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Laknath Gunathilake
1,860 Pointsunable to pass word_count challenge
I think I understand how to split the words, make it lower case as well to count it, but I can't seem to get everything together
def word_count(word):
val=list(word.lower().split())
num=val.count(word)
word_dict={}
for words in word:
return word_dict[word]=(num)
# 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.
1 Answer
jcorum
71,830 PointsLaknath,
def word_count(str):
word_list = str.lower().split()
word_dict = {} #create empty dictionary
for word in word_list: #loop through the words from the split
if word in word_dict: #see if word in already in the dictionary
num = word_dict[word] #if it is, get its current count
num += 1 #increment that count
word_dict.update({word: num}) #update the count in the dictionary
else: #otherwise
word_dict.update({word: 1}) #add the word to the dictionary with a count of 1
return word_dict #after looping through all the words return the dictionary