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

Posted by Ian Maina
Ian Maina
4,571 Points

What is wrong with my code?

It seems to be working well in workspaces but I only get "Bummer! Try again!"

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:
    counts = dict()
    words = str.split()

    for word in words:
        new_word = word.lower()
        if new_word in counts:
            counts[new_word] += 1
        else:
            counts[new_word] = 1

    return counts

1 Answer

Jennifer Nordell
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there! You're really close here, but you've forgotten to make it so that the function takes an argument. This also means that one of your other lines will need to be adjusted. Treehouse is going to send in a string to test it so it must accept information coming into it.

The declaration of your function should start with:

def word_count(words):

After you change this, you will receive an error about the split. If you modify that line just slightly, your code passes!

I'm going to leave a little debugging to you, but let me know if you're still stuck and I'll be glad to help! :sparkles:

Ian Maina
Ian Maina
4,571 Points

Thanks a lot.

Namrata Lamba
Namrata Lamba
1,347 Points

Hi Jennifer,

Can you tell whats wrong with my code?

def packer(**kwargs):
return (kwargs)

def word_count(string): string = string.lower()
list1 = string.split()
for item in list1: newlist.append(item = (newlist.count(item)) packer(",".join(newlist))