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yukcheong

PygLatin.py

Nov 13th, 2019
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Python 1.32 KB | None | 0 0
  1. pyg = 'ay'
  2.  
  3.  
  4.  
  5.  
  6. def runpyg() :
  7.   if len(original) > 0 and original.isalpha():
  8.     if original[0] in ('a' ,'A', 'e' ,'E', 'i' ,'I', 'o' ,'O', 'u','U'):
  9.       word = original.lower()
  10.       new_word = word + pyg
  11.       return new_word
  12.     else:  
  13.       word = original.lower()
  14.       first = word[0]
  15.       new_word = word + first + pyg
  16.       new_word = new_word[1:len(new_word)]
  17.       return new_word
  18.   else:
  19.     print 'empty'
  20.  
  21. def runeng() :
  22.   if len(original) > 0 and original.isalpha() and original[-1] in "y" and original[-2] in 'a':
  23.     if original[-3] in ('a' ,'A', 'e' ,'E', 'i' ,'I', 'o' ,'O', 'u','U'):
  24.       word = original.lower()
  25.       new_word =  word[:-2]
  26.       return new_word
  27.     else:  
  28.       word = original.lower()
  29.       new_word = word[-3] + word[:-3]
  30.       return new_word
  31.   else:
  32.     print 'Pls enter a valid PygLatin !'
  33.  
  34. def check_setting() :
  35.   if setting in ('p' , 'e'):
  36.     return True
  37.   else:
  38.     return False
  39.  
  40.  
  41.  
  42.  
  43. original = raw_input('Enter a word:')
  44. user_choice = raw_input("Please enter 'p' for pyglatin 'e' for English ")
  45. setting = user_choice.lower()
  46. while check_setting() !=True:
  47.   user_choice = raw_input("Please enter 'p' for pyglatin 'e' for English ")
  48.   setting = user_choice.lower()
  49. else:
  50.   pass
  51. if setting == "p":
  52.   print runpyg()
  53. else:
  54.   print runeng()
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