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

Bobby Tagget
Bobby Tagget
16,564 Points
Posted by Bobby Tagget
Bobby Tagget
16,564 Points

Apparently my solution is incorrect

I am not sure what I am doing wrong, I've tested my output through my Mac's terminal and it's showing me exactly what's expected of the 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(some_string):
    some_dict = {}
    string_split = some_string.split(" ")

    for word in string_split:
        some_dict[word.lower()] = 0

    for word in string_split:
        some_dict[word.lower()] = some_dict[word.lower()] + 1

    return some_dict

1 Answer

Jennifer Nordell
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STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there! You are so close here. If you'll take a good look at the Bummer! message you'll notice that it says you must split on all whitespace. Currently, you're only splitting on spaces. My guess is that your data sets that you tested did not include any tabs or new line characters. Which means that the results will not be what is expected. You are using split(" ") to split on spaces. To split on all whitespace you can use the split method with no arguments like so: .split(). This will split the string on all whitespace including tabs and new line characters.

Hope this helps! :sparkles:

Bobby Tagget
Bobby Tagget
16,564 Points
Bobby Tagget
Bobby Tagget
16,564 Points

This helps greatly! Thanks Jennifer