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

Tyler Shand
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Tyler Shand
Python Web Development Techdegree Student 15,043 Points
Posted by Tyler Shand
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Tyler Shand
Python Web Development Techdegree Student 15,043 Points

Code Works But Not Passing Challenge

When I run this code on my computer it runs as specified by the instructions but when I run it on Treehouse it fails. What am I doing 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(string):
    string = string.lower()
    word_counter = {}

    # Split 'string' into a list
    string_list = string.split(' ')

    # Give all words keys in 'word_count'
    for item in string_list:
        word_counter[item] = 0

    # Assign the number of times a word appears to it's dictionary key value
    for item in string_list:
        word_counter[item] += 1


    return word_counter

1 Answer

Chris Howell
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Chris Howell
Python Web Development Techdegree Graduate 49,702 Points
Chris Howell
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Chris Howell
Python Web Development Techdegree Graduate 49,702 Points

Hi Tyler Shand

So I ran your code, it looks like you are getting this error message:

Bummer! Hmm, didn't get the expected output. Be sure you're not splitting only on spaces!

this is inferring that in your code

def word_count(string):
    string = string.lower()
    word_counter = {}

    # Split 'string' into a list
    string_list = string.split(' ') # Problem here.

    # Give all words keys in 'word_count'
    for item in string_list:
        word_counter[item] = 0

    # Assign the number of times a word appears to it's dictionary key value
    for item in string_list:
        word_counter[item] += 1


    return word_counter

You are capturing spaces only, but need to capture all the different types of whitespace ("spaces, tabs, breaks, etc")

Python Docs: Split