turning a string in a dictionary

Question

I'm not real sure what the issue is with my code. I understand that these exercises look for something specific to be returned from the function, but I don't get why this isn't working.

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(x):
  x = "I am that I am".lower()
  count = 0
  word_dict = {}
  for y in x:
    count += 1
    word_dict[y] = count
    if y in word_dict:
      break
  return word_dict

Answer 1 · 2015-09-28T04:41:18Z

September 28, 2015 4:41am

As Jeffrey mentioned, the function needs to accept any input string. The other missing part is using split() to break the string into a list of words. Indentation increased to 4 spaces

def word_count(x):
    xlower = x.lower() #<-- lower case input string
    wordlist = xlower.split() #<-- split string into list of words
    count = 0
    word_dict = {}
    for y in wordlist:
        # Check if word already seen
        if y in word_dict:
            # Add to word count
            count += 1
            word_dict[y] += 1
        else:
            # Add new word to dict
            word_dict[y] = 1
    return word_dict

Answer 2 · 2015-09-25T02:46:11Z

September 25, 2015 2:46am

A few things about your code look off to me.

First, I would suggest changing the first part of the function so that you're not using a specific string. I'm pretty sure "I am that I am" is just an example. You might get test cases that use other strings.

Change it to something like:

def word_count(thestring):
    word_dict = {}
    lowercasestring = thestring.lower()
    for word in lowercasestring:
        #and then do stuff

Second, I'm not sure that your count variable is going to do what you want. It looks like it increases with every pass through the for loop regardless of whether the word is already in the dictionary.

Third, it looks to me like you're using break in the for loop incorrectly. I think the way you're using it means that the for loop will stop the first time it encounters a duplicate word. I don't think that's what you want. Maybe you meant to use continue instead.

Welcome to the Treehouse Community

Looking to learn something new?

Andrew Gordon

Andrew Gordon

turning a string in a dictionary

2 Answers

Chris Freeman

Chris Freeman

Jeffrey Ruder

Jeffrey Ruder