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Python Python Collections (Retired) Dictionaries Word Count

Posted by Imer Pacheco
Imer Pacheco
9,732 Points

Regarding "Word Count" challenge in Python Collections lesson

My code evaluates correctly in Workspaces but it does not pass the "Word Count" challenge in the Python Collections lesson. I am not clear on what I am doing wrong. Any guidance would be much appreciated. Thanks!

my_string = "I am that I am"

def word_count(a_string):
  low_str = a_string.lower()  # This converts all all letters to lower case
  spl_str = low_str.split()   # This splits each word into a list

  my_dict = {}
  a_list = []
  b_list = []
  c_list = []

  for word in spl_str: 
    if word == spl_str[0]:
      a_list.append(word)
      my_dict.update({word: len(a_list)})
      continue
    elif word == spl_str[1]:
      b_list.append(word)
      my_dict.update({word: len(b_list)})
      continue
    elif word == spl_str[2]:
      c_list.append(word)
      my_dict.update({word: len(c_list)})
      continue
    else:
      break


  print(my_dict)
  return(my_dict)

word_count(my_string)
word_count.py 
my_string = "I am that I am"

def word_count(a_string):
  low_str = a_string.lower()  # This converts all all letters to lower case
  spl_str = low_str.split()   # This splits each word into a list

  my_dict = {}
  a_list = []
  b_list = []
  c_list = []

  for word in spl_str: 
    if word == spl_str[0]:
      a_list.append(word)
      my_dict.update({word: len(a_list)})
      continue
    elif word == spl_str[1]:
      b_list.append(word)
      my_dict.update({word: len(b_list)})
      continue
    elif word == spl_str[2]:
      c_list.append(word)
      my_dict.update({word: len(c_list)})
      continue
    else:
      break

  print(my_dict)
  return(my_dict)

word_count(my_string)

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Your code can only handle strings of words where all words march one of the first three words in the string. It would fail if the input string were "I I am am that" and also fails for when there are more than three different words "I am all that I am".

Can you think of a solution that loops through the words and adds them to a dictionary: word_counts[this_word] += 1? You will need to check if the word hasn't been seen before: if this_word not in word_counts:

Post back if you need more hints. Good Luck!

Imer Pacheco
Imer Pacheco
9,732 Points

Mr. Freeman, Thank you very much! After taking your guidance into consideration I was able to formulate a better solution and, ultimately, pass the challenge!

Here is my code:

my_string = "How much wood could a woodchuck chuck if a woodchuck could chuck wood"

def word_count(a_string):
  case_low = a_string.lower()  # This converts all all letters to lower case
  words = case_low.split()   # This splits each word into a list

  my_dict = {}

  for word in words:
    value = 0
    if word not in my_dict:
      my_dict.update({word : value + 1})
      continue
    elif word in my_dict:
      my_dict[word] += 1
      continue
    else:
      break

  print(my_dict)

word_count(my_string)

I appreciate the time you took to answer my question, Mr. Freeman. :)

Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Congratulations on completing the challenge. If I could provide feedback on your solution. Once a solution is reached the code can be reviewed for simplification opportunities.

  • The function word_count should end with a return statement instead of a print statement
  • It's a good practice to use 4-space indentation instead of 2 spaces
  • Since the if and elif are complementary events (the word either is or is not in my_dict), the final else / break will never happen and can be removed
  • Similarly, due to the complementary conditions the elif can be replaced by else
  • Since value is reset each loop and only used as value + 1, the value + 1 can be replaced by 1
  • In a looping code block, there is an implied continue at the end. This makes the continue statements in the if statement unnecessary and they can be removed.

Here is your code simplified:

def word_count(a_string):
    case_low = a_string.lower()  # This converts all all letters to lower case
    words = case_low.split()   # This splits each word into a list 

    my_dict = {}

    for word in words:
        if word not in my_dict:
            my_dict.update({word : 1})
        else:
            my_dict[word] += 1

    return my_dict

Additionally, the first two statements could be combined, if so desired:

words = a_string.lower().split()

The dict.update can be replaced with an assignment:

`my_dict[word] = 1

Keep up the good work!

-- Chris

Imer Pacheco
Imer Pacheco
9,732 Points
Imer Pacheco
Imer Pacheco
9,732 Points

The code is so much cleaner and readable. Thanks Chris! I truly appreciate your input in my python education. I hope to communicate with you more in the future. Here is my revised code: 

def word_count(a_string):
    words = a_string.lower().split()   # Converts all letters to lower case then splits each word into a list

    my_dict = {}

    for word in words:
        if word not in my_dict:
          my_dict[word] = 1
        else:
          my_dict[word] += 1

    return(my_dict)