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Marya Kamali
580 Pointswhat's wrong ? int obj is not iterable?
Why my code doesn't work?
# 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, my_list):
numb=0
for key in my_dict:
for i in len(my_list):
if key == my_list[i]:
numb+=1
2 Answers
Seth Reece
32,867 PointsHi Mary,
In this challenge you want to loop through each item in the list, then check to see if it is a key in the dict. The integer object is not iterable comes from the line "for i in len(my_list):" len(my_list) will return an integer. e.g. 4. you need a list or dict etc to iterate. e.g. 1, 2, 3, 4. Also your function needs to return something. Don't forget to return numb at the end. To help with the challenge:
def members(my_dict, my_list):
numb=0
for i in my_list:
if i in my_dict:
numb+=1
return numb
William Laub
7,793 Pointslen(my_list) is just a single number so there's nothing to loop over. What you need is a list of numbers from 0 to len(my_list). Typically you would use xrange, but I haven't found that to be available here, so try
for i in range(0, len(my_list)):
Chris Freeman
Treehouse Moderator 68,423 Pointsxrange is Python 2. The site is based on Python 3 coding. range behaves as the Python2 xrange
William Laub
7,793 PointsI was not aware of that. Thanks for clarifying.
Marya Kamali
580 PointsMarya Kamali
580 PointsI thought that list are only searchable by their indices ... thanks