to choose the best answer so that your question receives a checkmark in forums.

Kind regards,

Leo

Brandon Bell
Brandon Bell
1,107 Points

Leonard Bode - The instructions literally tell you to use it.

"Write a function named string_factory that accepts a list of dictionaries as an argument. Return a new list of strings made by using ** for each dictionary in the list and the template string provided."

leonardbode
leonardbode
Courses Plus Student 4,011 Points

Brandon Bell I edited my answer.