如何让这个函数倒计时

Question

有一个getNextPlayer()函数可以向前计数，但我想将它应用于偶尔需要向后计数的纸牌游戏。你知道吗

def GetNextPlayer(self, p):
    """ Return the player to the left of the specified player, skipping players who have been knocked out
    """
    next = (p % self.numberOfPlayers) + 1
    # Skip any knocked-out players
    while next != p and self.knockedOut[next]:
        next = (next % self.numberOfPlayers) + 1
    return next

它来自于在http://www.aifactory.co.uk/newsletter/ISMCTS.txt找到的gard游戏脚本，是更大的montecarlo树搜索算法的一部分。我尝试了next=(p%self.numberOfPlayers)-1，但是它产生了无效的值

Answer 1

只需将+1更改为-1就会产生无效的值，因为modulo运算符在执行0 - 1 % self.numberOfPlayers时会忽略符号。例如-1 % 4 == 3

更新，感谢@pwnsauce，这将产生您需要的：

p - 1 if p >= 1 else self.numberOfPlayers - 1

这假设玩家索引从0开始，然后转到self.numberOfPlayers-1

Answer 2

你可以用这样的方法：

def GetNextPlayer(self, p, forward=True):
    """ Return the player to the left of the specified player, skipping players who have been knocked out
    """
    def get_next():
        ref = p if forward else p + self.numberOfPlayers - 1
        return p + 1

    next = get_next()
    # Skip any knocked-out players
    while next != p and self.knockedOut[next]:
        next = get_next()
    return next

我已经分离了引用（从0开始的计数）和周期性属性（模运算）。也使它通用于向后和向前