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Konstantin Kovar
16,003 PointsIs this challenge buggy? I have tried it locally and my solution works just fine.
I don't really have anything to add to my question there xD
# 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(string):
word_list = string.lower().split(' ')
sol = {}
for word in word_list:
if word in sol:
sol[word] += 1
else:
sol[word] = 1
return sol
2 Answers
Jason Anders
Treehouse Moderator 145,860 PointsHey there Konstantin,
You've pretty much right on, except for one thing... what you are splitting on. It isn't really defined in the original instructions, but if you run the code you have now, you get an error saying something about not splitting on "all whitespace." That's the big hint here.
Right now with .split(' ') you are splitting on whitespace, but only spaces. You'll want to include 'tabs,' 'returns,' etc. To accomplish this, you just leave the params passed into the split method empty.... .split().
Keep Coding!
:)
Konstantin Kovar
16,003 PointsAaaaah, yeah, that worked, thanks a million!