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 (Retired) Dictionaries Word Count

Challenge task evaluate seem like not working correctly for python . has any one faced similar issue?

I am getting Bummer! try again! for word_count.py .

The code i have written seems correct to me. at-least i was expecting some error reason. It is frustrating when you get this kind of error without any error code.

# 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.

def word_count(string_text):
    word_list_lower = string_text.lower()
    word_list = word_list_lower.split()
    for word in word_list:
        if word in word_dict:
            word_dict[word] += 1
            word_dict[word] = 1

return word_dict

the issue was return indentation....

2 Answers

Mark Christian Roma
Mark Christian Roma
5,232 Points

I also have another approach for this challenge ...

def word_count(string):
    new_dict = {}
    new_string = string.lower().split()
    for word in new_string:
        if word not in new_dict:
    return new_dict

I like it :D

Mrinil, One thing I found, the return is a tab too far to the left. I'll look and see if there's anything else

with that small fix, it passed fine :D