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

Seph Cordovano
Seph Cordovano
17,400 Points
Posted by Seph Cordovano
Seph Cordovano
17,400 Points

Python wordcount problem producing incorrect result?

I've gone over this quite a number of times, and pretty sure every angle is being covered as well as getting correct result in console. I know the return declaration isn't very pythonic but it's just for ensuring everything's correct for this specific problem.

The error is saying to make sure I'm lowercasing and splitting the words. Not only is that happening but I'm ensuring that it is a word and not a blank space with the if word: conditional.

I've emailed support and they assure me this isn't a bug but I'm not seeing that so I'm turning to you guys to help me see if that's true.

Meme: 'this is a bug, prove me wrong' haha

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(words_string) -> dict:
    words_dict = {}
    words_list = words_string.lower().split(" ")

    for word in words_list:
        if word:
            words_dict[word.strip()] = words_list.count(word)

    return words_dict

1 Answer

andren
andren
28,558 Points
andren
andren
28,558 Points

The error message you get asks you to make sure you are splitting on all whitespace. You are splitting on a space, which is a whitespace character, but not the only one. Tabs, line breaks, and various other characters also count as whitespace.

Conveniently enough the split method will actually split on all whitespace by default if you do not pass it any arguments. So if you take your current code and simply remove the argument you pass to split like this:

def word_count(words_string) -> dict:
    words_dict = {}
    words_list = words_string.lower().split()

    for word in words_list:
        if word:
            words_dict[word.strip()] = words_list.count(word)

    return words_dict

Then it will pass just fine.