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Start your free trialdima Dima
901 PointsHow do i do that, I am confused
Write a function named members that takes two arguments, a dictionary and a list of keys. Return a count of how many of the items in the list are also keys in the dictionary.
# 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(dict1,list1):
4 Answers
Matthew Hill
7,799 PointsHi Dima, here's my solution. I think it's easier if you just loop through the items in the list and check for membership in the dictionary.
def members(dict1, list1):
count = 0
for key in list1:
if key in dict1:
count+=1
return count
Let me know if that makes sense or not! Matt
Matthew Hill
7,799 PointsIt always helps to have another watch of the video, or maybe do a search online (sites such as stack overflow are usually very helpful), this way you can have some extra help figuring it out for yourself so you'll learn more than if I just gave you the answer.
As a hint try iterating through your list using:
for item in list:
# Do something here
Hope that helps! Matt
dima Dima
901 PointsI wrote this code but it still incorrect, what can I do?
count=0 index=0 def members(dict1,list1): for item in dict1: if item==list1[index]: count=count+1 index=index+1 else: index=index+1 continue print(count)
dima Dima
901 PointsThank you very much, it was far more simple then what I tried to do