方法1：使用裝飾器（decorator）

def singleton(cls, *args, **kw):    
    instances = {}    
    def _singleton():    
        if cls not in instances:    
            instances[cls] = cls(*args, **kw)    
        return instances[cls]    
    return _singleton    
  
@singleton    
class MyClass(object):    
    a = 1    
    def __init__(self, x=0):    
        self.x = x    
    
one = MyClass()    
two = MyClass()    
    
two.a = 3    
print one.a    
#3    
print id(one)    
#29660784    
print id(two)    
#29660784    
print one == two    
#True    
print one is two    
#True    
one.x = 1    
print one.x    
#1    
print two.x  

3
35607384
35607384
True
True
1
1

方法2：使用metaclass元類來實現(xiàn)

class Singleton2(type):    
    def __init__(cls, name, bases, dict):    
        super(Singleton2, cls).__init__(name, bases, dict)    
        cls._instance = None    
    def __call__(cls, *args, **kw):    
        if cls._instance is None:    
            cls._instance = super(Singleton2, cls).__call__(*args, **kw)    
        return cls._instance    
    
class MyClass(object):    
    __metaclass__ = Singleton2    
    
one = MyClass()    
two = MyClass()    
    
two.a = 3    
print one.a    
#3    
print id(one)    
#31495472    
print id(two)    
#31495472    
print one == two    
#True    
print one is two    

3
47911488
47911488
True
True

方法3：通過共享屬性來實現(xiàn)，所謂共享屬性，最簡單直觀的方法就是通過dict屬性指向同一個字典dict

class Borg(object):    
    _state = {}    
    def __new__(cls, *args, **kw):    
        ob = super(Borg, cls).__new__(cls, *args, **kw)    
        ob.__dict__ = cls._state    
        return ob    
    
class MyClass(Borg):    
    a = 1    
    
one = MyClass()    
two = MyClass()    
    
#one和two是兩個不同的對象,id, ==, is對比結(jié)果可看出    
two.a = 3    
print one.a    
#3    
print id(one)    
#28873680    
print id(two)    
#28873712    
print one == two    
#False    
print one is two    
#False    
#但是one和two具有相同的（同一個__dict__屬性）,見:    
print id(one.__dict__)    
#30104000    
print id(two.__dict__)