It returns values they aren't expecting... yes. Because you're appending your own information to theirs. Remember during challenges: try not to add variables and information they aren't looking for. Your output can be something they don't expect. Take a look at my solution.

def loopy(items):
    # Code goes here
    for item in items:
      if(item == 'STOP'):
        break
      else:
        print(item)

Hi Jennifer,

Again, thank you very much!

Hi Jennifer,

Can you please help me with the following (from someone else I get hints, but not the solution and I want to go on -> I think I am 99% there... :-) ).

See please my solution below.

It returns 0 in stead of 2.

It has maybe to do with the fact that a list-item is not the same datatype as the key of a dictionary...(?)

counts.py

You can check for dictionary membership using the

"key in dict" syntax from lists.

Example

my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}

my_list = ['apples', 'coconuts', 'grapes', 'strawberries']

members(my_dict, my_list) => 2

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

def members(dictionary, mylist): count = 0 for item in mylist: try: if dictionary[item] in [1,2,3]: count += 1 except: break

print(count) # Isn't necessary, but helps me with finding the correct solution. return(count)

members(my_dict, my_list) # Isn't necessary, but helps me with finding the correct solution.

If I delete the print and the memberscall, the solution is still not correct...

Strange thing is, that when I do this in a seperate workspace it does return 2. Strange...

I found the challenge you're working on! I'll show you what I did then walk you through what we're doing here.

def members(myList, myDict):
  count = 0;
  for item in myList:
    if item in myDict:
      count += 1
  return count

Ok so the nice fellows/gals over at Treehouse are going to send your members function some info and you need to return the count of the items where the item in the list is also a key in the dictionary. So we start by making a count variabe (you can name this anything) and setting it to 0. Then we look at every item in myList. Now, if that item is in myDict then we increment the count variable. At the very end when it's all said and done, we return the final count. Hope this clears things up!

Hi Jennifer, thanks again!

I made it harder then it should. But the strange thing of your solution is that item in myDict is apprendly good enough, although myDict excists of two parts (and I thought you had to be specific in naming the correct one)...