unable to pass word_count challenge

Question

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

word_count.py

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.

Answer 1 · 2016-06-01T03:15:07Z

June 1, 2016 3:15am

Laknath,

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

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Laknath Gunathilake

Laknath Gunathilake

unable to pass word_count challenge

1 Answer

jcorum

jcorum