Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Collections (Retired) Dictionaries Word Count

Posted by Paul Vitalis
Paul Vitalis
2,858 Points

Would you please teel me what's wrong with my word_count code?

I think my problem lies within the For loop

word_count.py 
# 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(st):
  my_dict = {}
  count = 0
  st = st.lower()
  st = st.split()
  for word in st:
    if word not in my_dict:
      my_dict[word] = count + 1
      count += 1
    else:
      my_dict[word] = count
      count += 1
  return my_dict
Paul Vitalis
Paul Vitalis
2,858 Points

Thank you! It worked

1 Answer

evertsbaard
evertsbaard
2,930 Points

Heyah, decided to run your script and saw that it outputs the required number times 2.

After snooping around in your code I saw that you used a counter variable which we don't really need.

Why not grab the value, edit it and instantly return it back to the dictionary like so:

def word_count(st):
  my_dict = {}
  st = st.lower()
  st = st.split()
  for word in st:
    if word not in my_dict:
      my_dict[word] = 1
    else:
      my_dict[word] += 1
  return my_dict