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 Todd Teese
Todd Teese
22,680 Points

I have the correct output, is there anything else to this?

I've tried this in a separate sandbox, and I'm getting the correct output dict, with words for keys and int's for values. Just wondering if there's something I'm doing wrong or weird in this code challenge.

Thanks in advance!

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(inputString):
    wordList = inputString.lower().split(" ")
    wordDict = dict()
    for word in wordList:
      if wordDict.has_key(word):
        wordDict[word] = wordDict[word] + 1
      else:
        wordDict[word] = 1
    return wordDict

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points

You are on the right path. Two errors to correct:

  • dict object wordDict has no attribute has_key. Instead use:
if word in wordDict.keys():
# or the sort version
if word in wordDict:  # .keys() is assumed
  • The split should be on whitespace not a literal " " (space). No argument means split on whitespace.

Post back if you need more help. Good luck!

Todd Teese
Todd Teese
22,680 Points

Awesome, thanks for the help!