Help on string_factory.py

Hey Tim, This is a tricky one for sure, and I agree deciding how to start can be mildly infuriating at times, haha. Here's how I completed this challenge: First, you need to create the function and set the parameters for the argument. As the instructions say, this function takes a single argument; a list of dictionaries. Name it whatever you want, I keep it simple and straightforward:

# 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(list_of_dicts):

Now that that's out of the way, we move on to the next stage. Here, before you do anything else, notice that the function needs to return a list. Perhaps you're familiar with this technique, but if not it's a very useful and necessary one- create an empty list that you can .append() later. Again, I like simple names:

# 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(list_of_dicts):
    new_list = []

Now that we have the list we can append, we can get to the meat of the function. Because the challenge wants our returned list to contain a string for each dictionary in list_of_dicts, we use a for loop, that loops through list_of_dicts and performs the commanded action the same number of times a dictionary appears in our list of dictionaries.

# 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(list_of_dicts):
    new_list = []
    for dict in list_of_dicts:

Now whatever we tell it to do will only happen once for every dictionary in list_of_dicts. Finally, we can use the template string provided and our new_list we made to both create the desired strings and simultaneously append the list. Afterwords, return new_list. This is where unpacking comes into play:

# 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(list_of_dicts):
    new_list = []
    for dict in list_of_dicts:
        new_list.append(template.format(**dict))
    return new_list

Unpacking simply takes the dictionary we give it and turns the keys and values into variables. So when it unpacks the dictionary {'name' : 'AJ', 'food' : 'enchiladas'}, it turns them into normal python variables (like writing name = 'AJ', or food = 'enchiladas') that can be used with .format(), for all intents and purposes. Unpacking took me a while to understand, so if you're struggling, I suggest taking a little break and coming back to it later. I think I covered most of this code challenge, but if I did something without explaining why, or you have any other questions, please feel free to ask, and I will answer to the best of my ability :)