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Ravi Kiran
2,196 PointsSolution is working in console. However its not getting accepted
Create a dictionary with word count
# 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):
pointer = 0
count_dict = {}
string_list = string.lower().split(" ")
for pointer in range(len(string_list)):
if string_list[pointer] in count_dict:
count_dict[string_list[pointer]] += 1
else:
count_dict[string_list[pointer]] = 1
return count_dict
2 Answers
Jennifer Nordell
Treehouse TeacherHi there, Ravi Kiran ! You're doing terrific and you are so close here! The problem isn't as much with the code as it is with your test data. The challenge asks you to split on all whitepsace. This includes things like tabs and newline characters. I would bet that you haven't tried any data that contains tabs and newline characters. Currently you are splitting only on spaces.
This splits on spaces:
.split(" ")
But this splits on all whitespace:
.split()
Simply removing the quotation marks from within the split() method causes this to pass. Good job!
Hope this helps!
Ravi Kiran
2,196 PointsThanks, I learnt a lot with just this problem and also from your comment.