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

Peter Tram
PLUS
Peter Tram
Courses Plus Student 6,898 Points
Posted by Peter Tram
Peter Tram
Courses Plus Student 6,898 Points

This code should work. Why do I receive a Bummer message?

d

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()
    string = string.split(" ")
    word_count_dict = {}

    for word in string:
        if word in word_count_dict.keys():
            word_count_dict[word] += 1
        else:
            word_count_dict[word] = 1

    return word_count_dict

2 Answers

Stuart Wright
Stuart Wright
41,118 Points
Stuart Wright
Stuart Wright
41,118 Points

You just need to make one small edit to pass the challenge:

    string = string.split()

The .split() method by default will split on all whitespace characters, which is slightly different from what your code does. For example, your code won't split on tabs or newline characters.