dict packing & unpacking (ref: Python Collections, favorite_food challenge)

First of all kwargs is just a common term for keyword arguments and is just a variable name. It can be any valid variable name, even dict.

By placing ** in front of the dictionary name in a function it breaks out the name/value pairs (hence unpacking).

So, all the challenge only needs 1 small change which is to unpack the passed parameter in the .format() function (almost too easy)

Here's a SO explanation that might be useful - https://stackoverflow.com/questions/1769403/understanding-kwargs-in-python

Hope that helps.

I finally passed this challenge but it felt like through lots of random guesses. Surprised to find out that **kwargs was not involved in the solution, even though it's used for unpacking

Thanks Dave. I resolved this, and found I had an incomplete understanding of packing and unpacking -- my incorrect understanding was that packing and unpacking always involves asterisks (** for dict and * for tuples), but I now know the assignment was asking for unpacking to be done without **. The difficult thing to grasp for me was that the use of "dict[name]" and "dict[food]" is considered unpacking -- I just thought of it as retrieving the value of a dict by passing the dict key through favorite_food(dict). So if that's incorrect, I welcome anyone's input. Alright, again, thank you!

Well, thank you sir. The link is correct, same challenge. I'm glad you replied because I learned a few things. Here is what I'm gathering from this conversation:

1) I can pass this challenge, by referencing rather than by unpacking. This means I can complete it without actually applying the concept from the lesson, because this is how I passed it: def favorite_food(dict): return "Hi, I'm {} and I love to eat {}!".format(dict['name'],dict['food'])

2) I see packing done in the parameters of the first "def" line, but I didn't know unpacking can be done after in line 2, as in the solution you've shown me. because my understanding is that unpacking is done in the function call, and packing is done in the function definition. Please correct me if I'm wrong about that.

I appreciated your hint that the solution was almost too simple. I don't know what to tell you, however. I suppose if I had known what you knew, then I would find it almost too simple as well :-)

Again thanks, now I know how I would have solved it for future reference.:-)

I guess there is something wrong with the link.

This is the code that I ran:

my_dict = {"name" : "Michelangelo", "food" : "PIZZA"}

def favorite_food(name = "None", food = "None"): return "Hi, I'm {} and I love to eat {}!".format(name, food)

favorite_food(**my_dict)

• It ran in the IDE but showed error in the challenge.