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Leo Marco Corpuz
18,975 Pointsdictionary word_count challenge
I couldn't think of any other way to count the number of key occurrences in a string other than the count method. Is this the right approach? Also, I'm not sure how to modify the for-loop in order to make the keys only appear once in the dictionary.
# 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(phrase):
phrase=phrase.split(" ")
dict_phrase={}
for word in phrase:
word=word.lower()
value_count= phrase.count(word)
dict_phrase[word]=value_count
return dict_phrase
1 Answer
Steven Parker
229,708 PointsThe error message contains some hints: "
It's suggesting that you convert the entire string to lowercase before you split it. Otherwise, when you apply the "count" method to "phrase", the word might not match.
And to split on "all whitespace" the "split" function should be given no argument. Giving it a space causes it to split on individual spaces and not combine them or consider other whitespace characters.
Happy coding! 
