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Python Python Collections (Retired) Dictionaries Word Count

Srikanth Srinivas
Srikanth Srinivas
1,465 Points
Posted by Srikanth Srinivas
Srikanth Srinivas
1,465 Points

Why doesn't this code pass?

Im sure that this code functions the way the prompt tells me it should, i checked it myself on an online compiler, yet it doesn't seem to pass. Whys that?

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.
anydict = {}
def word_count(anystring):

  anystring = anystring.lower()
  for word in anystring.split():
    anydict[word] = anystring.count(word)

  return anydict

1 Answer

Graham Mackenzie
Graham Mackenzie
2,747 Points

Hey Srikanth!

So, a lot about your code looks good, but a few things need fixing:

  • First of all, you need to put your anydict = {} declaration within the body of the function. Leaving it outside of the function that returns it could cause technical problems, and is also a bad practice aesthetically / clarity-wise.
  • You've got the right idea in calling the .lower() function on the string that is passed in, but assigning it to itself is also potentially problematic / inelegant. Better to assign it to a whole new variable.
  • Calling .split() as the next step is also the right idea, but you really want to do that outside of the for loop structure, and assign the result to its own new variable as well.
  • Notice that the instructions say to check for inclusion in the dict (with "if word in dict"-style syntax). Or add it to the dict... This is your clue that it's a good idea to use an if... else... statement here. Obviously, going this direction is going to require you to use a counting variable that instantiates or increases with each pass of the for loop, rather than using .count().

Your code may work in other environments, but for these Treehouse tests they're looking for very specific things, so it pays to slow down and spell things out explicitly. I hope this helps! Please mark this as the Best Answer if you found it most helpful.

Thanks and Be Well, Graham

Srikanth Srinivas
Srikanth Srinivas
1,465 Points
Srikanth Srinivas
Srikanth Srinivas
1,465 Points 
def word_count(anystring):
  anystring = anystring.lower()
  any_dict = {}
  splitword = anystring.split()
  for word in splitword:
    count = 1
    if word in anystring:
      x += 1
    any_dict[word] = count
  return any_dict

what about this code? it works in theory, but when i enter it on the challenge, it says "wheres word_count()?"

Graham Mackenzie
Graham Mackenzie
2,747 Points

So, this is much better along a lot of the lines I was suggesting, but there are still some issues:

  • You are still assigning anystring.lower() to itself. Assign it to a new variable, and then assign the .split() version of that variable to splitword.

  • Where did the variable x come from? Did you mean to use count?

  • Rather than checking if word in anystring, I think you mean to check if word in any_dict.

  • Let's go through the logic in your for loop: So, we start with a word that is in splitword. count gets set equal to 1. Then, we check to see if word in any_dict (let's say that it is.) Then (let's say we assume you meant count instead of x) count gets increased by 1. Then we apply count (which is now equal to 2) to the entry for that word in any_dict. But we only meant to increase the value by 1. Do you see the problem here? It would be better to set count = 0 and then use an if... else... statement to either increase the value of any_dict[word] by 1, or else set count = 1.

I hope this helps!

Be Well, Graham