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

Nicolas Baranowski
Nicolas Baranowski
4,813 Points

word_count bug?

for the word count challenge (https://teamtreehouse.com/library/python-collections-2/dictionaries/word-count) can anyone find why my code is not accepted?

Seems like the console is giving me the right results..

Thanks

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):

    dic = {}
    phraseL = ""
    word_list_clean = []

    for char in phrase:
        charL = char.lower()
        phraseL += charL

    #print(phraseL)

    word_list = phraseL.split(" ")

    #print(word_list)

    for word in word_list:
        if word != "":
            word_list_clean.append(word)
        else:
            continue

    #print(word_list_clean)

    for word in word_list_clean:
        if word not in dic:
            dic[word] = 1
        else:
            dic[word] += 1

    #print(dic)

    return dic

2 Answers

Steven Parker
Steven Parker
231,007 Points

The error message gives you a hint: "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" the argument of "split" should be left empty. Specifying a space causes it to split only on individual spaces.

Nicolas Baranowski
Nicolas Baranowski
4,813 Points

Thank you Steven.

I will try this. However I thought that this function:

for word in word_list: if word != "": word_list_clean.append(word) else: continue

would allow to compensate for this by cleaning word_list into a word_list_clean

Steven Parker
Steven Parker
231,007 Points

That would only remove empty strings from the list, it still wouldn't split on "white space" characters other than literal spaces.

Happy coding!