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

Alexander Bilton
Alexander Bilton
1,982 Points
Posted by Alexander Bilton
Alexander Bilton
1,982 Points

I can't see why this function is not accepted?

Hi,

I tried to run this code in workspaces, and it seems to me that it returns what is required. Can someone tell me why this is not accepted in 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(str1):
    word_lst = str1.lower().split(' ')
    word_dict = {}

    for item in word_lst:
        if not item in word_dict:
            word_dict.update({item:word_lst.count(item)})
        else:
            pass
    return word_dict

1 Answer