For the first challenge, I do not understand where to even start.

Finally did it.

some_dict = {'a': 1, 'b': 2, 'c': 3}

key_list = ['a', 'b', 'd']

def members(some_dict, key_list):

  counts = 0

  for item in key_list:

    if item in some_dict.keys():

      counts += 1

  return counts

Thank you for your help.

I added some more comments to the starter file, so hopefully we'll have less confusion on this in the future. Thanks for the input , everyone!

So I'm going to give a dictionary like:

some_dict = {'a': 1, 'b': 2, 'c': 3}`

And a list like:

key_list = ['a', 'b', 'd']

Tell me how many of the keys in key_list are in the some_dict dictionary.

Awesome, Kenneth! That's why you guys rock. All of the Treehouse teachers really make it known that they care about student success by being open and responsive to student feedback. Keep up the awesome sauce, now I'm back to your Python course (which is very well designed, btw)...

Hi David,

Try to break the problem down into the parts that you can do.

• Write a function named members that takes two arguments, a dictionary and a list of keys.
• Return a count of how many of the keys are in the dictionary.

It's asking how many of the keys that are in list, can also be found in the dictionary?

Try writing the code you know, and post it here inside backticks with the python language specified for proper syntax highlighting (reference the Markdown Cheatsheet below).

You did Python Basics, right? So you have to write a function. members = (some_dict, key_list) isn't a function, it's putting a tuple into a variable. Start with writing the function.

def members(some_dict, key_list):

OK, so now what? How would you hold onto a count of the number of keys that are good? How would you loop through the keys in the list? How would you check to see if they key is in the dictionary? Remember what keyword sends information back out of a function?

guys , i just found a bug in this example.

my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3} my_list = ['apples', 'coconuts', 'grapes', 'strawberries']

def members(my_dict, my_list) : return 2

! Congrats! You completed the challenge!

i ended up here to help understand the prompt for this challenge, and now that i have working code, i'm unable to get it to be accepted in the challenge prompt.

this code below runs fine in my workspace and returns 2 as the count (which is correct), but when i paste it into the code challenge i get the error "word_count() missing 1 required positional argument: 'key_list'"

my_dict = {'i': 1, 'am': 2, 'that': 3}
key_list = ['i', 'am', 'what']

def word_count(my_dict, key_list):
  count = 0
  for word in key_list:
    if word in my_dict.keys():
      count += 1
  print(count)
  return count

word_count(my_dict, key_list)

does anyone know why this would be? it doesn't make sense to me...

I can help get you started without spoiling too much.

• Write a function named members that takes two arguments, a dictionary and a list of keys

def members(some_dict, key_list):
  # your function logic goes here  

So I got...

some_dict = {'a': 1, 'b': 2, 'c': 3}

key_list = ['a', 'b', 'd']

members = (some_dict, key_list)

but I don't understand how to return a count of how many of the keys are in the dictionary.

Ya i was kinda confused on the wording of the question too. As soon as i saw what Kenneth put for some_dict and key_list it cleared it up.

Is it not possible to use the len method in this instance?

def members(my_dict, my_list):
  return len(my_dict.keys())

Getting "Expected 2, got 6." but would be a much more elegant solution (if it worked)!

I am wondering what other's opinions are on using list comprehensions as a solution? I came up with the following:

my_dict = {'a': 1, 'b': 2, 'c': 3}
my_list = ['a', 'b', 'd', 'e']

def members(my_dict, my_list):
  return len([dictkeys for dictkeys in my_dict.keys() if dictkeys in my_list])

Hello all, thanks for the questions in this forum. You comments and suggestions got me where I needed to be. To be honest, I was confused on the wording of the question: Write a function named members that takes two arguments, a dictionary and a list of keys. Return a count of how many of the keys are in the dictionary.

The noun "the keys" in the second sentence could refer to either the keys in the dictionary or the keys in my_list. Kenneth, the extra starter text is a bit helpful, but you might just consider revising the initial prompt to read something like this: Return a count of how many of the keys in my_list are ALSO in the dictionary.

That way it is clear that we are not just looping over my_dict to count those keys, but rather checking the keys in my_dict against the keys in my_list. Sorry for the nit-picky suggestion, but every once in awhile the English teacher in me makes an appearance. Thanks for the good stuff Treehouse team! Keep it up and happy coding! :)

I did it but, could you write - one more time - the easiest way to solve this problem, i can't say, that I understand it in 100 %

I'figured out another way of solving this problem, without using the keys()method:

def members(my_dict, my_keys):
    count = 0
    for item in my_keys:
        for key in my_dict:
        if key == item:
           count += 1
    return count 

Hopefully it is pythonically correct :-)

2

def members(some_dict, key_list):

  counts = 0

  for item in key_list:

    if item in some_dict.keys():

      counts += 1

  return counts

yaaah this one works for me

https://wiki.python.org/moin/DictionaryKeys

in addition to the explanations it may help to read over keys