2022 - 2023 Class 18

Class Slides

The solution to problem on slides 4-5: (HW)

(also found on slide)

import re
def subfunc(match):
  global i, x
  m = match.group(0)
  if m == 'AN':
    return 'UN'
  elif m == 'AU':
    return 'OO'
  elif m == 'A':
    if match.start() == len(x)-2:
      return 'A'
    return 'E'
  elif m == 'OW':
    return 'OO'
  elif m == 'O':
    return 'U'
  elif m == 'I':
    if match.start() == 1:
      return 'I'
    if i == False:
      i = True
      return 'EE'
    else:
      return 'I'
  elif m == 'EN\n':
    return 'EE\n'
  elif m == 'E\n':
    return 'E-A\n'
  elif m == '\nE':
    return '\nI'
  elif m == 'U':
    return 'OO'

for _ in range(5):
  i = False
  x = '\n'+input()+'\n'
  print(re.sub('(AN|AU|A|OW|O|I|EN\n|E\n|\nE|U)', subfunc, x)[1:-1])