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

Georgi Koemdzhiev
Georgi Koemdzhiev
12,610 Points
Posted by Georgi Koemdzhiev
Georgi Koemdzhiev
12,610 Points

Function that returns the word frequency in a string?

I am trying to solve a code challenge where I have to write a method that counts the words in a sentence. I have tested my code in pyCharm and it works. However, the online python interpreter does not want to accept my method as an answer to the challenge. Any ideas what I am missing?

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(myString):
    output_dict = {}
    for word in myString.lower():
        current_letter_counter = 0
        for word2 in myString:
            if word == word2.lower():
                current_letter_counter += 1
        if word == ' ':
            pass
        output_dict.update({word : current_letter_counter})
    return output_dict

1 Answer

Travis Howk
Travis Howk
7,169 Points

The code that you are running has a very high iteration rate. Part of the problem is that your first for loop is iterating over every single letter in the string (not the words). As such, you might be better off splitting your strings using something like the .split() method, which would allow your string to be broken up by a certain character like a space. It would look something like this:

my_string = "This is my sentence. I hope my code works."
my_array = my_string.split(" ")

Then you would have my_array = [This, is my, sentence., I, hope, my, code, works.] That would make it a bit easier to deal with.

Edit: Here is my code

def word_count(myString):
    output_dict = {}
    myarray = myString.split()
    for word in myarray:
        if word.lower() not in output_dict:
            output_dict[word.lower()] = 1
        else:
            output_dict[word.lower()] = output_dict[word.lower()] + 1
    return output_dict