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Mo Song
12,355 PointsChallenge Problem
This is my code, but I cannot figure out how to fix it.
my_dict = {} def word_count(string): string = string.split(" ") for word in string: if word in dict: my_dict[word] += 1 else: my_dict[word] =
# 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.
my_dict = {}
def word_count(string):
string = string.split(" ")
for word in string:
if word in dict:
my_dict[word] += 1
else:
my_dict[word] = 1
1 Answer
Carlos Federico Puebla Larregle
21,074 PointsI think your error is in your if statement, you are checking if "word in dict" and you dictionary variable is called my_dict instead of "dict" and you have to remember to actually return that dictionary. You can do this code challenge like this:
def word_count(string):
my_dict = {}
list_string = string.split(" ")
for word in list_string:
if word in my_dict.keys():
my_dict[word] += 1
else:
my_dict[word] = 1
return my_dict
I hope that helps a little bit.