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

Danilo Livassan
Danilo Livassan
2,199 Points
Posted by Danilo Livassan
Danilo Livassan
2,199 Points

I cant figure out why my code is wrong.

I have tested on my computer and thats return the same output. But not in the same order. {'do': 1, 'like': 1, 'sam': 1, 'i': 2, 'am': 1, 'it': 1, 'not': 1}

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):
    if type(phrase) is not str:
        return None
    phrase = phrase.lower().strip()
    phrase = phrase.replace(',', ' ')
    phrase = phrase.replace('.', ' ')
    dict_words = {}
    phrase = phrase.split(' ')
    for word in phrase[:]:
        if word != ' ' and word != '':
            if word in dict_words:
                dict_words[word] += 1
            else:
                dict_words[word] = 1
    return dict_words

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

You are SO close. The challenge asks you to split on WHITESPACE, instead of a literal SPACE. Use .split() with no arguments to default to whitespace. This will pass the challenge.

Additionally, you do not need to strip() or replace the commas and periods with spaces. The test data does not seem to contain any punctuation or newline characters.

Post back if you need more help. Good luck!

Arturo Barradas
PLUS
Arturo Barradas
Courses Plus Student 7,526 Points
Arturo Barradas
Arturo Barradas
Courses Plus Student 7,526 Points

Hello,

I just cleaned your code and tested it in the system and it has passed. 1) strip() is not needed your are not taken anything like return at the end 2) split will take care of taken each word, you do not need to specify (' ') (blank space) 3) in the for loop it is not needed the split range you are using. It is enough to take just phrase instead of phrase[:] 4) The if following the for is not really needed.