word_count.py

Question

help don't know what I am doing wrong says some words are missing

word_count.py

# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.
def word_count(s):
    s = s.lower()
    dictionary = {}
    count = 0
    list_s = s.split()
    for word in list_s:
        if word in dictionary:
            count = dictionary[word]
            count += 1
            dictionary.update({word: count})
    return dictionary

Answer 1 · 2016-11-30T05:45:30Z

November 30, 2016 5:45am

Hi Christopher,

It looks like you're going to be returning an empty dictionary.

Initially, your dictionary is empty. Then you only have in if condition in your loop. This condition is never going to be true since the dictionary doesn't have any words in it.

Your if block takes care of what the 2nd to last code comment is saying. The last code comment gives you a suggestion on what to do if the word wasn't in the dictionary. You can put that in an else statement.

Also, there's an easier way to increment the count in your if block.

dictionary[word] += 1

This will take the current count, add 1 to it, and then store it back as the count for that word.

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Christopher Wommack

Christopher Wommack

word_count.py

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Jason Anello

Jason Anello