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

Posted by Ben Bruder
Ben Bruder
9,130 Points

Why does this not pass?

I tested this code in a workspace and it seems to work correctly. However, it won't pass in the code challenge.

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):
    array = phrase.lower().split(' ')
    count = {}
    for word in array:
        count[word] = 0
    for word in array:
        count[word] += 1
    return count

1 Answer

Steven Parker
Steven Parker
229,657 Points
Steven Parker
Steven Parker
229,657 Points

The challenge uses a more rigorous test than the sample. But the error message you get contains a hint: " Be sure you're lowercasing the string and splitting on all whitespace!"

To split on "all whitespace" you must leave the argument to "split" empty. By providing a space you're telling it to split only on explicit space characters.