Subclassing tuple with multiple __init__ arguments

Because tuples are immutable, you have to override __new__ instead:

python docs

object.__new__(cls[, ...])

Called to create a new instance of class cls. __new__() is a static method (special-cased so you need not declare it as such) that takes the class of which an instance was requested as its first argument. The remaining arguments are those passed to the object constructor expression (the call to the class). The return value of __new__() should be the new object instance (usually an instance of cls).

Typical implementations create a new instance of the class by invoking the superclass’s __new__() method using super(currentclass, cls).__new__(cls[, ...]) with appropriate arguments and then modifying the newly-created instance as necessary before returning it.

If __new__() returns an instance of cls, then the new instance’s __init__() method will be invoked like __init__(self[, ...]), where self is the new instance and the remaining arguments are the same as were passed to __new__().

If __new__() does not return an instance of cls, then the new instance’s __init__() method will not be invoked.

__new__() is intended mainly to allow subclasses of immutable types (like int, str, or tuple) to customize instance creation. It is also commonly overridden in custom metaclasses in order to customize class creation.

To assign the tuple value you need to override the __new__ method:

class Foo(tuple):

    def __new__ (cls, a, b):
        return super(Foo, cls).__new__(cls, tuple(b))

The arguments seem to be ignored by the __init__ implementation of the tuple class, but if you need to do some init stuff you can do it as follows:

class Foo(tuple):

    def __new__ (cls, a, b):
        return super(Foo, cls).__new__(cls, tuple(b))

    def __init__(self, a, b):
        self.a=a
        self.b=b

if __name__ == '__main__':
    foo = Foo(None, [3, 4])
    print foo
    print foo.a
    print foo.b