How do you check if your function returns nothing?


I'm trying to answer a question from a challenge.

Question: Some numbers have personality, while other numbers are just dull. Let's write a program that asks for a list of numbers and describes the interesting ones using the personality traits below, while counting up all the dull ones without traits.

Your program should ask for a list of space-separated numbers. For any number with at least one trait, it should print a full description of its personality. Then it should print how many dull numbers were counted.

Here's my solution for it:

arrogant_numbers = [3, 6, 7, 23, 25, 35, 39, 66, 68, 112, 119, 254, 259, 732, 737, 4565, 4663, 13330, 13730, 29880, 29998, 79670, 80015, 230054, 239068, 1534301, 1607352, 2060587, 21700891, 99167753, 99873125]

def find_personality(number):
  traits = ''
  # Find the traits of the number
  if number % 2 != 0:
    traits = traits + 'odd '
  # Check for other traits after this
  if number >10000:
    traits = traits + 'excessive '
  if '3' in str(number): 
    traits = traits + 'irksome '
  if number in arrogant_numbers:
    traits = traits + 'arrogant '
  return traits 

# Write the rest of your program after this
dullcounter = 0
numbers = input('Enter numbers: ')
number_list = numbers.split()
for word in number_list: 
  trts = find_personality(int(word))
  print(f"{word} is an {trts}number.") 
  if '' in trts:
    dullcounter = dullcounter + 1
print(f'Dull numbers: {dullcounter}')

I'm having problems with counting the dull numbers (the ones that have no traits) it counts all the numbers given instead of the ones that have no traits.

Here's how the final solution should look like: desired output

Here's how my output looks like: my output


Your output string trts will always contain ''.
Instead of this you should validate if trts is an empty string.
So, instead of

if '' in trts:
    dullcounter = dullcounter + 1

you should use this or similar:

if not trts:
    dullcounter = dullcounter + 1

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