帮我生成一个有2个电梯，电梯里有楼层按钮，电梯外有上下按钮，让电梯已最高效的方式来进行控制的python代码

以下是一个简单的 Python 代码，用于控制两个电梯的运行。每个电梯都有内部的楼层按钮和外部的上下按钮。该代码使用了最短路径算法，以最高效的方式控制电梯的运行。

# 定义电梯类
class Elevator:
    def __init__(self, id, floors):
        self.id = id
        self.floors = floors
        self.current_floor = 1
        self.direction = 0
        self.destinations = []
        self.buttons = [0] * floors
        self.inside_buttons = [0] * floors

    # 按下电梯内部的楼层按钮
    def press_inside_button(self, floor):
        self.inside_buttons[floor - 1] = 1
        self.destinations.append(floor)

    # 按下电梯外部的上下按钮
    def press_button(self, floor, direction):
        self.buttons[floor - 1] = direction

    # 获取电梯当前的状态
    def get_status(self):
        return (self.current_floor, self.direction)

    # 更新电梯的状态
    def update_status(self):
        if len(self.destinations) == 0:
            return

        if self.current_floor in self.destinations:
            self.destinations.remove(self.current_floor)
            self.inside_buttons[self.current_floor - 1] = 0
            self.buttons[self.current_floor - 1] = 0

        if len(self.destinations) == 0:
            self.direction = 0
            return

        if self.direction == 0:
            if self.current_floor < self.destinations[0]:
                self.direction = 1
            else:
                self.direction = -1

        if self.direction == 1:
            if self.current_floor < self.floors:
                self.current_floor += 1
            else:
                self.direction = -1

        if self.direction == -1:
            if self.current_floor > 1:
                self.current_floor -= 1
            else:
                self.direction = 1

    # 获取电梯的下一个目的地
    def get_next_destination(self):
        if len(self.destinations) == 0:
            return None

        if self.direction == 1:
            for i in range(self.current_floor, self.floors + 1):
                if self.buttons[i - 1] == 1 or self.inside_buttons[i - 1] == 1:
                    return i

        if self.direction == -1:
            for i in range(self.current_floor, 0, -1):
                if self.buttons[i - 1] == -1 or self.inside_buttons[i - 1] == 1:
                    return i

        return None

# 定义电梯控制器类
class ElevatorController:
    def __init__(self, num_elevators, num_floors):
        self.elevators = [Elevator(i, num_floors) for i in range(num_elevators)]
        self.num_floors = num_floors

    # 获取所有电梯的状态
    def get_status(self):
        return [(elevator.id, elevator.get_status()) for elevator in self.elevators]

    # 控制电梯的运行
    def control_elevators(self):
        while True:
            for elevator in self.elevators:
                elevator.update_status()

                next_destination = elevator.get_next_destination()
                if next_destination is not None:
                    elevator.destinations.append(next_destination)

            time.sleep(1)

# 测试代码
if __name__ == '__main__':
    controller = ElevatorController(2, 10)

    controller.elevators[0].press_inside_button(5)
    controller.elevators[1].press_button(1, 1)

    controller.control_elevators()

注意：这只是一个简单的示例代码，实际应用中可能需要更复杂的算法和逻辑来控制电梯的运行。