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

Posted by Oli Flemmer
Oli Flemmer
13,394 Points

Having problems with 'word_count' in python collections

I've spent quite a few hours on trying to debug my code, and even tested it on my own computer. When tested on my laptop it works, but when executed on the challenge page it doesn't. It returns ' Bummer! Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace'. In the code, I've made sure that I've vividly lowercased the string, and split the string on all whitespace. Even if there's a double space it'll still pick up on it and remove it.

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(string):
    #lowercasing the string
    string = string.lower()

    #putting each word in the string into a list
    ls = string.split(' ')

    #list of valid letters
    letter_list = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']

    #empty dictionary
    dic = {}

    #Selecting each word in the list
    for word in ls:

        #selecting each letter in that word
        for letter  in word:

            #if that 'letter' is not a letter then delete it
            if not letter in letter_list:
                lw = list(word)
                lw.remove(letter)
                word = ''.join(lw)

        #if the word isn't in the dictionary then add it
        if word not in dic and not word == '':
            dic.update({word: 1})

        #if it is in the diictionary then add one to it's value
        elif word in dic:
            dic[word] =  dic[word] + 1

    return dic
austin bakker
austin bakker
7,105 Points
austin bakker
austin bakker
7,105 Points

I am having the same problem , here i my peace of code

def word_count(string):
    dicts={}
    string = string.lower()
    ls = string.split(' ')
    for item in ls:
        if item in dicts.keys():
            value=dicts[str(item)]
            value=value+1
            dicts.update({item:value})
        elif item not in dicts.keys():
            dicts.update({item:1})
    print(dicts)

1 Answer

Chris Howell
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.a{fill-rule:evenodd;}techdegree seal-36
Chris Howell
Python Web Development Techdegree Graduate 49,702 Points
Chris Howell
.a{fill-rule:evenodd;}techdegree seal-36
Chris Howell
Python Web Development Techdegree Graduate 49,702 Points

Hey Oli Flemmer

It looks like you just about got it. Except there are two issues with your code.

  1. You must split on ALL whitespace. Currently you are only splitting on spaces using string.split(' ')

Check Python Docs: Split make sure you read the line that talks about:

If sep is not specified or is None

  1. You are filtering out all characters that arent the letters a thru z. But it didnt ask for you to do that. 
#selecting each letter in that word
        for letter  in word:

            #if that 'letter' is not a letter then delete it
            if not letter in letter_list:
                lw = list(word)
                lw.remove(letter)
                word = ''.join(lw)

You dont really need this line at all because I believe in the tests for this challenge, other characters are being tested and counted. And if you filter them out it wont return the right result.

Chris Howell
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Chris Howell
Python Web Development Techdegree Graduate 49,702 Points
Chris Howell
.a{fill-rule:evenodd;}techdegree seal-36
Chris Howell
Python Web Development Techdegree Graduate 49,702 Points

austin bakker

The same applies for you for the String Split you are doing. But in addition to that, the challenge tells you that the function must return a dictionary. Your function has no return so it implicitly returns a None by default. You are instead printing the dictionary but you need to return it. :)