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

Benjamin Bradshaw
Benjamin Bradshaw
3,208 Points
Posted by Benjamin Bradshaw
Benjamin Bradshaw
3,208 Points

what am I doing wrong.

I have tried another way that i found cleaner and simpler but it would not work so now I am trying this method and I'm still failing to get past this challenge with no luck.

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):
    all_things = {}
    new_string = string.lower().split()

    for item in new_strings:
        if itme in all_things:
            all_things[item] += 1
        else:
            all_things[item] = 0 
    return all_things

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

You are very close!! A few typos to clean up:

  • new_string vs new_strings
  • the else code should init to 1 not 0
  • item not itme