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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Posted by Andrae Baker
Andrae Baker
5,563 Points

Stuck this "wordcount.py" challenge. I have, what I think to be a successful code, but I get "Bummer: Try again!"

I have my code that works in a different text editor but I cannot seem to figure out what the editor in Treehouse is looking for.

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_arg):
    mydict = {}
    split = string_arg.split()
    for string in split:
        mydict.update({string.lower():string_arg.count(string)})
    return mydict

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,457 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,457 Points

Try the test:

print(word_count("This is not fine"))

Another approach to try:

  • check if string is in the dict, if so, increment count, otherwise add to dict with count 1

Post back if you need more help. Good luck!!!

wilmington
wilmington
9,845 Points
wilmington
wilmington
9,845 Points

Hey there, I see the workspace for this task is giving a hard time. I was struggling with my code passing it too. I checked you code in a different editor with different strings and it worked just fine. I honestly, can not explain why the treehouse editor is not accepting the code, however, let me share my script, which passed and it is almost similar to yours, except I lowered values in my string earlier before the loop.

def word_count(word): my_list = word.lower().split() my_dict = {} for item in my_list: my_dict[item] = my_list.count(item) return my_dict

Good luck