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

Danilo Livassan
Danilo Livassan
2,199 Points
Posted by Danilo Livassan
Danilo Livassan
2,199 Points

I dont know why my code is wrong

I have tested on my computer and works fine

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):
    if type(phrase) is not str:
        return None
    phrase = phrase.lower().strip()
    phrase = phrase.replace(',', ' ')
    phrase = phrase.replace('.', ' ')
    dict_words = {}
    phrase = phrase.split(' ')
    for word in phrase[:]:
        if word != ' ' and word != '':
            if word in dict_words:
                dict_words[word] += 1
            else:
                dict_words[word] = 1

    return dict_words

1 Answer

Elad Ohana
Elad Ohana
24,456 Points
Elad Ohana
Elad Ohana
24,456 Points

Hi Danilo,

For some reason, the challenge doesn't seem to like that you were explicit with your split method. Since the default is already a space, it doesn't really matter if you add it or not. It seems to work when you use phrase.split() instead of phrase.split(' '). Sometimes the challenges are very sensitive.

Elad