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

Ken LaRose
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Ken LaRose
Python Web Development Techdegree Student 21,982 Points
Posted by Ken LaRose
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Ken LaRose
Python Web Development Techdegree Student 21,982 Points

Python word_count function isn't passing - what's wrong?

I wrote this word_count function to accept a string and return a dictionary filled with the lowercase version of each word and the integer count of each word in the original string. I can't pass the code challenge, but every string I send through this function comes out right... I think... Can anybody see what's wrong with this solution?

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(arg):
    words = arg.lower().split(' ')
    count = {}
    for word in words:
        if word in count:
            count[word] += 1
        else:
            count[word] = 1
    return count

1 Answer

eck
eck
43,038 Points

I think this line might be your problem: arg.lower().split(' '), which should probably just be arg.lower().split().

The reason for this is that splitting a string on spaces would treat a double space as a word, whereas calling .split without an argument will ensure empty spaces are all trimmed.

eck
eck
43,038 Points

Yeah, that was pretty subtle; glad it worked :D