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Start your free trialAryan Monga
4,061 PointsA quiz 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. I tried once but couldn't help it!
# 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":"Aryan", "food": "Tacos"}):
my_list=[]
my_list=template.format(["name"] , ["food"])
return my_list
1 Answer
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 PointsIn the values they give as an example, it is a list of dictionaries. It is a single item to import.
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
Aryan Monga
4,061 PointsAryan Monga
4,061 PointsThanks a ton!
ranga rao rayapati
443 Pointsranga rao rayapati
443 Pointsdef string_factory(values): # create empty list new_list = [] # Iterate dictionaries for val in values: # for visual reference, unpack dictionaries nameD = val['name'] foodD = val['food'] new_list.append(template.format(name = nameD, food = foodD)) return new_list