string_factory.py please help!!

Question

hy guys its me again. ive been stuck on this problem for a few days now. heres what ive tried and ive commented out the logic that ive used. can someone tell me what im doing wrong.

heres is their 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.

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(my_list_of_dicts):
    newlist = [] #this will create my new list
    for  dicts in my_list_of_dicts: #this will take each dict in the list of dicts
        names = dicts[name] #this will set the variable name from a name in a dict
        foods = dicts[food] #this will set the variable food from a food in a dict
        newlist.append(template.format(name = names, food = foods)) #this will append the new string into newlist
        return newlistlast #this will return the new list i have made

Answer 1 · 2017-03-08T19:23:35Z

March 8, 2017 7:23pm

You have a couple of errors in the return statement:

it is indented too far — as it is, it returns during the first pass through the loop
you return an undefined variable named newlistlast, but the list you create is named newlist

And the names of the keys should be in quotes, for example: dicts["name"]. But while that will fix it, I believe the challenge is really intending for you to make use of the ** unpacking operator. In that case, you would not need the individual variables and could just do this:

        newlist.append(template.format(**dicts))

Answer 2 · 2017-03-08T19:33:17Z

March 8, 2017 7:33pm

I see three errors:

the dict keys "name" and "food" should be in quotes
typo in return variable name. Should be newlist
return is indebted to far. It is currently inside the for and will return at the end of the first iteration

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Royce Sithole

Royce Sithole

string_factory.py please help!!

2 Answers

Steven Parker

Steven Parker

You have a couple of errors in the return statement:

Chris Freeman

Chris Freeman