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

Posted by Bassirou Rabo Hima
Bassirou Rabo Hima
4,665 Points

Slice error

Slice error

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):

    list_arg = arg.lower().split(' ')
    list_new = []

    for word in list_arg:
        if word not in list_new:
            list_new.append(word.lower())

    dict = {}

    for word in list_new:
        dict[word] = 0

    for word in list_new:
        for word2 in list_arg:
            if word == word2:
                dict[word] = int(dict[word]) + 1

    return dict

1 Answer

Nathan Burnham
Nathan Burnham
7,827 Points
Nathan Burnham
Nathan Burnham
7,827 Points

I think you are trying to do more than you need to here.

Your list_arg variable is a list already so you don't need to append all the words into a separate new list.

Then you just need an if/else sentence to add a word to your dictionary with count = 1 if it isn't in there already or to add 1 to the current count if it is.

You can do that with: dict[word] += 1