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

Jason Thompson
Jason Thompson
973 Points
Posted by Jason Thompson
Jason Thompson
973 Points

What am I missing?

This code produces the desired results on my Mac just fine. What am I missing?

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.

def word_count(strarg):
    strarg = strarg.lower()
    word_list = strarg.split() 
    dictionary = {}

    for base_word in word_list:
        word_count = 0
        for word in word_list:
            if word == base_word:
                word_count += 1
            dictionary[base_word] = word_count

    print(dictionary)

2 Answers

Jennifer Nordell
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there, Jason! You're not missing much at all. My guess here is that you used print so that you could see the results of your function, but the challenge asks you to return the dictionary... not print it.

Simply changing your last line to return dictionary, causes the challenge to pass! Good job! :sparkles:

Jason Thompson
Jason Thompson
973 Points
Jason Thompson
Jason Thompson
973 Points

I can't believe I missed that. Thanks, Jennifer.