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Python

Posted by Anthony Boardman
Anthony Boardman
4,672 Points

Create a function named word_count() that takes a string.

I can't figure this out for the life of me can someone help me!

Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key and the number of times it appears as the value.

This is what I've got:

# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.

[My Code]

string_sict = {}
for word in split_string():
    if word in string_dict[word]:
        string_dict[word] += 1
    else:
        string_dict[word] = 1

6 Answers

Kenneth Love
STAFF
Kenneth Love
Treehouse Guest Teacher
Kenneth Love
Kenneth Love
Treehouse Guest Teacher

Your code looks fine except you misspelled string_dict the first time you used it. Well, and split_string() isn't a built-in function. You'll need to split the string yourself using .split() on the string.

Anthony Boardman
Anthony Boardman
4,672 Points

Thanks that helped, there's a lot of words I spell wrong sometime with typing fast or because we spell the word differently in England! Ha.

durul
durul
62,690 Points

I could not understand this.

cheyenne hemmati
cheyenne hemmati
1,195 Points

Could I ask how this is separating keys and values. Thats the only part im not seeing in the answers

John Sanchez
John Sanchez
3,325 Points
John Sanchez
John Sanchez
3,325 Points

There's a really handy list method that I found in the docs when doing that code challenge. It's called [].count(x), Where x is the item that you want to check how many times it appears in the list.

For example: [1, 2, 3, 2, 1].count(1) = 2

Does that help?

mark cromer
mark cromer
7,840 Points
mark cromer
mark cromer
7,840 Points

This is the code that I wrote:

def word_count(sentence):
    new_dict = {}
    word_list = sentence.split()
    count = 0

    for original_word in word_list:
        if original_word not in new_dict:
            count = 0
            for comparison_word in word_list:
                if original_word == comparison_word:
                    count += 1
                    new_dict.update({original_word: count})
    return new_dict

I'm not sure the nested for loop is necessary; there may be a simpler way to do it... but this was the method I went with anyway.

john larson
john larson
16,594 Points

Hey Mark, I like how you used the .update(). When I first tried this challenge, I figured that update should be included, but had no clue how to do it. Just came across your code now. I'm gonna dissect it to see what makes it tick :D

john larson
john larson
16,594 Points

OK Mark, I took you idea and did it without the extra for loop. It didn't cut down on the lines of code like I thought it might. But your idea launched me into a new way of looking at it. Thanks :D

def word_count(s):
    dct_cnt = {}
    count = 0
    list_s = s.split()
    for word in list_s:
        if word in dct_cnt:
            count +=1
            dct_cnt.update({word: count})
        else:
            count = 1
            dct_cnt.update({word: count})
    return dct_cnt

I think your idea is more readable and makes more sense than the prevailing ideas. Everyone else went with:

count[item] += 1
# instead of
dct_cnt.update({word: count})
Arturo Langa
Arturo Langa
20,205 Points
Arturo Langa
Arturo Langa
20,205 Points

Thanks Kenneth! After 1 hr of feeling really dumb I got it.

def word_count(my_word):

my_dict = {} my_list = my_word.lower().split()

for word in my_list: my_dict[word] = my_list.count(word) print(my_dict)

word_count("I am that I am")

Enock Addey
Enock Addey
Courses Plus Student 10,420 Points

Simpler. :) Just that you would have to replace the print(my_dict) with return(my_dict).

Chad Duffey
Chad Duffey
16,531 Points
Chad Duffey
Chad Duffey
16,531 Points

My code works outside of workspaces, but i get an error when trying to submit it as a solution in workspaces.

(Error is: "Hmm, didn't get the expected output").

Can anyone suggest what i've done wrong here?

def word_count(sentance):
    words = sentance.lower().split(' ')
    returndict = {}
    for word in words:
        if word in returndict:
            returndict[word] += 1
        else:
            returndict[word] = 1
    return returndict

thanks a million.

durul
durul
62,690 Points
durul
durul
62,690 Points

what is my fault. I could not find any resolve . 

word_count = {}
for word in word_count.split():
    if word in word_count[word]:
        word_count[word] += 1
    else:
        word_count[word] = 1
Kenneth Love
Kenneth Love
Treehouse Guest Teacher
Kenneth Love
Kenneth Love
Treehouse Guest Teacher

You're only using a single variable,word_count for everything. First it's an empty dictionary. Then you try to .split() it (which can only be done on strings). Then you're back to using it like a dictionary.