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Question

我试图实现一个函数，其中我有两个玩家，他们的收益取决于他们的行动。你知道吗

def game(action1,action2):

  if action1 == "a" and action2 == "a":
     payoff1 = 1
     payoff2 = 1
  elif action1 == "a" and action2 == "b":
     payoff1 = -5
     payoff2 = 3
  elif action1 == "b" and action2 == "a":
     payoff1 = 3
     payoff2 = -5
  elif action1 == "b" and action2 == "b":
     payoff1 = 2
     payoff2 = 2
 return payoff1 , payoff2

然后我会有这个游戏的策略（例子）：

def TitForTat(round_num, previous_action):
    if round_num == 0:
       action = "a"
    else:
       action = previous_action
  return action

def AlwaysDefect():
  return "b"

action1 = TitForTat (0,'c')
action2 = AlwaysDefect()

game (action1,action2)

这将返回一个错误：

local variable 'payoff1' referenced before assignment

我试图将它们初始化为“0”，但都是一样的。 如果我有所有的正值，那么精确的函数工作得很好。你知道吗

编辑：

很抱歉出现了打字错误。函数AlwaysDefect（）返回“b”和“d”。你知道吗

Answer 1

你的意思是AlwaysDefect函数返回'd'还是'b'？你知道吗

既然它返回了'd'，您就不必使用任何if语句，而且您也没有payoff1或payoff2的定义

也是一个打字错误：payyoff2->；payoff2

您可以设置默认语句来修复它（在else部分）：

def game(action1,action2):

  if action1 == "a" and action2 == "a":
     payoff1 = 1
     payoff2 = 1
  elif action1 == "a" and action2 == "b":
     payoff1 = -5
     payoff2 = 3
  elif action1 == "b" and action2 == "a":
     payoff1 = 3
     payoff2 = -5
  elif action1 == "b" and action2 == "b":
     payoff1 = 2
     payoff2 = 2
  else:
     payoff1 = 0 # or any other value
     payoff2 = 0 # or any other value
  return payoff1, payoff2

Answer 2

函数调用中的错误太多。使用字典：

def game(action1, action2):
    payoffs = {
        ('a', 'a'): (1, 1),  ('a', 'b'): (-5, 3),
        ('b', 'a'): (3, -5), ('b', 'b'): (2, 2)
    }
    return payoffs.get((action1, action2), (0, 0))

Answer 3

如果操作是c或d，则if语句无法捕获，这将导致payoff1和payoff2未初始化。你知道吗