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

Leo Marco Corpuz
Leo Marco Corpuz
18,975 Points
Posted by Leo Marco Corpuz
Leo Marco Corpuz
18,975 Points

word_count challenge update

Here's my code updated. I can't figure out what's wrong.

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(phrase):
    phrase=phrase.lower()
    phrase=phrase.split(" ")
    dict_phrase={}
    for word in phrase:
       if word in dict_phrase:
           continue
       else: 
           value_count= phrase.count(word)
           dict_phrase[word]=value_count
    return dict_phrase

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

The error message contains a hint: ":x:Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To split on "all whitespace" the "split" function should be given no argument. Giving it a space causes it to split on individual spaces and not combine them or consider other whitespace characters.

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