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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Attila Kószó
PLUS
Attila Kószó
Courses Plus Student 9,312 Points
Posted by Attila Kószó
Attila Kószó
Courses Plus Student 9,312 Points

The output works, but the workspace doesn't accept it.

I've made the code in pycharm and it seemingly works, appereanly I have no idea what's the problem in the workspace. The example output matches with mine.

wordcount.py 
# 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):
    result = {}
    string = string.lower()
    list_of_keys = string.split(' ')
    for key in list_of_keys:
        if key in result.keys():
            result[key] += 1
        else:
            result[key] = 1
    return result

1 Answer

Taylor Schimek
Taylor Schimek
19,318 Points
Taylor Schimek
Taylor Schimek
19,318 Points

Howdy Attila, You are super close to getting this right. The only thing is you delimiter in the .split() method. You have it set like this: .split(' '). That means that the method is looking for a single white space. If you remove the quotes and make it .split(), then if there happens to be two white spaces together in the string, the split() method will consider them one delimiter.
That should pass the challenge. Hope that helps.