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

Word Count Challenge...

I'm having a lot of trouble understanding how to do this challenge - I tried to solve it on my own with the code below being my latest attempt - So, I would appreciate a breakdown of how I should proceed with this challenge from someone - Thank you in advance, and here is the exercise as well that I'm currently working on:

I need you to make a function named word_count. It should accept a single argument which will be a string. The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string, lowercased. The values will be how many times that particular word appears in the string.

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(Axanar):
    Axanar_dict = {}
    list_of_words = re.split(Axanar.lower())
    for a in list_of_words:
        try:
            Axanar_dict[a] += 1
        except KeyError:
            Axanar_dict[a] = 1

    print(Axanar_dict)
    return Axanar_dict

2 Answers

Stuart Wright
Stuart Wright
41,120 Points

You have the right idea. You just have the wrong syntax for splitting the string into a list of words.

The correct way is:

list_of_words = Axanar.lower().split()

Change that one line and your code should pass.

I changed the line, but my code still doesn't pass the challenge:

def word_count(Axanar):
    Axanara_dict = {}
    list_of_words = Axanar.lower().split()
    for a in list_of_words:
        try:
            Axanar_dict[a] += 1
        except KeyError:
            Axanar_dict[a] = 1

    print(Axanar_dict)
    return Axanar_dict
Stuart Wright
Stuart Wright
41,120 Points

Your latest version contains a typo. Fix that and you're good to go:

Axanara_dict = {}
      ^

Whew, Jesus - Thank you, Stuart... P :)