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

Posted by Michal Krasnodebski
Michal Krasnodebski
7,224 Points

This works

this code works when tested with the example. Why am I failing the task?

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(s):
    s = s.lower()
    l = s.split(' ')
    d = {}
    for w in l:
        if w in d:
            d[w] +=1
        else:
            d[w] = 1
    return d
Jaxon Gonzales
Jaxon Gonzales
3,562 Points

I am having this same issue. Here is my code:

def word_count(string):
    dict_to_return = {}
    list_of_words = list(string.lower().split(" "))

    for word in list_of_words:
        if word in dict_to_return.keys():
            dict_to_return[word] += 1
        else:
            dict_to_return[word] = 1

    return dict_to_return

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,460 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,460 Points

Be sure to split on all whitespace and not just a literal SPACE.

:point_right: call split with no argument to default to all whitespace