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Evandro Luis Lima Pastor
1,061 PointsWhere's the word_count error...
This error message is not helping at all. What I am doing wrong here?
Thanks for any help :)
# 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(word):
lower_word = word.lower()
split_word = lower_word.split()
count=0
if value in split_word:
count+=1
result.update({'value':count})
return result
2 Answers
Chase Swanson
Courses Plus Student 8,886 PointsEvandro,
You are missing the FOR loop to iterate through the list. You are going straight to conditional if statement.
for each word in the list, count how many times it appears.
Hope that helps nudge you in the right direction.
Evandro Luis Lima Pastor
1,061 PointsHey Chase!
Thanks for the tip. I've tried to create a counter miself but I've found the collection.counter
It did the work. But I will try to make it without using this class.
Thanks for your help!
Chase Swanson
Courses Plus Student 8,886 PointsChase Swanson
Courses Plus Student 8,886 PointsForgot to mention, I don't believe you will need an if statement at all. You already know each word occurs in the list at least once.