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

Chaltu Oli
Chaltu Oli
4,915 Points
Posted by Chaltu Oli
Chaltu Oli
4,915 Points

I think my code should work

It works on my workspace but it keeps saying ": Hmm, didn't get the expected output. Be sure you're lowercasing the stri" on the code challenge, I don't know why?

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(string):
    dictionary = {}
    string = string.lower()
    for total in string.lower().split():
        count = 1

        if total in dictionary:
            count += 1
            dictionary[total] = count
        else:
            count = 1
            dictionary[total] = count
    return dictionary

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 Ulfina,

You are making a logic error with your count variable.

Consider the test string ("test test test"), what do you expect to see as your output and what do you actually get?

Cheers

Alex