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otarie
984 PointsWhy is my members() function returning 0?
The challenge is as follows: Write a function named members that takes 2 arguments, a dictionary and a list of keys. Return a count of how many items in the list are also keys in the dictionary.
Can someone please tell me what is wrong with my code? thanks!
# 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(listy, dicto):
answer = 0
for key in dicto:
count = 0
while count < len(listy):
if key == listy[count]:
answer +=1
count +=1
return answer
3 Answers
Dan Johnson
40,533 PointsYour code is fine, you just have your positional arguments flipped. The challenge expects the dictionary first and the list second.
Also if you want to avoid having nested loops you can use the in keyword to check the contents of the list:
if key in listy:
answer +=1
otarie
984 PointsThanks for the quick response. Why are my positional arguments flipped?
Dan Johnson
40,533 PointsThey're flipped since the challenge is going to be passing in the dictionary first, and then the list. The last line of the comments shows how it'll be called:
# members(my_dict, my_list) => 2
otarie
984 PointsRight ... just flipped it and it worked. thanks!!!