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Brian Hartley
3,115 PointsLoop returning 3x the number that it should
Something is strange with this challenge. The 'for key in dict' loop is seeing all items in a dictionary as keys. For example, the following functions return 9 in the Challenge workspace, which is incorrect. On my local machine (using Python 2 and Python 3), they return the 3, which is correct, as there are 3 keys in the provided dictionary being passed to the function.
c = {'a': 'val1', 'b': 'val2', 'c': 'val3'}
# should return 3
def members(someDic):
count = 0
for key in someDic:
count +=1
return count
# also should return 3
def members(someDic):
count = 0
for key, value in someDic.items():
count +=1
return count
My workaround was to return:
int(count / 3)
in the Challenge workspace.
3 Answers
Robert Richey
Courses Plus Student 16,352 PointsHi Brian,
That is strange! I can, however, point out that the function members "takes two arguments". Here is how I tend to solve this challenge:
def members(my_dict, my_list):
count = 0
# for each key in my_list:
# if key in my_dict:
# increment count
return count
Brian Hartley
3,115 PointsRobert, you're right, and thanks for the reply. I should have noted that the code I posted was an example of the strange loop behavior, taken from my troubleshooting efforts on my local machine, and not directly from the challenge. I hope I didn't further confuse anyone who has come across this post while dealing with the same issue. The workaround I posted completes the challenge either way.
However, I will also try your if statement tomorrow as well. Perhaps I am going about the loop incorrectly.
Kenneth Love
Treehouse Guest Teacherfor key in my_dict is always going to only see the keys, that's how looping through a dictionary works.