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Python Python Collections (Retired) Dictionaries Word Count

Posted by Andy Christie
Andy Christie
1,361 Points

What does KeyError: mean and how is it fixed?

I am getting KeyError:"I"

I've been trying to figure it out for hours now

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.

arguement = ("I am that I am")

def word_count(arguement):
  my_dict = {}
  m_string = arguement.split()
  for each_item in m_string:
    my_dict[each_item]

    if each_item in my_dict:
      my_dict[each_item] += 1
    else:
      my_dict[each_item] = 0

  return my_dict
my_dicts = word_count(arguement)
print(my_dicts)

1 Answer

Vittorio Somaschini
Vittorio Somaschini
33,371 Points
Vittorio Somaschini
Vittorio Somaschini
33,371 Points

Hello Andy.

Having a look at your code I see that you have hardcoded the initial string, which is something that we don't want to do in the code challenges.

The initial string will be auto-inputted into the compiler for testing your code, there is no need to create it manually.

Also we are not required to print anything.

All we need to code here is the word_count function.

Let's try again with this hints in mind and just focus on the function itself, I think you are not too far from the solution: you have initialized the dictionary, and I see you on the right track.

One last hint: if the word is in the dictionary only 1 time the count should be 1, not 0!

Let me know if you need further help and please share the new code you will come up with.

Thanks

Vittorio