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Python Python Collections (Retired) Dictionaries Membership

Willy Meier
Willy Meier
6,926 Points
Posted by Willy Meier
Willy Meier
6,926 Points

I 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)

counts.py 
# 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
Maciej Walczak
5,042 Points
Maciej Walczak
Maciej Walczak
5,042 Points

I 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
Willy Meier
6,926 Points
Willy Meier
Willy Meier
6,926 Points

Hi Maciej, Thanks! Good to know that I can compare items in list and in a dictionary directly. Willy