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Ian Cole
454 PointsDictionary Word_Count Problems
So I don't understand how this is not passing. I run this outside of the challenge and it does exactly what it's supposed to, separates at all the whitespace and lowercases the output.
# 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):
lower_string = string.lower()
split_string = lower_string.split(" ")
word_dict = {}
for word in split_string:
if word in word_dict:
word_dict[word] += 1
else:
word_dict[word] = 1
word_count = word_dict
return word_count
1 Answer
Alex Koumparos
Python Development Techdegree Student 36,887 PointsHi Ian,
The problem is with how you are splitting the string. You are using the argument " " which is fine if you know that the string is going to have single whitespace spacing. If you don't know that, and there is the possibility that there could be double spaces (or some other multiple) or other whitespace characters (e.g., tabs) then you could end up with an array that's not what you expect.
For example, your use of split would turn "hello world" into ["hello", "", "world"] and "hello\tworld" into ["hello\tworld"]. Although the text of the challenge doesn't expressly state that you might get strings of words that are spaced with something other than single space characters, the parser does give a hint when it tells you, "didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!".
To split on all whitespace instead of single spaces, you just use the split method without any arguments. So "hello world".split() and "hello\tworld".split() will both produce ["hello", "world"].
Hope this clears everything up for you.
Cheers
Alex
Ian Cole
454 PointsIan Cole
454 PointsThanks, it worked! I actually didn't know that was how white-space was handled.