Reputation: 21
I'm working on a project where I manipulate a lot of sorted lists of elements, and I need to be able to remove any of these quickly. Since I don't need any kind of indexation, I thought that the doubly linked list structure would be best. I couldn't find any good pre-made module so I made my own:
class Node: # nodes for doubly-linked lists
def __init__(self, val, dll):
self.val = val
self.next = None
self.prev = None
self.dll = dll
class DLList: # doubly-linked lists
def __init__(self):
self.first = None
self.last = None
self.len = 0
# def __iter__(self):
# self.curr = self.first
# return self
#
# def __next__(self):
# if self.curr == None:
# raise StopIteration
# self.curr = self.curr.next
# if self.curr == None:
# raise StopIteration
# return self.curr
def append(self, val): # add a node with value val at the end of the list
node = Node(val, self)
node.prev = self.last
self.last = node
if self.first == None: # <=> if self was empty
self.first = node
self.len += 1
def appendleft(self, val): # same as previous, but at the beginning of the list
node = Node(val, self)
node.next = self.first
self.first = node
if self.last == None:
self.last = node
self.len += 1
def nodeat(self, i): # gives the ith node (starting at 0)
if i == -1:
return None
if i > self.len or i < -1:
raise IndexError('index out of range')
curr = self.first
for j in range(i):
curr = curr.next
return curr
def remove(self, node): # remove a given node in the list
if node.dll != self: #cannot remove a node that is not in the list
raise ValueError('node not in list')
p = node.prev
n = node.next
v = node.val
node.dll = None
if p != None:
p.next = n
else:
self.first = n
if n != None:
n.prev = p
else:
self.last = p
self.len -= 1
return v
def add(self, val, i): # add a node at the ith place in the list
node = Node(val, self)
if i > self.len:
raise IndexError('index out of range')
self.len += 1
previ = self.nodeat(i)
node.prev = previ.prev
node.next = previ
previ.prev = node
def clear(self): # empty the list
self.first = None
self.last = None
self.len = 0
def extend(self, iterable): # add the elements of iterable in order at the end of the list
for i in iterable:
self.append(i)
self.len += 1
def extendleft(self, iterable): # same as previous, but at the beginning (and in reverse order)
for i in iterable:
self.appendleft(i)
self.len += 1
def dll_to_list(self): # return a python list with the elements of the doubly-linked list
res = []
curr = self.first
while curr != None:
res.append(curr.val)
curr = curr.next
return res
def is_empty(self): # check whether the list is empty
return self.len == 0
Since I would lose time checking that the item I want to remove is in the list by browsing it, I added a pointer toward the list a Node is in inside the node, so that I can check that I'm not removing things from the wrong list.
Those lists are stocked in a Python dictionary, and at some point I started getting 'node not in list' errors. Does anyone know how it could appear? I never use anything but the methods listed here to manipulate the lists...
Otherwise, does anyone know about a well-coded module that I could use in place of this one?
Thanks!
Upvotes: 2
Views: 6690
Reputation: 51643
A doubly linked list has links going both directions.
Example:
def append(self, val): # add a node with value val at the end of the list
node = Node(val, self) # new node, ok
node.prev = self.last # ok, the new nodes prev is the last node of your list
self.last = node # ok, your new node is now the last of your list
if self.first == None: # yeah, ok, if its empty its also the first one now
self.first = node
self.len += 1
but ... you do not set the back-direction:
node.prev.next = node # before node.prev = self.last
Similar in your other appends. You have to always clear/reset/set all the links in both directions if you add/remove things into a doubly-linked-list:
( Red are all the changed variabels on append )
Essentially your list is not complete - if you operate / iterate on it, things will go missing in unexpected ways
Upvotes: 3