Why is the Borg pattern better than the Singleton pattern in Python

Why is the Borg pattern better than the Singleton pattern in Python

The real reason that borg is different comes down to subclassing.

If you subclass a borg, the subclass objects have the same state as their parents classes objects, unless you explicitly override the shared state in that subclass. Each subclass of the singleton pattern has its own state and therefore will produce different objects.

Also in the singleton pattern the objects are actually the same, not just the state (even though the state is the only thing that really matters).

In python if you want a unique object that you can access from anywhere just create a class Unique that only contains static attributes, @staticmethods, and @classmethods; you could call it the Unique Pattern. Here I implement and compare the 3 patterns:

Unique

#Unique Pattern
class Unique:
#Define some static variables here
    x = 1
    @classmethod
    def init(cls):
        #Define any computation performed when assigning to a new object
        return cls

Singleton

#Singleton Pattern
class Singleton:

    __single = None 

    def __init__(self):
        if not Singleton.__single:
            #Your definitions here
            self.x = 1 
        else:
            raise RuntimeError(A Singleton already exists) 

    @classmethod
    def getInstance(cls):
        if not cls.__single:
            cls.__single = Singleton()
        return cls.__single

Borg

#Borg Pattern
class Borg:

    __monostate = None

    def __init__(self):
        if not Borg.__monostate:
            Borg.__monostate = self.__dict__
            #Your definitions here
            self.x = 1

        else:
            self.__dict__ = Borg.__monostate

Test

#SINGLETON
print nSINGLETONn
A = Singleton.getInstance()
B = Singleton.getInstance()

print At first B.x = {} and A.x = {}.format(B.x,A.x)
A.x = 2
print After A.x = 2
print Now both B.x = {} and A.x = {}n.format(B.x,A.x)
print  Are A and B the same object? Answer: {}.format(id(A)==id(B))


#BORG
print nBORGn
A = Borg()
B = Borg()

print At first B.x = {} and A.x = {}.format(B.x,A.x)
A.x = 2
print After A.x = 2
print Now both B.x = {} and A.x = {}n.format(B.x,A.x)
print  Are A and B the same object? Answer: {}.format(id(A)==id(B))


#UNIQUE
print nUNIQUEn
A = Unique.init()
B = Unique.init()

print At first B.x = {} and A.x = {}.format(B.x,A.x)
A.x = 2
print After A.x = 2
print Now both B.x = {} and A.x = {}n.format(B.x,A.x)
print  Are A and B the same object? Answer: {}.format(id(A)==id(B))

Output:

SINGLETON

At first B.x = 1 and A.x = 1
After A.x = 2
Now both B.x = 2 and A.x = 2

Are A and B the same object? Answer: True

BORG

At first B.x = 1 and A.x = 1
After A.x = 2
Now both B.x = 2 and A.x = 2

Are A and B the same object? Answer: False

UNIQUE

At first B.x = 1 and A.x = 1
After A.x = 2
Now both B.x = 2 and A.x = 2

Are A and B the same object? Answer: True

In my opinion, Unique implementation is the easiest, then Borg and finally Singleton with an ugly number of two functions needed for its definition.

Why is the Borg pattern better than the Singleton pattern in Python

It is not. What is generally not recommended is a pattern like this in python:

class Singleton(object):

 _instance = None

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

 @classmethod
 def instance(cls):
  if cls._instance is None:
   cls._instance = cls(...)
  return cls._instance

where you use a class method to get the instance instead of the constructor. Pythons metaprogramming allows much better methods, e.g. the one on Wikipedia:

class Singleton(type):
    def __init__(cls, name, bases, dict):
        super(Singleton, cls).__init__(name, bases, dict)
        cls.instance = None

    def __call__(cls, *args, **kw):
        if cls.instance is None:
            cls.instance = super(Singleton, cls).__call__(*args, **kw)

        return cls.instance

class MyClass(object):
    __metaclass__ = Singleton

print MyClass()
print MyClass()