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

Michal Krasnodebski
Michal Krasnodebski
7,224 Points
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
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 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 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