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Start your free trialKourosh G
1,520 PointsIn the workspaces I get the required result, however I get "Bummer! Didn't get all of the results in the output!" msg..
Am I missing something?
# 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):
new_list = [template.format(**values[0]),template.format(**values[1])]
return new_list
1 Answer
Vidhya Sagar
1,568 PointsThe problem with your code is , that you assumed that there are only two dictonary items in the list.I think That was just a blueprint of what the function should do. So go for a for loop, so that no matter how many dictonaries are there in the list ,the compiler will do it explicitly. Try it on your own ,i have also added my code for your reference .
template = "Hi, I'm {name} and I love to eat {food}!"
def string_factory(values):
op=[]
for item in values:
op.append(template.format(**item))
return op