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! While you're at it, check out some resources Treehouse students have shared here.

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 (2016, retired 2019) Dictionaries Word Count

Posted by mamadou diene
.a{fill-rule:evenodd;}techdegree
mamadou diene
Python Web Development Techdegree Student 1,971 Points

Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it. I need

i need help on thisone

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):
    new_dict = {}
    word_count = []
    string_list = string.lower().split()


    for word in string_list():
        word_count = 1
        if word in word_count:
            new_dict[word] +=1                        
        else:
            new_dict
            word= word_count

    return new_dict

1 Answer

Wade Williams
Wade Williams
24,476 Points
Wade Williams
Wade Williams
24,476 Points

You're thinking on how to do this is close, but you need to breakdown the steps a little more to clarify each step.

def word_count(string):

    # lowercase string and split by whitespace
    words = string.lower().split()

    # Create dictionary to hold our word counts
    word_dict = {}

    for word in words:

        # If word is in word_dict just add 1 to the count
        if word in word_dict:
            word_dict[word] += 1

        # word is NOT in word_dict, so add it and set the count to 1
        else:
            word_dict[word] = 1

    # return word_dict with word counts
    return word_dict