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Michael Hansen
Courses Plus Student 962 PointsMy solutions works for me, but fails the internal test.
def word_count(string): dict = {} my_list = string.lower().split(' ') for word in my_list: if word not in dict: dict[word] = 0 dict[word] += 1 return dict
print(word_count('I do not like it Sam I Am'))
# 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):
dict = {}
my_list = string.lower().split(' ')
for word in my_list:
if word not in dict:
dict[word] = 0
dict[word] += 1
return dict
print(word_count('I do not like it Sam I Am'))
2 Answers
Alexander Davison
65,469 PointsYour code pretty much works fine, but notice that the challenge responded with:
Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!
It says to "split on all whitespace", but it seems that you are only splitting on only spaces! Instead of calling .split(' '), try .split(). (If you provide no arguments to .split(), it will by default split on all whitespace)
Michael Hansen
Courses Plus Student 962 PointsThank you! Worked like a charm.