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

wordcount

I have tested my code and everything seems to work fine, I get back a dictionary with the words in lower case with a key value of the correct number of times the word appears in the sentence but the test still says it's incorrect.

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(input):
    input_split = list(input.split(" "))
    results = {}
    for x in range(len(input_split)):
        count = 0
        for y in range(len(input_split)):
            if input_split[x].lower() == input_split[y].lower():
                count += 1
        results[input_split[x].lower()] = count
    return results

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 67,989 Points

You are very close! The challenge is looking for the input to be split on whitespace instead of a literal space. To split on whitespace is the default for split()

lol, yep. I removed the " " from split and it worked just fine. Was driving me nuts I tested like a 100 times. Thanks :)