在python的同一类中创建类的新对象？

Question

Is it possible to create an new object of a class in itself (in python) ?

In Order to explain the idea more i wrote this code which i don't think that it works.

The new object should be clearly independent from the current object(new attributes etc..)

class LinkedList:

    def __init__(self):
        """ Construct an empty linked list. """
        self.first = None
        self.last = None

   def insert_before(self, new_item, next_item):
        """ Insert new_item before next_item. If next_item is None, insert
        new_item at the end of the list. """

        # First set the (two) new prev pointers (including possibly last).
        if next_item is not None:
            new_item.prev = next_item.prev
            next_item.prev = new_item
        else:
            new_item.prev = self.last
            self.last = new_item
        # Then set the (two) new next pointers (including possibly first).
        if new_item.prev is not None:
            new_item.next = next_item
            new_item.prev.next = new_item
        else:
            new_item.next = self.first
            self.first = new_item

    # assuming l1, l2 obj of class node with prev and next (attributes)
    def slice(self, l1, l2):
        curr = l1

        new = LinkedList()
        new.first = l1
        new.last = l2
        if l1.prev is not None:
           l2.prev = l1.prev
        else:
           self.first = l2.next
           self.first.prev = None

        if l2.next is not None:
           l1.next = l2.next
        else:
            self.last = l2.next
        Return new

class Node:

    def __init__(self, value):
    """ Construct an item with given value. Also have an id for each item,
    so that we can simply show pointers as ids. """

        Node.num_items += 1
        self.id = Node.num_items
        self.prev = None
        self.next = None
        self.value = value

    def __repr__(self):
    """ Item as human-readable string. In Java or C++, use a function like
    toString(). """

        return "[id = #" + str(self.id) \
           + ", prev = #" + str(0 if self.prev is None else self.prev.id) \
           + ", next = #" + str(0 if self.next is None else self.next.id) \
           + ", val = " + str(self.value) + "]"

Answer 1

Is it possible to create an new object of a class in itself (in python) ?

Yes. It is completely possible.

Here is an example:

>>> class A:
    def __init__(self, value):
        self.value = value

    def make_a(self, value):
        return A(value)

    def __repr__(self):
        return 'A({})'.format(self.value)


>>> a = A(10)
>>> a.make_a(15)
A(15)
>>> 

Now you're probably thinking, "But the A class hasn't been defined yet. Why isn't Python raising a NameError ?" The reason this works it because of the way Python executes your code.

When Python is creating the A class, specifically the make_a method, it sees that the identifier A is being called. It doesn't know whether A is a function or class, or even if A is defined . But it doesn't need to know.

What A is exactly is determined at run-time. That is the only time Python checks whether A is really defined or not.

This is also the reason why you are able to compile a function that references undefined variables:

>>> def foo():
    a + b


>>> foo # Python compiled foo...
<function foo at 0x7fb5d0500f28>
>>> foo() # but we can't call it.
Traceback (most recent call last):
  File "<pyshell#9>", line 1, in <module>
    foo() # but we can't call it.
  File "<pyshell#7>", line 2, in foo
    a + b
NameError: name 'a' is not defined
>>> 

Answer 2

That looks like a deque (double-ended queue) and the slice operation isn't quit right. You need to fixup the prev/next links on the item before l1 and the one after l2.

def slice(self, l1, l2):
    new = LinkedList()
    new.first = l1
    new.last = l2
    if l1.prev:
       l1.prev.next = l2.next
    if l2.next:
       l2.next.prev = l1.prev
    l1.prev = l2.next = None
    return new