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Python Python Collections (Retired) Dictionaries Membership

simon bao
simon bao
5,522 Points

Why doesn't this work? If i insert print(count) into the code it shows 2, which is the correct number.

def members():
    my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}
    my_list = ['apples', 'coconuts', 'grapes', 'strawberries']
    count = 0
    for word in my_dict:
        if word in my_list:
            count += 1
    return count
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

def members():
    my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}
    my_list = ['apples', 'coconuts', 'grapes', 'strawberries']
    count = 0
    for word in my_dict:
        if word in my_list:
            count += 1
    return count

1 Answer

Chris Jones
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Chris Jones
Java Web Development Techdegree Graduate 23,933 Points

Hey simon,

You're very close. Because you're writing a function, the dictionary and list will be passed into the function. So, you don't need to define those as my_dict and my_list. Try deleting those two variables and you should pass the code challenge.

def members(d, l):
    count = 0
    for word in d:
        if word in l:
            count += 1
    return count

Let me know if you have any questions or any more issues.