有没有简单，优雅的方式来定义单身人士？
似乎有很多方法可以在Python中定义单例。 关于Stack Overflow是否有共识？
我真的没有看到需要，因为具有函数（而不是类）的模块可以很好地用作单例。 它的所有变量都将绑定到模块，无论如何都不能重复实例化。
如果你想使用一个类，就没有办法在Python中创建私有类或私有构造函数，所以你不能保护多个实例，除了通过使用API的惯例。 我仍然只是将方法放在模块中，并将模块视为单例。
这是我自己实现的单身人士。 你所要做的就是装饰课堂; 要获得单身人士，则必须使用Instance方法。 这是一个例子：
@Singleton
class Foo:
def __init__(self):
print 'Foo created'
f = Foo() # Error, this isn't how you get the instance of a singleton
f = Foo.instance() # Good. Being explicit is in line with the Python Zen
g = Foo.instance() # Returns already created instance
print f is g # True
代码如下：
class Singleton:
"""
A non-thread-safe helper class to ease implementing singletons.
This should be used as a decorator -- not a metaclass -- to the
class that should be a singleton.
The decorated class can define one `__init__` function that
takes only the `self` argument. Also, the decorated class cannot be
inherited from. Other than that, there are no restrictions that apply
to the decorated class.
To get the singleton instance, use the `instance` method. Trying
to use `__call__` will result in a `TypeError` being raised.
"""
def __init__(self, decorated):
self._decorated = decorated
def instance(self):
"""
Returns the singleton instance. Upon its first call, it creates a
new instance of the decorated class and calls its `__init__` method.
On all subsequent calls, the already created instance is returned.
"""
try:
return self._instance
except AttributeError:
self._instance = self._decorated()
return self._instance
def __call__(self):
raise TypeError('Singletons must be accessed through `instance()`.')
def __instancecheck__(self, inst):
return isinstance(inst, self._decorated)
你可以像这样覆盖__new__方法：
class Singleton(object):
_instance = None
def __new__(cls, *args, **kwargs):
if not cls._instance:
cls._instance = super(Singleton, cls).__new__(
cls, *args, **kwargs)
return cls._instance
if __name__ == '__main__':
s1 = Singleton()
s2 = Singleton()
if (id(s1) == id(s2)):
print "Same"
else:
print "Different"