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Atul Jha
3,592 Pointsnot able to go through the task....but code working on workspace!
unable to figure out the error in m code
# 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
output = []
def members(my_dict, my_list):
count = 0
for key in my_dict:
for word in my_list:
if word == key:
count += 1
output.append(word)
else:
continue
print(count)
return(count, output)
2 Answers
Chris Freeman
Treehouse Moderator 68,423 PointsYou want to avoid looping over both the dict and the list as this has len(dict)*len(list) comparisons. If the dict was huge that would lead to many unnecessary comparisons. Instead, loop over the small list and do a simple check for the key:
def members(my_dict, my_list):
count = 0
for word in my_list:
if word in my_dict:
count += 1
return count
Afrid Mondal
6,255 PointsHi Atul, You are doing just some extra work I think. What yu have to do just return the number counts.
def members(my_dict, my_list):
count = 0
for key in my_dict:
for word in my_list:
if word == key:
count += 1
return count
Atul Jha
3,592 PointsThank you!
Atul Jha
3,592 PointsAtul Jha
3,592 PointsThank you!..It makes a lot of sense.!