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

Posted by Dylan Martin
Dylan Martin
10,399 Points

My solution works in visual studio but not here. What's the problem?

I get 'Bummer!'

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(words):
    words = words.split(" ")
    mydict = {}
    for word in words:
        word = word.lower()
        try:
            mydict[word] += 1
        except:
            mydict[word] = 1

    return mydict

2 Answers

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

You're probably not testing with the same input the challenge uses (it does not only use the sample in comments).

But the error message contained a clue: "Bummer! Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To cause "split" to use "all whitespace", leave the argument empty. Supplying a space tells it to split only on spaces.

Dylan Martin
Dylan Martin
10,399 Points
Dylan Martin
Dylan Martin
10,399 Points

Thanks. split(" ") was the problem. I don't remember whitespace being covered in this course.