如果已经存在，则返回类实例而不是创建一个新实例

Question

我为我正在进行的一些实验室实验的结果定义了一个名为 Experiment 的类。这个想法是创建一种数据库：如果我添加一个实验，它将在退出之前被腌制到一个数据库，并在启动时重新加载（并添加到类注册表中）。

我的班级定义是：

class IterRegistry(type):
    def __iter__(cls):
        return iter(cls._registry)


class Experiment(metaclass=IterRegistry):
    _registry = []
    counter = 0

    def __init__(self, name, pathprotocol, protocol_struct, pathresult, wallA, wallB, wallC):
        hashdat = fn.hashfile(pathresult)
        hashpro = fn.hashfile(pathprotocol)
        chk = fn.checkhash(hashdat)
        if chk:
            raise RuntimeError("The same experiment has already been added")
        self._registry.append(self)
        self.name = name
        [...]

虽然fn.checkhash 是一个检查包含结果的文件的哈希值的函数：

def checkhash(hashdat):
    for exp in cl.Experiment:
        if exp.hashdat == hashdat:
            return exp
    return False

这样，如果我添加以前添加的实验，它就不会被覆盖。

如果已经存在，是否有可能以某种方式返回现有实例而不是引发错误？ （我知道在__init__ 块是不可能的）

Answer 1

如果你想自定义创建而不是仅仅在新创建的对象中初始化，你可以使用__new__：

class Experiment(metaclass=IterRegistry):
    _registry = []
    counter = 0

    def __new__(cls, name, pathprotocol, protocol_struct, pathresult, wallA, wallB, wallC):
        hashdat = fn.hashfile(pathresult)
        hashpro = fn.hashfile(pathprotocol)
        chk = fn.checkhash(hashdat)
        if chk:                      # already added, just return previous instance
            return chk
        self = object.__new__(cls)   # create a new uninitialized instance
        self._registry.append(self)  # register and initialize it
        self.name = name
        [...]
        return self                  # return the new registered instance

Answer 2

尝试这样做（非常简单的示例）：

class A:
    registry = {}

    def __init__(self, x):
        self.x = x

    @classmethod
    def create_item(cls, x):
        try:
            return cls.registry[x]
        except KeyError:
            new_item = cls(x)
            cls.registry[x] = new_item
            return new_item


A.create_item(1)
A.create_item(2)
A.create_item(2)  # doesn't add new item, but returns already existing one

Answer 3

经过四年的问题，我来到这里，Serge Ballesta 的回答帮助了我。我用更简单的语法创建了这个例子。

如果base 是None，它将始终返回创建的第一个对象。

class MyClass:
    instances = []

    def __new__(cls, base=None):
        if len(MyClass.instances) == 0:
            self = object.__new__(cls)
            MyClass.instances.append(self)

        if base is None:
            return MyClass.instances[0]
        else:
            self = object.__new__(cls)
            MyClass.instances.append(self)
            # self.__init__(base)
            return self

    def __init__(self, base=None):
        print("Received base = %s " % str(base))
        print("Number of instances = %d" % len(self.instances))
        self.base = base


R1 = MyClass("apple")
R2 = MyClass()
R3 = MyClass("banana")
R4 = MyClass()
R5 = MyClass("apple")

print(id(R1), R1.base)
print(id(R2), R2.base)
print(id(R3), R3.base)
print(id(R4), R4.base)
print(id(R5), R5.base)
print("R2 == R4 ? %s" % (R2 == R4))
print("R1 == R5 ? %s" % (R1 == R5))

它给了我们结果

Received base = apple 
Number of instances = 2
Received base = None 
Number of instances = 2
Received base = banana 
Number of instances = 3
Received base = None 
Number of instances = 3
Received base = apple 
Number of instances = 4
2167043940208 apple
2167043940256 None
2167043939968 banana
2167043940256 None
2167043939872 apple
R2 == R4 ? True
R1 == R5 ? False

很高兴知道__init__ 总是在__new__ 的return 之前被调用，即使你不调用它（在注释部分）或者你返回一个已经存在的对象。