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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Jared Barr
Jared Barr
1,911 Points

word_count Bummer: Try again!

Hello, The error I am receiving is not very detailed, which hopefully means that I am close to the answer. This code seems to work in my workspace while not in the challenge:

wordcount.py
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.

#split sentence into list of words

#count how many words in list of words


def word_count(sentence):
    word_list = sentence.lower().split()
    count_list = []
    for x in word_list:
        count_list.append(word_list.count(x))
    word_count_dict = dict(zip(word_list, count_list))
#    for key,val in word_count_dict.items():
#        print(key, ":", val)

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,457 Points

You are SO close. The function needs to return a value. Return word_count_dict and you'll pass the challenge!

Jared Barr
Jared Barr
1,911 Points

/)_-)

of course in the next exercise I have been utilizing return...

Thanks!