是否可以在自身(在python中(创建一个类的新对象?
为了更多地解释这个想法,我写了这段代码,我认为它不起作用。
新对象应该明显独立于当前对象(新属性等(。
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) + "]"
是否可以在自身(在 python 中(创建一个类的新对象?
是的。这是完全可能的。
下面是一个示例:
>>> 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)
>>>
现在你可能会想,"但是A
类还没有定义。为什么Python不提高NameError
? 之所以这样做,是因为Python执行代码的方式。
当 Python 创建 A
类(特别是 make_a
方法(时,它会看到标识符A
正在被调用。它不知道A
是函数还是类,甚至不知道是否定义了A
。但它不需要知道。
A
究竟是什么在运行时确定。这是Python唯一一次检查A
是否真正定义。
这也是您能够编译引用未定义变量的函数的原因:
>>> 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
>>>
这看起来像一个双端队列,切片操作没有退出。您需要修复 l1 之前的项目和 l2 之后的项目上的上一个/下一个链接。
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