Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community!

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Collections (Retired) Dictionaries Word Count

Laknath Gunathilake
Laknath Gunathilake
1,860 Points

unable to pass word_count challenge

I think I understand how to split the words, make it lower case as well to count it, but I can't seem to get everything together

def word_count(word):
    for words in word:
        return word_dict[word]=(num) 

# 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.

1 Answer


def word_count(str):  
  word_list = str.lower().split()
  word_dict = {} #create empty dictionary
  for word in word_list:  #loop through the words from the split
    if word in word_dict:  #see if word in already in the dictionary
      num = word_dict[word]  #if it is, get its current count
      num += 1  #increment that count
      word_dict.update({word: num})  #update the count in the dictionary
    else: #otherwise
      word_dict.update({word: 1}) #add the word to the dictionary with a count of 1

  return word_dict #after looping through all the words return the dictionary