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

Posted by ahmed salah
ahmed salah
9,195 Points

what is wrong with this code ?

def word_count(words_string): count = {} words = words_string.split(" ")

for word in words:
    try:
        count[word.lower()]+=1

    except:
        count[word.lower()]=1

return count
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(words_string):
    count = {}
    words = words_string.split(" ")

    for word in words:
        try:
            count[word.lower()]+=1

        except:
            count[word.lower()]=1

    return count

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,426 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,426 Points

So Close! You need to split on WHITESPACE instead of SPACE. Whitespace is the default split when no arguments are used: split()

ahmed salah
ahmed salah
9,195 Points

i wouldn't ever notice thank you a lot .. i got confused as it seemed working on my PC