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 Ryan Hallett
Ryan Hallett
5,796 Points

word_count code challenge

I run this in workspaces and it gives me exactly what I'm looking for. Why is this not passing?

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(st):
    my_dict = {}
    words = st.lower().split(" ")
    for word in words:
        my_dict.update({word: words.count(word)})
    return my_dict

2 Answers

Steven Parker
Steven Parker
231,248 Points
Steven Parker
Steven Parker
231,248 Points

When I tried your code I got this hint message: "Bummer! Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!"

To split on all whitespace, the argument to split should be empty. Supplying an argument of a space causes it to split only on spaces.

Ryan Hallett
Ryan Hallett
5,796 Points

Ah awesome, thanks. I thought you needed to specify what to split on, ie " " being a single space, but I guess that doesn't cover two spaces etc etc. Nice!

Dawit Tsegaye
Dawit Tsegaye
950 Points
Dawit Tsegaye
Dawit Tsegaye
950 Points

def word_count(string): word_count = {} string = string.lower().split()

 for word in string:
     if word in word_count:
         word_count[word] += 1
     else:
         word_count[word] = 1

retrun word_count