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Joaquin Cardenas
13,822 Pointsneed help with python word_count challenge.
I dont know why my code isn't working i tested it on windows and works fine.
# 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(" ")
r_dict = dict()
for word in word_list:
if word in r_dict.keys():
r_dict[word] += 1
else:
r_dict[word] = 1
return r_dict
1 Answer
Chris Freeman
Treehouse Moderator 68,441 PointsYou are VERY close! The there is a difference between splitting on a literal space (" ") and splitting on whitespace which includes spaces and tabs. And in unicode there are many characters that could be considered whitespace. Splitting on whitespace is the usual convention unless explictly splitting on a literal space is required.
Splitting on whitespace is the default if no argument is given: .split()
Post back if you need more help. Good luck!
Joaquin Cardenas
13,822 PointsIt worked! Thanks :)
Joaquin Cardenas
13,822 PointsJoaquin Cardenas
13,822 PointsSolution:
change line: word_list = string.lower().split(" ") to word_list = string.lower().split()
Credits: Chris Freeman