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qifeng sun
6,274 PointsWhy this is wrong
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.
# 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(values):
# create empty list
new_list = []
# Iterate dictionaries
for val in values:
# for visual reference, unpack dictionaries
name = val['name']
food = val['food']
new_list.append(template.format(name, food))
return new_list
1 Answer
taejooncho
Courses Plus Student 3,923 PointsHi, there.
I am going to say probably because of KeyError... A KeyError generally means the key doesn't exist. So when Python is looking at the dictionary variable: value, and is trying to find a key called name. Which obviously does not exist. There is key "name", but not key name.
A better way to solve this challenge would be using **.
For example, using your code.
def string_factory(values):
new_list = []
for val in values:
new_list.append(template.format(**val))
return new_list