0xD34F

import re

def has_consecutive_characters(s, count):
  return bool(re.search(rf'([\dA-Z])\1{{{count - 1}}}', s, re.IGNORECASE))


arr = [
  'PAaAssword',
  'Paaaassword',
  'P111ssword',
  'Password',
]

print([ has_consecutive_characters(n, 3) for n in arr ])

count = {}

for n in sum([ m[::2] for m in arr ], []):
  count[n] = count.get(n, 0) + 1

re.search(f'{нужный_символ}(\d+)', string).group(1)

def sum_items(arr):
  result = {}

  for item in arr:
    v, k = int(item[:-1]), item[-1]
    result[k] = result.get(k, 0) + v

  return [ str(v) + k for k, v in result.items() ]

arr = sorted(obj.values(), key=lambda n: n['count'])

print('\n'.join(f'{i}. {n}' for i, n in enumerate(arr, 1)))

chunk_size = 3
arr = [ int(data[i:i + chunk_size]) for i in range(0, len(data), chunk_size) ]

Как это вообще работает?

Как мне при срезе включать последний элемент списка?

user = next((n for n in users if n['id'] == id), None)

[ f'{arr[i]} : {arr[i + 1]}' for i in range(0, len(arr), 2) ]

bs = ' '.join(n.upper() if n in cms else n for n in bs.split(' '))

for n in cms:
  if n in bs:
    bs = re.sub(f'\\b{n}\\b', n.upper(), bs)

for i, n in enumerate(result):
  if n[3] == 10640:
    n = list(n)
    n[3] = 'Вход'
    result[i] = tuple(n)

result = [ (*n[:3], 'Вход') if n[3] == 10640 else n for n in result ]

chunk_size = 3
newString = ' '.join(string[i:i + chunk_size] for i in range(0, len(string), chunk_size))