Python中对象初始化时引用未初始化对象的优化方案问询

针对你遇到的对象互相引用初始化顺序问题，这里有几种更简洁的实现方式：

1. 给Room类添加邻接房间的方法 #

给Room类加一个add_adj方法，创建完房间后直接调用方法关联，代码更连贯，还能避开可变默认参数的陷阱：

class Room(object):
    def __init__(self, name="The Room"):
        self.name = name
        self.adj = {}  # 内部初始化空字典，避免所有实例共享同一字典的问题
    
    def add_adj(self, direction, room):
        self.adj[direction] = room

# 创建房间实例
room1 = Room("Room 1")
room2 = Room("Room 2")
room3 = Room("Room 3")

# 直接调用方法关联邻接关系
room1.add_adj('n', room2)
room1.add_adj('s', room3)
room2.add_adj('s', room1)
room3.add_adj('n', room1)

2. 使用字符串名称延迟绑定 #

如果需要定义大量房间，可先用字符串指代邻接房间，最后统一解析绑定，不用纠结创建顺序：

class Room(object):
    def __init__(self, name="The Room", adj=None):
        self.name = name
        self.adj = adj or {}
        self._unresolved = adj or {}  # 暂存未解析的字符串名称

    def resolve_adj(self, room_map):
        # 把字符串名称替换为实际房间对象
        for dir, name in self._unresolved.items():
            self.adj[dir] = room_map[name]
        self._unresolved = {}

# 创建时用字符串指代邻接房间
room1 = Room("Room 1", {'n': 'Room 2', 's': 'Room 3'})
room2 = Room("Room 2", {'s': 'Room 1'})
room3 = Room("Room 3", {'n': 'Room 1'})

# 构建房间名称到实例的映射
room_map = {r.name: r for r in [room1, room2, room3]}

# 统一解析所有邻接关系
for room in room_map.values():
    room.resolve_adj(room_map)

3. 用工厂函数封装创建逻辑 #

把房间创建和关联的逻辑封装到一个函数里，外部调用一步到位：

class Room(object):
    def __init__(self, name="The Room"):
        self.name = name
        self.adj = {}

def create_linked_rooms():
    # 在函数内部完成创建和关联
    room1 = Room("Room 1")
    room2 = Room("Room 2")
    room3 = Room("Room 3")
    
    room1.adj = {'n': room2, 's': room3}
    room2.adj = {'s': room1}
    room3.adj = {'n': room1}
    
    return room1, room2, room3

# 直接获取所有关联好的房间
room1, room2, room3 = create_linked_rooms()

内容的提问来源于stack exchange，提问作者Natty