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

Christopher Porter
Christopher Porter
2,148 Points
Posted by Christopher Porter
Christopher Porter
2,148 Points

My function seems correct in work space but it won't pass me on the challenge.

The challenge error: "Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

I have run my function in the workspace with multiple different sentences, which all return expected results.

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(sentence):
    sentence = sentence.lower()
    split_sentence = sentence.split(" ")
    word_count_dict = {}
    for word in split_sentence:
        word_count_dict[word] = split_sentence.count(word)
    return word_count_dict

1 Answer

Alex Koumparos
seal-mask
.a{fill-rule:evenodd;}techdegree
Alex Koumparos
Python Development Techdegree Student 36,887 Points
Alex Koumparos
.a{fill-rule:evenodd;}techdegree
Alex Koumparos
Python Development Techdegree Student 36,887 Points

Hi Christopher,

Nice work trying to debug your code using different sentences. However, it looks like all the sentences you tried just had single spaces. Let's look again at the error message you're getting, and the very subtle clue that it's giving you about where you're going wrong:

Be sure you're ... splitting on all whitespace!

In order to properly test your code, you'll need to try some sentences that have different configurations of whitespace than the example provided. Consider testing with strings like:

"hello  world"

(two consecutive spaces)

"hello\tworld"

(tab character (considered whitespace))

"hello\nworld"

(newline character (considered whitespace))

Then check out the docs for str.split() and see if there's an alternative argument for sep that might give you the result you want.

Cheers

Alex