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Python Python Collections (Retired) Dictionaries Word Count

Posted by Homa Shayan
Homa Shayan
2,038 Points

don't understand the error

I have word_count() in my code. I don't understand why the error says 'where's word_count()".

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(my_str):
  my_str = my_str.lower()
  my_list = my_str.split()
  for word in my_list:
    if word in my_dict:
      my_dict[word] += 1
    else:
      my_dict[word] = 1
  return my_dict

1 Answer

Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

Hi Homa

The reason you are getting that error is because there is an error in the script. my_dict is not declared as a dictionary.Try the code below and notice line 4.

def word_count(my_str):
  my_str = my_str.lower()
  my_list = my_str.split()
  my_dict={}
  for word in my_list:
    if word in my_dict:
      my_dict[word] += 1
    else:
      my_dict[word] = 1
  return my_dict
Homa Shayan
Homa Shayan
2,038 Points

I just defined my_dicts as you suggested (which i did forget) but it didn't fix the problem. It still says "where's is word_count()"

Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

Hi Homa

I just copy pasted the code from above in the code challenge and it passed.

Homa Shayan
Homa Shayan
2,038 Points

Hi Andres

I had my_dicts instead of my_dict. It worked thank you :)