如何通过字典修改Python类实例成员变量？代码修改无效求方案

你当前的代码里，slotDictionary存储的是self.slot1到self.slot5的字符串值副本——因为Python的字符串是不可变类型，当你把self.slot1存入字典时，实际存的是当时空字符串的值，而非对self.slot1这个属性的引用。所以修改字典里的元素，只会改变字典本身的内容，不会同步到类实例的属性上。

方法1：通过属性名字符串+setattr修改 #

把字典的value设为属性的名称字符串，然后用setattr()函数直接修改实例属性：

class Solution(object):
    def __init__(self):
        self.slot1 = ""
        self.slot2 = ""
        self.slot3 = ""
        self.slot4 = ""
        self.slot5 = ""
        # 存储属性名称而非属性值
        self.slotDictionary = {1: "slot1", 2: "slot2", 3: "slot3", 4: "slot4", 5: "slot5"}
    
    def checkAnswer(self):
        choice = int(input("What is the answer?"))
        # 用setattr修改对应属性
        setattr(self, self.slotDictionary[choice], "something")

solution = Solution()
solution.checkAnswer()
# 验证：输入1后，print(solution.slot1)会输出"something"

方法2：用可变对象包装属性值 #

可变类型（比如列表）在字典中存储的是引用，修改可变对象的内容会同步到原属性。可以把每个slot改成列表类型：

class Solution(object):
    def __init__(self):
        # 用列表包装值，列表是可变类型
        self.slot1 = [""]
        self.slot2 = [""]
        self.slot3 = [""]
        self.slot4 = [""]
        self.slot5 = [""]
        self.slotDictionary = {1: self.slot1, 2: self.slot2, 3: self.slot3, 4: self.slot4, 5: self.slot5}
    
    def checkAnswer(self):
        choice = int(input("What is the answer?"))
        # 修改列表的第一个元素，会同步到原属性
        self.slotDictionary[choice][0] = "something"

solution = Solution()
solution.checkAnswer()
# 验证：输入1后，print(solution.slot1[0])会输出"something"

方法3：直接操作实例的__dict__（不推荐但可行） #

Python实例的属性存储在__dict__字典中，你可以直接利用这个特性，不需要额外定义slotDictionary：

class Solution(object):
    def __init__(self):
        self.slot1 = ""
        self.slot2 = ""
        self.slot3 = ""
        self.slot4 = ""
        self.slot5 = ""
    
    def checkAnswer(self):
        choice = int(input("What is the answer?"))
        # 直接通过__dict__修改属性，key是属性名的字符串
        self.__dict__[f"slot{choice}"] = "something"

solution = Solution()
solution.checkAnswer()

内容的提问来源于stack exchange，提问作者Massimo Genna