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Python Python Collections (Retired) Dictionaries Word Count

Matthew Roser
Matthew Roser
3,220 Points
Posted by Matthew Roser
Matthew Roser
3,220 Points

"Bummer! Where's "word_count()"" Not sure what this means.

Not sure what I'm missing here, and I'm not sure what this error means.

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(string):
    dictionary = {}
    string.lower()
    string_list = string.split()
    for word in string_list:
        dictionary[word] = x.count(word)
    return dictionary

2 Answers

andren
andren
28,558 Points
andren
andren
28,558 Points

That error message is quite misleading. There is an error in your code but for some reason the code checker does not report the actual error unless you try to call your method within the challenge workspace. If you do you get this error message:

NameError: name 'x' is not defined

Which points toward the fact that you use a variable named "x" inside your for loop to refer to your list, even though you named your list "string_list". Fixing that typo, like this:

def word_count(string):
    dictionary = {}
    string.lower()
    string_list = string.split()
    for word in string_list:
        dictionary[word] = string_list.count(word)
    return dictionary

Will allow you to pass the challenge.

Matthew Roser
Matthew Roser
3,220 Points
Matthew Roser
Matthew Roser
3,220 Points

I can't believe I missed that! Thank you!