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Jessica Barron
1,215 PointsI have already figured out a way to do this but I am quirious why this code doesn't work
please explain why this code doesnt work and a then explain a way that is close to this that WOULD work if possible. THANK YOU!
# 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):
word_list = string.split()
word_dict = {}
count = 1
for word in word_list:
if word in word_dict:
count += 1
continue
word_dict[word] = count
return word_dict
word_count('i am that i am')
1 Answer
Nick Osborne
4,612 PointsHi, I think the problem is that the code is incrementing 1 whenever any word is in the dict already.
Something like this just increments the count for the particular word:
def word_count(string):
word_list = string.split()
word_dict = {}
for word in word_list:
if word in word_dict:
word_dict[word] += 1
continue
word_dict[word] = 1
return word_dict