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

uday kiran
uday kiran
3,222 Points
Posted by uday kiran
uday kiran
3,222 Points

My code is not working

def word_count (x):

l=x.lower().split(" ")
dic={}

for i in l:
    try:
        if dic[i]:
            dic[i] +=1
    except KeyError:
        dic[i]=1

return dic
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 (x):

    l=x.lower().split(" ")
    dic={}

    for i in l:
        try:
            if dic[i]:
                dic[i] +=1
        except KeyError:
            dic[i]=1

    return dic

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points

So Close! split() on all whitespace (with no arguments) instead of splitting on a literal SPACE, will pass the challenge!