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

Posted by rob bert
rob bert
5,041 Points

Code that seems to be fine not passing.

So when i run the code on my local machine with i get the return value of: {'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}

which seems to be pretty darn close to what the comment says i should be getting. {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}

yet the task fails.

wordcount.py 
def word_count(string):
    out = {}
    x = string.lower()
    x = x.split(' ')
    for item in x:
        if item not in out:
            out[item] = 1
        else:
            out[item] += 1
    return out

1 Answer

Alex Koumparos
seal-mask
.a{fill-rule:evenodd;}techdegree
Alex Koumparos
Python Development Techdegree Student 36,887 Points
Alex Koumparos
.a{fill-rule:evenodd;}techdegree
Alex Koumparos
Python Development Techdegree Student 36,887 Points

Hi Rob,

The problem is with how you're splitting the string. It's not clear from the challenge description (although it is hinted by the error message) but some of the sample strings that Treehouse will pass into your function have whitespace that isn't just single spaces (maybe double spaces, maybe tabs).

When your function gets passed a string with a double space, e.g., "hello world", your split will create ['hello', '', 'world']. When it gets passed a string with a tab, e.g., "hello\tworld", your split will create ['hello\tworld"].

To split on all whitespace, rather than just single spaces, you use the split method with no arguments.

Cheers

Alex

rob bert
rob bert
5,041 Points

Thanks, that worked. I feel a bit silly that i didn't try that.