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Willy Meier
6,926 PointsI tried almost same code on workspace and it works, but I still get keyerror 3 related to my_dict[x].
Code I used on workspace and works: def members(my_dict, my_list): x = 3 z = 0 while x: for item in my_list: if item == my_dict[x]: z += 1 x -= 1 print(z)
my_list = ['apples', 'coconuts', 'grapes', 'strawberries'] my_dict = {1 : 'apples', 2 : 'bananas', 3 : 'coconuts'}
members(my_dict, my_list)
# 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):
x = 3
z = 0
while x:
for item in my_list:
if item == my_dict[x]:
z += 1
x -= 1
return z
2 Answers
Maciej Walczak
5,042 PointsI redid your code into this since i think you dont need a while loop. You just need to go through the list and update answer variable.
def member(my_dict, my_list):
answer = 0
for item in my_list:
if item in my_dict: #you need to check if item is in my_dict not if item is [x] index in my_dict
answer += 1
return answer
Willy Meier
6,926 PointsHi Maciej, Thanks! Good to know that I can compare items in list and in a dictionary directly. Willy
Haider Ali
Python Development Techdegree Graduate 24,728 PointsHaider Ali
Python Development Techdegree Graduate 24,728 PointsHi Maciej, I felt that your comment was a very good solution so I have changed it to an answer. Well done :).