Question

我已经制作了一个寻宝游戏，我需要它，这样一旦一个包含宝箱的广场被访问了3次，它就变成了一个“土匪”。为了统一访问宝箱的次数，我在TreasureChestCoordinates列表中添加了许多相同的坐标。然后我计划有一个if语句，这样如果任何一个坐标写入3次，所有3个将被删除并放入“强盗”列表中。这是该部分的草案代码：

            # Treasure Chests Collecting Coins #
        if (x,y) in TreasureCoords[0:(TCNumber -1)]:
            print("You have landed on a treasure chest! You can visit it 2 more times before it becomes a bandit.")
            CoinCollection = CoinCollection + 10
            TreasureCoords.append(x,y)

        #if TreasureCoords.count(any item) > 3:
            #remove all copies of item from list
            #add item to bandit list

x，y表示用户当前所在的坐标.CoinCollection是一个单独的变量，当用户登陆宝箱或强盗时会发生变化。顺便说一下，坐标的数量是无限的，因为用户先前决定了在游戏里。那么，我如何复制我的坐标列表，一旦它们在列表中3次删除它们并将它们放入新的列表中？我认为这是最简单的方法来“计算”用户在广场上的次数，并将宝箱更改为强盗。

Answer 1

首先......您不需要在列表中搜索三次出现的任何坐标。这太浪费了。如果在发生这种情况后将其删除，那么唯一可能出现三次的坐标就是您刚刚访问过的坐标。

如果确实希望按照您的问题所示进行操作，请执行while (x,y) in TreasureCoords: TreasureCoords.remove((x,y))之类的操作。但是，使用dict可能会更容易。你甚至可以避免使用单独的变量。假设ItemCoords最初填充了代表未经访问的宝箱的坐标：

ItemCoords[(x,y)] = 0 # add a treasure

...然后你可以看看宝藏被访问了多少次：

visits = ItemCoords.get((x,y), None)
if visits is None:
    pass # nothing there
else if visits > 3:
    fightBandit() # ...or whatever
else:
    print("You have landed on a treasure chest! You can visit it %i more times before it becomes a bandit." % (2 - visits))
    CoinCollection += 10
    ItemCoords[(x,y)] += 1

这仍然过于简单化。你可能想要一种更加面向对象的方法，正如Christian所暗示的那样：

class Treasure:
    def __init__(x,y):
        self.x = x
        self.y = y
        self.visits = 0

    def visit():
        print("You have landed on a treasure chest! You can visit it %i more times before it becomes a bandit." % (2 - visits))    
        CoinCollection += 10

        if visits == 3:
            world[(self.x, self.y)] = Bandit(self.x, self.y)

class Bandit:
    def __init__(x,y):
        self.x = x
        self.y = y

    def visit():
        # Replace this with whatever happens when the player meets a bandit
        print("You have been eaten by a grue.")
        sys.exit(0)

# Initialize the world
for n in range(NumberOfTreasure):
    # get x,y somehow, e.g. random
    world[(x,y)] = Treasure(x,y)

# ...code for player wandering around...
here = world.get((x,y), None)
if here is not None:
    here.visit()

Answer 2

我认为这是一个对象有益的情况 - 而不是跟踪有关同一事物的事实的单独列表，使用存储所有相关信息的对象列表。

看起来像这样：

class Monster(object):
    #define like you like it
    pass

class Treasure(object):
   def init(x, y):
      self.x = x
      self.y = y
      self.visits = 0

    def visit():
        self.visits += 1

        if self.visits >= 3:
            return Monster()
        else:
            return 10


calling code:
    treasure = Treasure(1,1)
    rv = treasure.visit()
    if isinstance(rv, Monster):
        # lets the monster rip him to shreds
    else:
        CoinCollection +=treasure.visit()

Python将未知值添加到列表中

2 个答案: