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csr13
33,250 PointsI think this works, what do you think?
Idk if this works. It sure outputs the dictionary with the word count on my interpreter, unless there is a comma here and there Idk.
# 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(to_split):
beta = to_split.lower().split(' ')
d = dict.fromkeys(beta, 0)
for _ in beta:
if _ in d.keys():
d[_] += 1
return d
1 Answer
Jason Anders
Treehouse Moderator 145,858 PointsHi carlaos91
That will work if you only want to split on spaces and not all whitespace. But what happens if what's being passed in used, for example, tabs and returns? These are considered whitespace, but not spaces, so in that instance, the split wouldn't work.
You just need to change what is being passed into the split() method to split on all whitespace and not just spaces.
Otherwise... Nice work!! :)
