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

Idai Yaffe
Idai Yaffe
20,429 Points
Posted by Idai Yaffe
Idai Yaffe
20,429 Points

Works perfectly in an IDE but wrong in treehouse for some reason

Can someone spot my problem / Give me his solution

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):
    string = string.lower()
    words = string.split(' ')
    output = {}
    for word in words:
        if word in output.keys():
            output[word] += 1
        else:
            output[word] = 1
    return output
Idai Yaffe
Idai Yaffe
20,429 Points
Idai Yaffe
Idai Yaffe
20,429 Points

I tried this too:

def word_count(string):
    output = {}
    string = string.lower()
    string = string.split(' ')
    for word in string:
        output[word] = string.count(word)
    return output

1 Answer

Stuart Wright
Stuart Wright
41,119 Points
Stuart Wright
Stuart Wright
41,119 Points

Use simply:

string = string.split()

When you don't pass any parameters to .split(), it will split on all whitespace characters, which is what the challenge wants.