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

Enzie Riddle
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Enzie Riddle
Front End Web Development Techdegree Graduate 19,278 Points
Posted by Enzie Riddle
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Enzie Riddle
Front End Web Development Techdegree Graduate 19,278 Points

Dictionaries, Packing, and Unpacking

Here's the question:

Let's test unpacking dictionaries in keyword arguments. You've used the string .format() method before to fill in blank placeholders. If you give the placeholder a name, though, like in template below, you fill it in through keyword arguments to .format(), like this: template.format(name="Kenneth", food="tacos") 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.

What am I doing wrong? I've been following along with the last video, but I am honestly quite confused on packing and unpacking and the what's and why's of when we use them. Thanks!

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 {name} and I love to eat {food}!"

def string_factory(name=None, food=None, second_name=None, second_food=None, **kwargs):
    print(kwargs)

def unpacker(name=None, food=None, second_name=None, second_food=None):
    if name and food and second_name and second_food:
        print("Hi, I'm {} and I love to eat {}!".format(name, food) + " Hi, I'm {} and I love to eat {}!".format(second_name, second_food))
    else:
        print("Name and favorite food please!")
        return unpacker(**{name: "Michelangelo", food: "PIZZA", second_name: "Garfield", second_food: "Lasagna"})

string_factory(name="Michelangelo", food="PIZZA", second_name="Garfield", second_food="Lasagna")
Juan Suarez
Juan Suarez
1,406 Points

I did no look closely to the code but it seems that you are printing instead of returning, the code checker does not like that.

2 Answers

Dario Bahena
Dario Bahena
10,697 Points
Dario Bahena
Dario Bahena
10,697 Points 
# 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!"]

#disclaimer - not sure if this is what is expected but it is the solution I came up with that works. 

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

def string_factory(val):
    tempelateList = []   #create an empty array
    for i in val:    # loop through the list provided like the one in the example of values
        tempelateList.append(template.format(**i))  # **i is the keyword argument {"name": "Michelangelo", "food": "PIZZA"} on the first run
    return tempelateList # returns an array with the correct strings

#string_factory(values)
Kamontat swasdikulavath
Kamontat swasdikulavath
2,447 Points
Kamontat swasdikulavath
Kamontat swasdikulavath
2,447 Points

I don't understand why does the 'i' work? Please explain

Afloarei Andrei
Afloarei Andrei
5,163 Points
Afloarei Andrei
Afloarei Andrei
5,163 Points

I need some help. I've written this peace of coade

values = [{"name": "Andrei", "food": "pizza"}, {"name": "Ioana", "food": "hamburger"}] template = "Hi, I'm {} and I love to eat {}!"

def string_factory(name = None, food = None): if name and food: print(template.format(name, food)) else: print("Something went wrong!")

string_factory(values[0]) string_factory(values[1])

and if I try to add the result to a list I get this TypeError: string_factory() argument after ** must be a mapping, not list I don't know how to solve this one.