首页 文章

对于循环不“刷新”Python

提问于
浏览 次
1

我在Python中运行一个石头剪刀蜥蜴Spock游戏 . 我希望它用不同的机器人玩5轮 . 我遇到的问题是，当我尝试在我的主函数中循环回合时，即使机器人随机选择其移动，也会打印出相同的匹配5次 .

for循环：

for x in range(0, 5):
    p1move = p1.play()
    p2move = p2.play()
    print(p1move.compareTo(p2move))

游戏方法：

class Human(Player):
    if decisionint == 1:
       def play(hum1):
           hum1 = Rock
           return hum1

机器人的例子：

class RandomBot(Player):
num = randint(1, 5)
if num == 1:
    def play(rando1):
        rando1 = Rock
        return rando1
else:
    if num == 2:
        def play(rando2):
            rando2 = Paper
            return rando2

摇滚课：

class Rock(Element):
def compareTo(_name):
    if _name == Lizard:
        global p1wins
        p1wins += 1
        print("Rock crushes Lizard, Win")

主要方法询问你想要两个玩家是谁，然后将你的第一选择设置为 p1 ，将你的第二选择设置为 p2

如果您需要更多代码，请告诉我们 . 我认为这应该足够但我并不积极 . 谢谢

python loops

2 回答

  • 1

    虽然我喜欢你在这里使用物品，但我认为你可以做的比你更多的继承，并进一步滥用干 . 试试这个：

    import random

class Selection(object):
    def __init__(self, value, wins_against, name=None):
        self.value = value
        self.wins_against = wins_against
        self.name = name
    def compare(self,other):
        if other.value == self.value: return 0
        elif other.value in self.wins_against: return 1
        else: return -1

Rock = Selection(0, (2,3), "Rock")         # beats scissors+lizard
Paper = Selection(1, (0,4), "Paper")       # beats rock+spock
Scissors = Selection(2, (1,3), "Scissors") # beats paper+lizard
Lizard = Selection(3, (1,4), "Lizard")     # beats paper+spock
Spock = Selection(4, (0,2), "Spock")       # beats rock+scissors

class Player(object):
    def __init__(self,name=None):
        self.choice = None
        if name is None: self.name = "Computer"
        else: self.name = name
    def choose(self):
        self.choice = random.choice([Rock,Paper,Scissors,Lizard,Spock])
    def compare(self,other):
        """1 = win, 0 = tie, -1 = lose"""
        return self.choice.compare(other.choice)

class Human(Player):
    def choose(self):
        self.choice = None
        while self.choice is None:
            print("""1. Rock
2. Paper
3. Scissors
4. Lizard
5. Spock""")
            in_ = input(">> ")
            try:
                self.choice = {"1":Rock,"2":Paper,"3":Scissors,"4":Lizard,"5":Spock}[in_]
            except KeyError:
                pass

def play(player1,player2):
    for player in (player1,player2):
        player.choose()
    print("{} vs {}".format(player1.choice.name,player2.choice.name))
    result = player1.compare(player2)
    if result:
        print("{0.name} wins!".format(
            player1 if result > 0 else player2)) 
    else:
        print("Tie game!")
    print("\n\n")


Player1 = Human("Joe Schmoe")
Player2 = Player() # bot

for _ in range(5):
    play(Player1,Player2)
    回复于
  • 2

    当你调用你的游戏方法时，你将Rock分配给实例名称，然后返回它，但基本上相当于：

    class Human(Player):
    if decisionint == 1:
       def play(self):
           return Rock

    如果您希望它变化以便您可以正确建模，则需要每次都返回随机选择的元素 .

    例如靠近代码顶部：

    import random

CHOICES = (Rock, Paper, Scissors, Lizard, Spock)

    在你的 play 方法中：

    return random.choice(CHOICES)
    回复于

相关问题