Return to Answer

• you could count values instead of adding them to a list

  nb_valid = 0

  for current in address: for valid in valid_characters: if valid == current: nb_valid += 1

  ...
  return len(address) == nb_valid and len(address_list) == 8 and invalid_segment == False

• you could break out of the valid_characters loops as soon as you've found a match.

• in Python, for loops accept a else part meaning "this loop exited the normal way: it didn't encounter a break". You could use this to return False as soon as a character is invalid.

  for current in address: for valid in valid_characters: if valid == current: nb_valid += 1 break else: # nobreak - invalid character return False

• you could count values instead of adding them to a list

   nb_valid = 0
  
   for current in address:
       for valid in valid_characters:
           if valid == current:
               nb_valid += 1
  
   ...    
   return len(address) == nb_valid and len(address_list) == 8 and invalid_segment == False
  

• you could break out of the valid_characters loops as soon as you've found a match.

• in Python, for loops accept a else part meaning "this loop exited the normal way: it didn't encounter a break". You could use this to return False as soon as a character is invalid.

  for current in address: for valid in valid_characters: if valid == current: nb_valid += 1 break else: # nobreak - invalid character return False

TODO

def ipv6_addr_is_valid(address):
    address_list = address.split(":")
    return len(address_list) == 8 and all(c in valid_characters for c in address) and all(len(c) <= 4 for c in address_list)

To be continued

You could use all/any to make some parts of your code more concise. It doesn't help too much in your case because you are using return already.

def ipv6_addr_is_valid(address):
    if any(c not in valid_characters for c in address):
        return False
    address_list = address.split(":")

    if any(len(c) > 4 for c in address_list):
        return False  # Invalid segment

    return len(address_list) == 8

Which can be re-organised:

TODO