Python中的石头、布、剪刀（带大于和小于）

Question

有没有办法我可以做一个石头-布-剪刀游戏，有大于和小于的符号？在

我知道做RPS有不同的方法，但我想具体了解一下“大于”和“小于”。在

这是我的代码：

"Rock" > "Scissors" and 'Rock' < 'Paper'
"Paper" > "Rock" and 'Scissors' < 'Rock' 
"Sissors" > "Paper" and 'Paper' < 'Scissors'
choose1 = input("Player One, Enter your answer ")
choose2 = input("Player Two, Enter your answer ")

if choose1 == "Paper":
    "Paper" > "Rock"
if choose1 == "Scissor":
    "Scissor" > "Rock"


if choose1 != "Rock" and choose1 != "Paper" and choose1 != "Scissors":
    print("Player one please chose Rock, Paper, or Scissors")

if choose2 != "Rock" and choose2 != "Paper" and choose2 != "Scissors":
    print("Player two please chose Rock, Paper, or Scissors")

if choose1 > choose2:
    print('Player1 ({}) beats ({})'.format (choose1, choose2))
else:
    print('Player2 ({}) beats ({})'.format (choose2, choose1))

这个游戏起作用了，然而，它认为石头打败了一切，纸打败了剪刀，剪刀却什么也没打。在

如何修复此代码以使其正确执行？在

Answer 1

一个纯粹的“数学”形式使用的运算符lt和gt，而不需要黑客攻击的运算符可以实现。在

#ask the choices and map to P=3,R=2,S=1, using a dictionary for instance
#any difference between options will work as long as simmetric 
#and the rest is adapted with the step different from one

result = abs(choose1 - choose2)
if not result:
    print "Tie"

if result == 1:
    #choose max
    if choose1 > choose2:
         print "Player 1 wins"
    else:
         print "Player 2 wins"
else: 
    #choose min:
    if choose1 < choose2:
         print "Player 1 wins"
    else:
         print "Player 2 wins"