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

J.R. Wright
J.R. Wright
18,003 Points

This code works when I run it in my own Python shell...why not here?

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(some_string):
    word_list = some_string.lower().split(' ')
    cdict = {}
    for word in word_list:
        if word in cdict:
            cdict[word] += 1
        else:
            cdict[word] = 1
    return cdict

2 Answers

Steven Parker
Steven Parker
231,275 Points

The challenges test your code with more complex data than shown in sample. The issue is hinted in the message "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", leave the argument to "split" empty. Providing a space causes it to split only on individual spaces.

J.R. Wright
J.R. Wright
18,003 Points

Aha, that does make sense, thanks for the reply!