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

J.R. Wright
J.R. Wright
18,003 Points
Posted by 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
229,744 Points
Steven Parker
Steven Parker
229,744 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
J.R. Wright
J.R. Wright
18,003 Points

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