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

Duarte Reis
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.a{fill-rule:evenodd;}techdegree
Duarte Reis
Python Web Development Techdegree Student 2,395 Points

Getting the right result in Pycharm, but not passing the test

Hi! This code works when I run it in Pycharm and I get the same exact result as the example. But I can't pass the test here.

Not sure of what is wrong.

Thanks a lot for any help in advance!

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(palavra):
    dicio = {}
    palavra_list=(palavra.lower()).split(" ")
    for word in palavra_list:
        if word in dicio:
            continue
        else:
            variavel = word
            times = palavra_list.count(word)
            dicio.update({variavel : times})

    return(dicio)

2 Answers

Steven Parker
Steven Parker
230,248 Points

When you run this, the system gives you some hints: "Bummer: Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!".

The rules of "split" say to split on "all whitespace" the argument should be left empty or set to None.