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

Posted by Cal Mitchell
Cal Mitchell
3,494 Points

Please help me understand why my word_count() function works well in IDLE, but doesn't pass the code 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(string):
    dictionary = {}
    string_lower = string.lower()
    word_list = string_lower.split(" ")
    for word in word_list:
        number = word_list.count(word)
        dictionary.update({word:number})
    return(dictionary)

1 Answer

juliansteffen
juliansteffen
9,802 Points
juliansteffen
juliansteffen
9,802 Points

Dear Cal,

yes your solution is "totally" fine. There is just a little thing,

word_list = string_lower.split(" ")
#should be
word_list = string_lower.split()

#and just a PEP 8 improvement
return(dictionary)
#should be
return dictionary

Hope this helps

Greetings Julian