Quelle est la façon la plus simple d'utiliser une liste chaînée en python? Dans le schéma, une liste chaînée est définie simplement par '(1 2 3 4 5). Les listes, [1, 2, 3, 4, 5]et les tuples de Python (1, 2, 3, 4, 5)ne sont pas, en fait, des listes liées, et les listes liées ont quelques propriétés intéressantes telles que la concaténation à temps constant et la possibilité d'en référencer des parties distinctes. Rendez-les immuables et ils sont vraiment faciles à utiliser!

Voici quelques fonctions de liste basées sur la représentation de Martin contre Löwis :

cons   = lambda el, lst: (el, lst)
mklist = lambda *args: reduce(lambda lst, el: cons(el, lst), reversed(args), None)
car = lambda lst: lst[0] if lst else lst
cdr = lambda lst: lst[1] if lst else lst
nth = lambda n, lst: nth(n-1, cdr(lst)) if n > 0 else car(lst)
length  = lambda lst, count=0: length(cdr(lst), count+1) if lst else count
begin   = lambda *args: args[-1]
display = lambda lst: begin(w("%s " % car(lst)), display(cdr(lst))) if lst else w("nil\n")

où w = sys.stdout.write

Bien que les listes à double chaînage soient utilisées dans la recette d'ensemble ordonnée de Raymond Hettinger , les listes à liaison simple n'ont aucune valeur pratique en Python.

Je n'ai jamais utilisé une seule liste chaînée en Python pour aucun problème sauf éducatif.

Thomas Watnedal a suggéré une bonne ressource pédagogique Comment penser comme un informaticien, Chapitre 17: Listes liées :

Une liste chaînée est soit:

• la liste vide, représentée par Aucun, ou

• un nœud qui contient un objet cargo et une référence à une liste liée.

  class Node: 
    def __init__(self, cargo=None, next=None): 
      self.car = cargo 
      self.cdr = next    
    def __str__(self): 
      return str(self.car)
  
  def display(lst):
    if lst:
      w("%s " % lst)
      display(lst.cdr)
    else:
      w("nil\n")
  

Pour certains besoins, un deque peut également être utile. Vous pouvez ajouter et supprimer des éléments aux deux extrémités d'un deque au coût O (1).

from collections import deque
d = deque([1,2,3,4])

print d
for x in d:
    print x
print d.pop(), d

J'ai écrit ça l'autre jour

#! /usr/bin/env python

class Node(object):
    def __init__(self):
        self.data = None # contains the data
        self.next = None # contains the reference to the next node


class LinkedList:
    def __init__(self):
        self.cur_node = None

    def add_node(self, data):
        new_node = Node() # create a new node
        new_node.data = data
        new_node.next = self.cur_node # link the new node to the 'previous' node.
        self.cur_node = new_node #  set the current node to the new one.

    def list_print(self):
        node = self.cur_node # cant point to ll!
        while node:
            print node.data
            node = node.next



ll = LinkedList()
ll.add_node(1)
ll.add_node(2)
ll.add_node(3)

ll.list_print()

La réponse acceptée est plutôt compliquée. Voici une conception plus standard:

L = LinkedList()
L.insert(1)
L.insert(1)
L.insert(2)
L.insert(4)
print L
L.clear()
print L

C'est une LinkedListclasse simple basée sur la conception C ++ simple et le chapitre 17: Listes liées , comme recommandé par Thomas Watnedal .

class Node:
    def __init__(self, value = None, next = None):
        self.value = value
        self.next = next

    def __str__(self):
        return 'Node ['+str(self.value)+']'

class LinkedList:
    def __init__(self):
        self.first = None
        self.last = None

    def insert(self, x):
        if self.first == None:
            self.first = Node(x, None)
            self.last = self.first
        elif self.last == self.first:
            self.last = Node(x, None)
            self.first.next = self.last
        else:
            current = Node(x, None)
            self.last.next = current
            self.last = current

    def __str__(self):
        if self.first != None:
            current = self.first
            out = 'LinkedList [\n' +str(current.value) +'\n'
            while current.next != None:
                current = current.next
                out += str(current.value) + '\n'
            return out + ']'
        return 'LinkedList []'

    def clear(self):
        self.__init__()

Les listes immuables sont mieux représentées par deux tuples, aucun représentant NIL. Pour permettre la formulation simple de telles listes, vous pouvez utiliser cette fonction:

def mklist(*args):
    result = None
    for element in reversed(args):
        result = (element, result)
    return result

Pour travailler avec de telles listes, je préfère fournir toute la collection de fonctions LISP (c'est-à-dire premier, deuxième, nième, etc.), plutôt que d'introduire des méthodes.

Voici une version légèrement plus complexe d'une classe de liste chaînée, avec une interface similaire aux types de séquence de python (c'est-à-dire prend en charge l'indexation, le découpage, la concaténation avec des séquences arbitraires, etc.). Il devrait avoir O (1) en préfixe, ne copie pas les données sauf si nécessaire et peut être utilisé de manière interchangeable avec les tuples.

Ce ne sera pas aussi efficace en termes d'espace ou de temps que les cellules lisp contre, car les classes python sont évidemment un peu plus lourdes (vous pouvez améliorer légèrement les choses avec " __slots__ = '_head','_tail'" pour réduire l'utilisation de la mémoire). Il aura cependant les caractéristiques de performances souhaitées en gros O.

Exemple d'utilisation:

>>> l = LinkedList([1,2,3,4])
>>> l
LinkedList([1, 2, 3, 4])
>>> l.head, l.tail
(1, LinkedList([2, 3, 4]))

# Prepending is O(1) and can be done with:
LinkedList.cons(0, l)
LinkedList([0, 1, 2, 3, 4])
# Or prepending arbitrary sequences (Still no copy of l performed):
[-1,0] + l
LinkedList([-1, 0, 1, 2, 3, 4])

# Normal list indexing and slice operations can be performed.
# Again, no copy is made unless needed.
>>> l[1], l[-1], l[2:]
(2, 4, LinkedList([3, 4]))
>>> assert l[2:] is l.next.next

# For cases where the slice stops before the end, or uses a
# non-contiguous range, we do need to create a copy.  However
# this should be transparent to the user.
>>> LinkedList(range(100))[-10::2]
LinkedList([90, 92, 94, 96, 98])

La mise en oeuvre:

import itertools

class LinkedList(object):
    """Immutable linked list class."""

    def __new__(cls, l=[]):
        if isinstance(l, LinkedList): return l # Immutable, so no copy needed.
        i = iter(l)
        try:
            head = i.next()
        except StopIteration:
            return cls.EmptyList   # Return empty list singleton.

        tail = LinkedList(i)

        obj = super(LinkedList, cls).__new__(cls)
        obj._head = head
        obj._tail = tail
        return obj

    @classmethod
    def cons(cls, head, tail):
        ll =  cls([head])
        if not isinstance(tail, cls):
            tail = cls(tail)
        ll._tail = tail
        return ll

    # head and tail are not modifiable
    @property  
    def head(self): return self._head

    @property
    def tail(self): return self._tail

    def __nonzero__(self): return True

    def __len__(self):
        return sum(1 for _ in self)

    def __add__(self, other):
        other = LinkedList(other)

        if not self: return other   # () + l = l
        start=l = LinkedList(iter(self))  # Create copy, as we'll mutate

        while l:
            if not l._tail: # Last element?
                l._tail = other
                break
            l = l._tail
        return start

    def __radd__(self, other):
        return LinkedList(other) + self

    def __iter__(self):
        x=self
        while x:
            yield x.head
            x=x.tail

    def __getitem__(self, idx):
        """Get item at specified index"""
        if isinstance(idx, slice):
            # Special case: Avoid constructing a new list, or performing O(n) length 
            # calculation for slices like l[3:].  Since we're immutable, just return
            # the appropriate node. This becomes O(start) rather than O(n).
            # We can't do this for  more complicated slices however (eg [l:4]
            start = idx.start or 0
            if (start >= 0) and (idx.stop is None) and (idx.step is None or idx.step == 1):
                no_copy_needed=True
            else:
                length = len(self)  # Need to calc length.
                start, stop, step = idx.indices(length)
                no_copy_needed = (stop == length) and (step == 1)

            if no_copy_needed:
                l = self
                for i in range(start): 
                    if not l: break # End of list.
                    l=l.tail
                return l
            else:
                # We need to construct a new list.
                if step < 1:  # Need to instantiate list to deal with -ve step
                    return LinkedList(list(self)[start:stop:step])
                else:
                    return LinkedList(itertools.islice(iter(self), start, stop, step))
        else:       
            # Non-slice index.
            if idx < 0: idx = len(self)+idx
            if not self: raise IndexError("list index out of range")
            if idx == 0: return self.head
            return self.tail[idx-1]

    def __mul__(self, n):
        if n <= 0: return Nil
        l=self
        for i in range(n-1): l += self
        return l
    def __rmul__(self, n): return self * n

    # Ideally we should compute the has ourselves rather than construct
    # a temporary tuple as below.  I haven't impemented this here
    def __hash__(self): return hash(tuple(self))

    def __eq__(self, other): return self._cmp(other) == 0
    def __ne__(self, other): return not self == other
    def __lt__(self, other): return self._cmp(other) < 0
    def __gt__(self, other): return self._cmp(other) > 0
    def __le__(self, other): return self._cmp(other) <= 0
    def __ge__(self, other): return self._cmp(other) >= 0

    def _cmp(self, other):
        """Acts as cmp(): -1 for self<other, 0 for equal, 1 for greater"""
        if not isinstance(other, LinkedList):
            return cmp(LinkedList,type(other))  # Arbitrary ordering.

        A, B = iter(self), iter(other)
        for a,b in itertools.izip(A,B):
           if a<b: return -1
           elif a > b: return 1

        try:
            A.next()
            return 1  # a has more items.
        except StopIteration: pass

        try:
            B.next()
            return -1  # b has more items.
        except StopIteration: pass

        return 0  # Lists are equal

    def __repr__(self):
        return "LinkedList([%s])" % ', '.join(map(repr,self))

class EmptyList(LinkedList):
    """A singleton representing an empty list."""
    def __new__(cls):
        return object.__new__(cls)

    def __iter__(self): return iter([])
    def __nonzero__(self): return False

    @property
    def head(self): raise IndexError("End of list")

    @property
    def tail(self): raise IndexError("End of list")

# Create EmptyList singleton
LinkedList.EmptyList = EmptyList()
del EmptyList

llist - types de données de listes liées pour Python

Le module llist implémente des structures de données de listes liées. Il prend en charge une liste doublement liée, c'est dllist-à- dire et une structure de données liée individuellementsllist .

objets dllist

Cet objet représente une structure de données de liste doublement liée.

first

Premier dllistnodeobjet de la liste. Nonesi la liste est vide.

last

Dernier dllistnode objet de la liste. Aucun si la liste est vide.

Les objets dllist prennent également en charge les méthodes suivantes:

append(x)

Ajouter xà droite de la liste et retourner inséré dllistnode.

appendleft(x)

Ajouter xau côté gauche de la liste et retourner inséré dllistnode.

appendright(x)

Ajouter xà droite de la liste et retourner inséré dllistnode.

clear()

Supprimez tous les nœuds de la liste.

extend(iterable)

Ajoutez des éléments du iterablecôté droit de la liste.

extendleft(iterable)

Ajoutez des éléments du iterablecôté gauche de la liste.

extendright(iterable)

Ajoutez des éléments du iterablecôté droit de la liste.

insert(x[, before])

Ajouter xà droite de la liste si beforen'est pas spécifié, ou insérer xà gauche de dllistnode before. Retour inséré dllistnode.

nodeat(index)

Nœud de retour (de type dllistnode) à index.

pop()

Supprimez et renvoyez la valeur d'un élément du côté droit de la liste.

popleft()

Supprimez et renvoyez la valeur d'un élément du côté gauche de la liste.

popright()

Supprimer et renvoyer la valeur d'un élément du côté droit de la liste

remove(node)

Supprimer nodede la liste et renvoyer l'élément qui y était stocké.

dllistnode objets

classe llist.dllistnode([value])

Renvoie un nouveau nœud de liste doublement lié, initialisé (éventuellement) avec value.

dllistnode les objets fournissent les attributs suivants:

next

Nœud suivant dans la liste. Cet attribut est en lecture seule.

prev

Noeud précédent dans la liste. Cet attribut est en lecture seule.

value

Valeur stockée dans ce nœud. Compilé à partir de cette référence

sllist

class llist.sllist([iterable]) Renvoie une nouvelle liste liée individuellement initialisée avec des éléments de iterable. Si iterable n'est pas spécifié, le nouveau sllistest vide.

Un ensemble similaire d'attributs et d'opérations est défini pour cet sllistobjet. Consultez cette référence pour plus d'informations.

class Node(object):
    def __init__(self, data=None, next=None):
        self.data = data
        self.next = next

    def setData(self, data):
        self.data = data
        return self.data

    def setNext(self, next):
        self.next = next

    def getNext(self):
        return self.next

    def hasNext(self):
        return self.next != None


class singleLinkList(object):

    def __init__(self):
        self.head = None

    def isEmpty(self):
        return self.head == None

    def insertAtBeginning(self, data):
        newNode = Node()
        newNode.setData(data)

        if self.listLength() == 0:
            self.head = newNode
        else:
            newNode.setNext(self.head)
            self.head = newNode

    def insertAtEnd(self, data):
        newNode = Node()
        newNode.setData(data)

        current = self.head

        while current.getNext() != None:
            current = current.getNext()

        current.setNext(newNode)

    def listLength(self):
        current = self.head
        count = 0

        while current != None:
            count += 1
            current = current.getNext()
        return count

    def print_llist(self):
        current = self.head
        print("List Start.")
        while current != None:
            print(current.getData())
            current = current.getNext()

        print("List End.")



if __name__ == '__main__':
    ll = singleLinkList()
    ll.insertAtBeginning(55)
    ll.insertAtEnd(56)
    ll.print_llist()
    print(ll.listLength())

J'ai basé cette fonction supplémentaire sur Nick Stinemates

def add_node_at_end(self, data):
    new_node = Node()
    node = self.curr_node
    while node:
        if node.next == None:
            node.next = new_node
            new_node.next = None
            new_node.data = data
        node = node.next

La méthode qu'il a ajoute le nouveau nœud au début alors que j'ai vu beaucoup d'implémentations qui ajoutent généralement un nouveau nœud à la fin mais quoi qu'il en soit, c'est amusant à faire.

Voici ce que j'ai trouvé. C'est similaire à celui de Riccardo C. , dans ce fil, sauf qu'il imprime les nombres dans l'ordre au lieu de l'inverse. J'ai également fait de l'objet LinkedList un itérateur Python afin d'imprimer la liste comme vous le feriez pour une liste Python normale.

class Node:

    def __init__(self, data=None):
        self.data = data
        self.next = None

    def __str__(self):
        return str(self.data)


class LinkedList:

    def __init__(self):
        self.head = None
        self.curr = None
        self.tail = None

    def __iter__(self):
        return self

    def next(self):
        if self.head and not self.curr:
            self.curr = self.head
            return self.curr
        elif self.curr.next:
            self.curr = self.curr.next
            return self.curr
        else:
            raise StopIteration

    def append(self, data):
        n = Node(data)
        if not self.head:
            self.head = n
            self.tail = n
        else:
            self.tail.next = n
            self.tail = self.tail.next


# Add 5 nodes
ll = LinkedList()
for i in range(1, 6):
    ll.append(i)

# print out the list
for n in ll:
    print n

"""
Example output:
$ python linked_list.py
1
2
3
4
5
"""

Je viens de faire ça comme un jouet amusant. Il doit être immuable tant que vous ne touchez pas aux méthodes avec préfixe de soulignement et implémente un tas de magie Python comme l'indexation et len.

Lorsque vous utilisez des listes liées immuables, pensez à utiliser directement le tuple de Python.

ls = (1, 2, 3, 4, 5)

def first(ls): return ls[0]
def rest(ls): return ls[1:]

C'est vraiment cette facilité, et vous pouvez garder les fonctions supplémentaires comme len (ls), x dans ls, etc.

class LL(object):
    def __init__(self,val):
        self.val = val
        self.next = None

    def pushNodeEnd(self,top,val):
        if top is None:
            top.val=val
            top.next=None
        else:
            tmp=top
            while (tmp.next != None):
                tmp=tmp.next        
            newNode=LL(val)
            newNode.next=None
            tmp.next=newNode

    def pushNodeFront(self,top,val):
        if top is None:
            top.val=val
            top.next=None
        else:
            newNode=LL(val)
            newNode.next=top
            top=newNode

    def popNodeFront(self,top):
        if top is None:
            return
        else:
            sav=top
            top=top.next
        return sav

    def popNodeEnd(self,top):
        if top is None:
            return
        else:
            tmp=top
            while (tmp.next != None):
                prev=tmp
                tmp=tmp.next
            prev.next=None
        return tmp

top=LL(10)
top.pushNodeEnd(top, 20)
top.pushNodeEnd(top, 30)
pop=top.popNodeEnd(top)
print (pop.val)

J'ai mis une classe de liste Python 2.x et 3.x à un seul lien sur https://pypi.python.org/pypi/linked_list_mod/

Il est testé avec CPython 2.7, CPython 3.4, Pypy 2.3.1, Pypy3 2.3.1 et Jython 2.7b2, et est livré avec une belle suite de tests automatisés.

Il comprend également des classes LIFO et FIFO.

Ils ne sont cependant pas immuables.

class LinkedStack:
'''LIFO Stack implementation using a singly linked list for storage.'''

_ToList = []

#---------- nested _Node class -----------------------------
class _Node:
    '''Lightweight, nonpublic class for storing a singly linked node.'''
    __slots__ = '_element', '_next'     #streamline memory usage

    def __init__(self, element, next):
        self._element = element
        self._next = next

#--------------- stack methods ---------------------------------
def __init__(self):
    '''Create an empty stack.'''
    self._head = None
    self._size = 0

def __len__(self):
    '''Return the number of elements in the stack.'''
    return self._size

def IsEmpty(self):
    '''Return True if the stack is empty'''
    return  self._size == 0

def Push(self,e):
    '''Add element e to the top of the Stack.'''
    self._head = self._Node(e, self._head)      #create and link a new node
    self._size +=1
    self._ToList.append(e)

def Top(self):
    '''Return (but do not remove) the element at the top of the stack.
       Raise exception if the stack is empty
    '''

    if self.IsEmpty():
        raise Exception('Stack is empty')
    return  self._head._element             #top of stack is at head of list

def Pop(self):
    '''Remove and return the element from the top of the stack (i.e. LIFO).
       Raise exception if the stack is empty
    '''
    if self.IsEmpty():
        raise Exception('Stack is empty')
    answer = self._head._element
    self._head = self._head._next       #bypass the former top node
    self._size -=1
    self._ToList.remove(answer)
    return answer

def Count(self):
    '''Return how many nodes the stack has'''
    return self.__len__()

def Clear(self):
    '''Delete all nodes'''
    for i in range(self.Count()):
        self.Pop()

def ToList(self):
    return self._ToList

Classe de liste liée

class LinkedStack:
# Nested Node Class
class Node:
    def __init__(self, element, next):
        self.__element = element
        self.__next = next

    def get_next(self):
        return self.__next

    def get_element(self):
        return self.__element

def __init__(self):
    self.head = None
    self.size = 0
    self.data = []

def __len__(self):
    return self.size

def __str__(self):
    return str(self.data)

def is_empty(self):
    return self.size == 0

def push(self, e):
    newest = self.Node(e, self.head)
    self.head = newest
    self.size += 1
    self.data.append(newest)

def top(self):
    if self.is_empty():
        raise Empty('Stack is empty')
    return self.head.__element

def pop(self):
    if self.is_empty():
        raise Empty('Stack is empty')
    answer = self.head.element
    self.head = self.head.next
    self.size -= 1
    return answer

Usage

from LinkedStack import LinkedStack

x = LinkedStack()

x.push(10)
x.push(25)
x.push(55)


for i in range(x.size - 1, -1, -1):

    print '|', x.data[i].get_element(), '|' ,
    #next object

    if x.data[i].get_next() == None:
        print '--> None'
    else:
        print  x.data[i].get_next().get_element(), '-|---->  ',

Production

| 55 | 25 -|---->   | 25 | 10 -|---->   | 10 | --> None

Voici ma mise en œuvre simple:

class Node:
    def __init__(self):
        self.data = None
        self.next = None
    def __str__(self):
        return "Data %s: Next -> %s"%(self.data, self.next)

class LinkedList:
    def __init__(self):
        self.head = Node()
        self.curNode = self.head
    def insertNode(self, data):
        node = Node()
        node.data = data
        node.next = None
        if self.head.data == None:
            self.head = node
            self.curNode = node
        else:
            self.curNode.next = node
            self.curNode = node
    def printList(self):
        print self.head

l = LinkedList()
l.insertNode(1)
l.insertNode(2)
l.insertNode(34)

Production:

Data 1: Next -> Data 2: Next -> Data 34: Next -> Data 4: Next -> None

Voici ma solution:

la mise en oeuvre

class Node:
  def __init__(self, initdata):
    self.data = initdata
    self.next = None

  def get_data(self):
    return self.data

  def set_data(self, data):
    self.data = data

  def get_next(self):
    return self.next

  def set_next(self, node):
    self.next = node


# ------------------------ Link List class ------------------------------- #
class LinkList:

  def __init__(self):
    self.head = None

  def is_empty(self):
    return self.head == None

  def traversal(self, data=None):
    node = self.head
    index = 0
    found = False
    while node is not None and not found:
      if node.get_data() == data:
        found = True
      else:
        node = node.get_next()
        index += 1
    return (node, index)

  def size(self):
    _, count = self.traversal(None)
    return count

  def search(self, data):
    node, _ = self.traversal(data)
    return node

  def add(self, data):
    node = Node(data)
    node.set_next(self.head)
    self.head = node

  def remove(self, data):
    previous_node = None
    current_node = self.head
    found = False
    while current_node is not None and not found:
      if current_node.get_data() == data:
        found = True
        if previous_node:
          previous_node.set_next(current_node.get_next())
        else:
          self.head = current_node
      else:
        previous_node = current_node
        current_node = current_node.get_next()
    return found

Usage

link_list = LinkList()
link_list.add(10)
link_list.add(20)
link_list.add(30)
link_list.add(40)
link_list.add(50)
link_list.size()
link_list.search(30)
link_list.remove(20)

Idée de mise en œuvre originale

http://interactivepython.org/runestone/static/pythonds/BasicDS/ImplementinganUnorderedListLinkedLists.html

Je pense que la mise en œuvre ci-dessous remplit le projet de loi avec grâce.

'''singly linked lists, by Yingjie Lan, December 1st, 2011'''

class linkst:
    '''Singly linked list, with pythonic features.
The list has pointers to both the first and the last node.'''
    __slots__ = ['data', 'next'] #memory efficient
    def __init__(self, iterable=(), data=None, next=None):
        '''Provide an iterable to make a singly linked list.
Set iterable to None to make a data node for internal use.'''
        if iterable is not None: 
            self.data, self.next = self, None
            self.extend(iterable)
        else: #a common node
            self.data, self.next = data, next

    def empty(self):
        '''test if the list is empty'''
        return self.next is None

    def append(self, data):
        '''append to the end of list.'''
        last = self.data
        self.data = last.next = linkst(None, data)
        #self.data = last.next

    def insert(self, data, index=0):
        '''insert data before index.
Raise IndexError if index is out of range'''
        curr, cat = self, 0
        while cat < index and curr:
            curr, cat = curr.next, cat+1
        if index<0 or not curr:
            raise IndexError(index)
        new = linkst(None, data, curr.next)
        if curr.next is None: self.data = new
        curr.next = new

    def reverse(self):
        '''reverse the order of list in place'''
        current, prev = self.next, None
        while current: #what if list is empty?
            next = current.next
            current.next = prev
            prev, current = current, next
        if self.next: self.data = self.next
        self.next = prev

    def delete(self, index=0):
        '''remvoe the item at index from the list'''
        curr, cat = self, 0
        while cat < index and curr.next:
            curr, cat = curr.next, cat+1
        if index<0 or not curr.next:
            raise IndexError(index)
        curr.next = curr.next.next
        if curr.next is None: #tail
            self.data = curr #current == self?

    def remove(self, data):
        '''remove first occurrence of data.
Raises ValueError if the data is not present.'''
        current = self
        while current.next: #node to be examined
            if data == current.next.data: break
            current = current.next #move on
        else: raise ValueError(data)
        current.next = current.next.next
        if current.next is None: #tail
            self.data = current #current == self?

    def __contains__(self, data):
        '''membership test using keyword 'in'.'''
        current = self.next
        while current:
            if data == current.data:
                return True
            current = current.next
        return False

    def __iter__(self):
        '''iterate through list by for-statements.
return an iterator that must define the __next__ method.'''
        itr = linkst()
        itr.next = self.next
        return itr #invariance: itr.data == itr

    def __next__(self):
        '''the for-statement depends on this method
to provide items one by one in the list.
return the next data, and move on.'''
        #the invariance is checked so that a linked list
        #will not be mistakenly iterated over
        if self.data is not self or self.next is None:
            raise StopIteration()
        next = self.next
        self.next = next.next
        return next.data

    def __repr__(self):
        '''string representation of the list'''
        return 'linkst(%r)'%list(self)

    def __str__(self):
        '''converting the list to a string'''
        return '->'.join(str(i) for i in self)

    #note: this is NOT the class lab! see file linked.py.
    def extend(self, iterable):
        '''takes an iterable, and append all items in the iterable
to the end of the list self.'''
        last = self.data
        for i in iterable:
            last.next = linkst(None, i)
            last = last.next
        self.data = last

    def index(self, data):
        '''TODO: return first index of data in the list self.
    Raises ValueError if the value is not present.'''
        #must not convert self to a tuple or any other containers
        current, idx = self.next, 0
        while current:
            if current.data == data: return idx
            current, idx = current.next, idx+1
        raise ValueError(data)

class LinkedList:
    def __init__(self, value):
        self.value = value
        self.next = None

    def insert(self, node):
        if not self.next:
            self.next = node
        else:
            self.next.insert(node)

    def __str__(self):
        if self.next:
            return '%s -> %s' % (self.value, str(self.next))
        else:
            return ' %s ' % self.value

if __name__ == "__main__":
    items = ['a', 'b', 'c', 'd', 'e']    
    ll = None
    for item in items:
        if ll:
            next_ll = LinkedList(item)
            ll.insert(next_ll)
        else:
            ll = LinkedList(item)
    print('[ %s ]' % ll)

Tout d'abord, je suppose que vous voulez des listes chaînées. En pratique, vous pouvez utiliser collections.deque, dont l'implémentation CPython actuelle est une liste de blocs doublement liés (chaque bloc contient un tableau de 62 objets cargo). Il englobe la fonctionnalité de la liste liée. Vous pouvez également rechercher une extension C appelée llistsur pypi. Si vous voulez une implémentation pure-Python et facile à suivre de la liste chaînée ADT, vous pouvez jeter un œil à mon implémentation minimale suivante.

class Node (object):
    """ Node for a linked list. """
    def __init__ (self, value, next=None):
        self.value = value
        self.next = next

class LinkedList (object):
    """ Linked list ADT implementation using class. 
        A linked list is a wrapper of a head pointer
        that references either None, or a node that contains 
        a reference to a linked list.
    """
    def __init__ (self, iterable=()):
        self.head = None
        for x in iterable:
            self.head = Node(x, self.head)

    def __iter__ (self):
        p = self.head
        while p is not None:
            yield p.value
            p = p.next

    def prepend (self, x):  # 'appendleft'
        self.head = Node(x, self.head)

    def reverse (self):
        """ In-place reversal. """
        p = self.head
        self.head = None
        while p is not None:
            p0, p = p, p.next
            p0.next = self.head
            self.head = p0

if __name__ == '__main__':
    ll = LinkedList([6,5,4])
    ll.prepend(3); ll.prepend(2)
    print list(ll)
    ll.reverse()
    print list(ll)

Exemple de liste doublement liée (enregistrer sous linkedlist.py):

class node:
    def __init__(self, before=None, cargo=None, next=None): 
        self._previous = before
        self._cargo = cargo 
        self._next  = next 

    def __str__(self):
        return str(self._cargo) or None 

class linkedList:
    def __init__(self): 
        self._head = None 
        self._length = 0

    def add(self, cargo):
        n = node(None, cargo, self._head)
        if self._head:
            self._head._previous = n
        self._head = n
        self._length += 1

    def search(self,cargo):
        node = self._head
        while (node and node._cargo != cargo):
            node = node._next
        return node

    def delete(self,cargo):
        node = self.search(cargo)
        if node:
            prev = node._previous
            nx = node._next
            if prev:
                prev._next = node._next
            else:
                self._head = nx
                nx._previous = None
            if nx:
                nx._previous = prev 
            else:
                prev._next = None
        self._length -= 1

    def __str__(self):
        print 'Size of linked list: ',self._length
        node = self._head
        while node:
            print node
            node = node._next

Test (enregistrer sous test.py):

from linkedlist import node, linkedList

def test():

    print 'Testing Linked List'

    l = linkedList()

    l.add(10)
    l.add(20)
    l.add(30)
    l.add(40)
    l.add(50)
    l.add(60)

    print 'Linked List after insert nodes:'
    l.__str__()

    print 'Search some value, 30:'
    node = l.search(30)
    print node

    print 'Delete some value, 30:'
    node = l.delete(30)
    l.__str__()

    print 'Delete first element, 60:'
    node = l.delete(60)
    l.__str__()

    print 'Delete last element, 10:'
    node = l.delete(10)
    l.__str__()


if __name__ == "__main__":
    test()

Sortie :

Testing Linked List
Linked List after insert nodes:
Size of linked list:  6
60
50
40
30
20
10
Search some value, 30:
30
Delete some value, 30:
Size of linked list:  5
60
50
40
20
10
Delete first element, 60:
Size of linked list:  4
50
40
20
10
Delete last element, 10:
Size of linked list:  3
50
40
20

J'ai également écrit une liste liée unique basée sur un didacticiel, qui contient les deux classes de base Node et Linked List, et quelques méthodes supplémentaires pour l'insertion, la suppression, l'inversion, le tri, etc.

Ce n'est ni le meilleur ni le plus simple, mais ça marche bien.

"""
🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏

Single Linked List (SLL):
A simple object-oriented implementation of Single Linked List (SLL) 
with some associated methods, such as create list, count nodes, delete nodes, and such. 

🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏
"""

class Node:
    """
    Instantiates a node
    """
    def __init__(self, value):
        """
        Node class constructor which sets the value and link of the node

        """
        self.info = value
        self.link = None

class SingleLinkedList:
    """
    Instantiates the SLL class
    """
    def __init__(self):
        """
        SLL class constructor which sets the value and link of the node

        """
        self.start = None

    def create_single_linked_list(self):
        """
        Reads values from stdin and appends them to this list and creates a SLL with integer nodes

        """
        try:
            number_of_nodes = int(input("👉   Enter a positive integer between 1-50 for the number of nodes you wish to have in the list: "))
            if number_of_nodes <= 0 or number_of_nodes > 51:
                print("💛 The number of nodes though must be an integer between 1 to 50!")
                self.create_single_linked_list()

        except Exception as e:
            print("💛 Error: ", e)
            self.create_single_linked_list()


        try:
            for _ in range(number_of_nodes):
                try:
                    data = int(input("👉   Enter an integer for the node to be inserted: "))
                    self.insert_node_at_end(data)
                except Exception as e:
                    print("💛 Error: ", e)
        except Exception as e:
            print("💛 Error: ", e)

    def count_sll_nodes(self):
        """
        Counts the nodes of the linked list

        """
        try:
            p = self.start
            n = 0
            while p is not None:
                n += 1
                p = p.link

            if n >= 1:
                print(f"💚 The number of nodes in the linked list is {n}")
            else:
                print(f"💛 The SLL does not have a node!")
        except Exception as e: 
            print("💛 Error: ", e)

    def search_sll_nodes(self, x):
        """
        Searches the x integer in the linked list
        """
        try:
            position =  1
            p = self.start
            while p is not None:
                if p.info == x:
                    print(f"💚 YAAAY! We found {x} at position {position}")
                    return True

                #Increment the position
                position += 1 
                #Assign the next node to the current node
                p = p.link
            else:
                print(f"💔 Sorry! We couldn't find {x} at any position. Maybe, you might want to use option 9 and try again later!")
                return False
        except Exception as e:
            print("💛 Error: ", e)

    def display_sll(self):
        """
        Displays the list
        """
        try:
            if self.start is None:
                print("💛 Single linked list is empty!")
                return

            display_sll = "💚 Single linked list nodes are: "
            p = self.start
            while p is not None:
                display_sll += str(p.info) + "\t"
                p = p.link

            print(display_sll)

        except Exception as e:
            print("💛 Error: ", e) 

    def insert_node_in_beginning(self, data):
        """
        Inserts an integer in the beginning of the linked list

        """
        try:
            temp = Node(data)
            temp.link = self.start
            self.start = temp
        except Exception as e:
            print("💛 Error: ", e)

    def insert_node_at_end(self, data):
        """
        Inserts an integer at the end of the linked list

        """
        try:            
            temp = Node(data)
            if self.start is None:
                self.start = temp
                return

            p = self.start  
            while p.link is not None:
                p = p.link
            p.link = temp
        except Exception as e:
            print("💛 Error: ", e)


    def insert_node_after_another(self, data, x):
        """
        Inserts an integer after the x node

        """
        try:
            p = self.start

            while p is not None:
                if p.info == x:
                    break
                p = p.link

            if p is None:
                print(f"💔 Sorry! {x} is not in the list.")
            else:
                temp = Node(data)
                temp.link = p.link
                p.link = temp
        except Exception as e: 
            print("💛 Error: ", e)


    def insert_node_before_another(self, data, x):
        """
        Inserts an integer before the x node

        """

        try:

            # If list is empty
            if self.start is None:
                print("💔 Sorry! The list is empty.")
                return 
            # If x is the first node, and new node should be inserted before the first node
            if x == self.start.info:
                temp = Node(data)
                temp.link = p.link
                p.link = temp

            # Finding the reference to the prior node containing x
            p = self.start
            while p.link is not None:
                if p.link.info == x:
                    break
                p = p.link

            if p.link is not None:
                print(f"💔 Sorry! {x} is not in the list.")
            else:
                temp = Node(data)
                temp.link = p.link
                p.link = temp           

        except Exception as e:
            print("💛 Error: ", e)

    def insert_node_at_position(self, data, k):
        """
        Inserts an integer in k position of the linked list

        """
        try:
            # if we wish to insert at the first node
            if k == 1:
                temp = Node(data)
                temp.link = self.start
                self.start = temp
                return

            p = self.start
            i = 1

            while i < k-1 and p is not None:
                p = p.link
                i += 1

            if p is None:
                print(f"💛 The max position is {i}") 
            else:    
                temp = Node(data)
                temp.link = self.start
                self.start = temp

        except Exception as e:
            print("💛 Error: ", e)

    def delete_a_node(self, x):
        """
        Deletes a node of a linked list

        """
        try:
            # If list is empty
            if self.start is None:
                print("💔 Sorry! The list is empty.")
                return

            # If there is only one node
            if self.start.info == x:
                self.start = self.start.link

            # If more than one node exists
            p = self.start
            while p.link is not None:
                if p.link.info == x:
                    break   
                p = p.link

            if p.link is None:
                print(f"💔 Sorry! {x} is not in the list.")
            else:
                p.link = p.link.link

        except Exception as e:
            print("💛 Error: ", e)

    def delete_sll_first_node(self):
        """
        Deletes the first node of a linked list

        """
        try:
            if self.start is None:
                return
            self.start = self.start.link

        except Exception as e:
            print("💛 Error: ", e)


    def delete_sll_last_node(self):
        """
        Deletes the last node of a linked list

        """
        try:

            # If the list is empty
            if self.start is None:
                return

            # If there is only one node
            if self.start.link is None:
                self.start = None
                return

            # If there is more than one node    
            p = self.start

            # Increment until we find the node prior to the last node 
            while p.link.link is not None:
                p = p.link

            p.link = None   

        except Exception as e:
            print("💛 Error: ", e)


    def reverse_sll(self):
        """
        Reverses the linked list

        """

        try:

            prev = None
            p = self.start
            while p is not None:
                next = p.link
                p.link = prev
                prev = p
                p = next
            self.start = prev

        except Exception as e:
            print("💛 Error: ", e)


    def bubble_sort_sll_nodes_data(self):
        """
        Bubble sorts the linked list on integer values

        """

        try:

            # If the list is empty or there is only one node
            if self.start is None or self.start.link is None:
                print("💛 The list has no or only one node and sorting is not required.")
            end = None

            while end != self.start.link:
                p = self.start
                while p.link != end:
                    q = p.link
                    if p.info > q.info:
                        p.info, q.info = q.info, p.info
                    p = p.link
                end = p

        except Exception as e:
            print("💛 Error: ", e)


    def bubble_sort_sll(self):
        """
        Bubble sorts the linked list

        """

        try:

            # If the list is empty or there is only one node
            if self.start is None or self.start.link is None:
                print("💛 The list has no or only one node and sorting is not required.")
            end = None

            while end != self.start.link:
                r = p = self.start
                while p.link != end:
                    q = p.link
                    if p.info > q.info:
                        p.link = q.link
                        q.link = p
                    if  p != self.start:
                        r.link = q.link
                    else:
                        self.start = q
                    p, q = q, p
                    r = p
                    p = p.link
                end = p

        except Exception as e:
            print("💛 Error: ", e)


    def sll_has_cycle(self):
        """
        Tests the list for cycles using Tortoise and Hare Algorithm (Floyd's cycle detection algorithm)
        """

        try:

            if self.find_sll_cycle() is None:
                return False
            else:
                return True


        except Exception as e:
            print("💛 Error: ", e)


    def find_sll_cycle(self):
        """
        Finds cycles in the list, if any
        """

        try:

            # If there is one node or none, there is no cycle
            if self.start is None or self.start.link is None:
                return None

            # Otherwise, 
            slowR = self.start
            fastR = self.start

            while slowR is not None and fastR is not None:
                slowR = slowR.link
                fastR = fastR.link.link
                if slowR == fastR: 
                    return slowR

            return None

        except Exception as e:
            print("💛 Error: ", e)


    def remove_cycle_from_sll(self):
        """
        Removes the cycles
        """

        try:

            c = self.find_sll_cycle()

            # If there is no cycle
            if c is None:
                return

            print(f"💛 There is a cycle at node: ", c.info)

            p = c
            q = c
            len_cycles = 0
            while True:
                len_cycles += 1
                q = q.link

                if p == q:
                    break

            print(f"💛 The cycle length is {len_cycles}")

            len_rem_list = 0
            p = self.start

            while p != q:
                len_rem_list += 1
                p = p.link
                q = q.link

            print(f"💛 The number of nodes not included in the cycle is {len_rem_list}")

            length_list = len_rem_list + len_cycles

            print(f"💛 The SLL length is {length_list}")

            # This for loop goes to the end of the SLL, and set the last node to None and the cycle is removed. 
            p = self.start
            for _ in range(length_list-1):
                p = p.link
            p.link = None


        except Exception as e:
            print("💛 Error: ", e)


    def insert_cycle_in_sll(self, x):
        """
        Inserts a cycle at a node that contains x

        """

        try:

            if self.start is None:
                return False

            p = self.start
            px = None
            prev = None


            while p is not None:
                if p.info == x:
                    px = p
                prev = p
                p = p.link

            if px is not None:
                prev.link = px
            else:
                print(f"💔 Sorry! {x} is not in the list.")


        except Exception as e:
            print("💛 Error: ", e)


    def merge_using_new_list(self, list2):
        """
        Merges two already sorted SLLs by creating new lists
        """
        merge_list = SingleLinkedList()
        merge_list.start = self._merge_using_new_list(self.start, list2.start)
        return merge_list

    def _merge_using_new_list(self, p1, p2):
        """
        Private method of merge_using_new_list
        """
        if p1.info <= p2.info:
            Start_merge = Node(p1.info)
            p1 = p1.link
        else:
            Start_merge = Node(p2.info)
            p2 = p2.link            
        pM = Start_merge

        while p1 is not None and p2 is not None:
            if p1.info <= p2.info:
                pM.link = Node(p1.info)
                p1 = p1.link
            else:
                pM.link = Node(p2.info)
                p2 = p2.link
            pM = pM.link

        #If the second list is finished, yet the first list has some nodes
        while p1 is not None:
            pM.link = Node(p1.info)
            p1 = p1.link
            pM = pM.link

        #If the second list is finished, yet the first list has some nodes
        while p2 is not None:
            pM.link = Node(p2.info)
            p2 = p2.link
            pM = pM.link

        return Start_merge

    def merge_inplace(self, list2):
        """
        Merges two already sorted SLLs in place in O(1) of space
        """
        merge_list = SingleLinkedList()
        merge_list.start = self._merge_inplace(self.start, list2.start)
        return merge_list

    def _merge_inplace(self, p1, p2):
        """
        Merges two already sorted SLLs in place in O(1) of space
        """
        if p1.info <= p2.info:
            Start_merge = p1
            p1 = p1.link
        else:
            Start_merge = p2
            p2 = p2.link
        pM = Start_merge

        while p1 is not None and p2 is not None:
            if p1.info <= p2.info:
                pM.link = p1
                pM = pM.link
                p1 = p1.link
            else:
                pM.link = p2
                pM = pM.link
                p2 = p2.link

        if p1 is None:
            pM.link = p2
        else:
            pM.link = p1

        return Start_merge

    def merge_sort_sll(self):
        """
        Sorts the linked list using merge sort algorithm
        """
        self.start = self._merge_sort_recursive(self.start)


    def _merge_sort_recursive(self, list_start):
        """
        Recursively calls the merge sort algorithm for two divided lists
        """

        # If the list is empty or has only one node
        if list_start is None or list_start.link is None:
            return list_start

        # If the list has two nodes or more
        start_one = list_start
        start_two = self._divide_list(self_start)
        start_one = self._merge_sort_recursive(start_one)
        start_two = self._merge_sort_recursive(start_two)
        start_merge = self._merge_inplace(start_one, start_two)

        return start_merge

    def _divide_list(self, p):
        """
        Divides the linked list into two almost equally sized lists
        """

        # Refers to the third nodes of the list
        q = p.link.link

        while q is not None and p is not None:
            # Increments p one node at the time
            p = p.link
            # Increments q two nodes at the time
            q = q.link.link

        start_two = p.link
        p.link = None

        return start_two

    def concat_second_list_to_sll(self, list2):
        """
        Concatenates another SLL to an existing SLL
        """

        # If the second SLL has no node
        if list2.start is None:
            return

        # If the original SLL has no node
        if self.start is None:
            self.start = list2.start
            return

        # Otherwise traverse the original SLL
        p = self.start
        while p.link is not None:
            p = p.link

        # Link the last node to the first node of the second SLL
        p.link = list2.start



    def test_merge_using_new_list_and_inplace(self):
        """

        """

        LIST_ONE = SingleLinkedList()
        LIST_TWO = SingleLinkedList()

        LIST_ONE.create_single_linked_list()
        LIST_TWO.create_single_linked_list()

        print("1️⃣  The unsorted first list is: ")
        LIST_ONE.display_sll()

        print("2️⃣  The unsorted second list is: ")
        LIST_TWO.display_sll()


        LIST_ONE.bubble_sort_sll_nodes_data()
        LIST_TWO.bubble_sort_sll_nodes_data()

        print("1️⃣  The sorted first list is: ")
        LIST_ONE.display_sll()

        print("2️⃣  The sorted second list is: ")
        LIST_TWO.display_sll()

        LIST_THREE = LIST_ONE.merge_using_new_list(LIST_TWO)

        print("The merged list by creating a new list is: ")
        LIST_THREE.display_sll()


        LIST_FOUR = LIST_ONE.merge_inplace(LIST_TWO)

        print("The in-place merged list is: ")
        LIST_FOUR.display_sll()     


    def test_all_methods(self):
        """
        Tests all methods of the SLL class
        """

        OPTIONS_HELP = """
📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗
    Select a method from 1-19:                                                          
🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒🍒
        ℹ️   (1)    👉  Create a single liked list (SLL).
        ℹ️   (2)    👉  Display the SLL.                            
        ℹ️   (3)    👉  Count the nodes of SLL. 
        ℹ️   (4)    👉  Search the SLL.
        ℹ️   (5)    👉  Insert a node at the beginning of the SLL.
        ℹ️   (6)    👉  Insert a node at the end of the SLL.
        ℹ️   (7)    👉  Insert a node after a specified node of the SLL.
        ℹ️   (8)    👉  Insert a node before a specified node of the SLL.
        ℹ️   (9)    👉  Delete the first node of SLL.
        ℹ️   (10)   👉  Delete the last node of the SLL.
        ℹ️   (11)   👉  Delete a node you wish to remove.                           
        ℹ️   (12)   👉  Reverse the SLL.
        ℹ️   (13)   👉  Bubble sort the SLL by only exchanging the integer values.  
        ℹ️   (14)   👉  Bubble sort the SLL by exchanging links.                    
        ℹ️   (15)   👉  Merge sort the SLL.
        ℹ️   (16)   👉  Insert a cycle in the SLL.
        ℹ️   (17)   👉  Detect if the SLL has a cycle.
        ℹ️   (18)   👉  Remove cycle in the SLL.
        ℹ️   (19)   👉  Test merging two bubble-sorted SLLs.
        ℹ️   (20)   👉  Concatenate a second list to the SLL. 
        ℹ️   (21)   👉  Exit.
📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗📗
        """


        self.create_single_linked_list()

        while True:

            print(OPTIONS_HELP)

            UI_OPTION = int(input("👉   Enter an integer for the method you wish to run using the above help: "))

            if UI_OPTION == 1:
                data = int(input("👉   Enter an integer to be inserted at the end of the list: "))
                x = int(input("👉   Enter an integer to be inserted after that: "))
                self.insert_node_after_another(data, x)
            elif UI_OPTION == 2:
                self.display_sll()
            elif UI_OPTION == 3:
                self.count_sll_nodes()
            elif UI_OPTION == 4:
                data = int(input("👉   Enter an integer to be searched: "))
                self.search_sll_nodes(data)
            elif UI_OPTION == 5:
                data = int(input("👉   Enter an integer to be inserted at the beginning: "))
                self.insert_node_in_beginning(data)
            elif UI_OPTION == 6:
                data = int(input("👉   Enter an integer to be inserted at the end: "))
                self.insert_node_at_end(data)
            elif UI_OPTION == 7:
                data = int(input("👉   Enter an integer to be inserted: "))
                x = int(input("👉   Enter an integer to be inserted before that: "))
                self.insert_node_before_another(data, x)
            elif UI_OPTION == 8:
                data = int(input("👉   Enter an integer for the node to be inserted: "))
                k = int(input("👉   Enter an integer for the position at which you wish to insert the node: "))
                self.insert_node_before_another(data, k)
            elif UI_OPTION == 9:
                self.delete_sll_first_node()
            elif UI_OPTION == 10:
                self.delete_sll_last_node()
            elif UI_OPTION == 11:
                data = int(input("👉   Enter an integer for the node you wish to remove: "))
                self.delete_a_node(data)
            elif UI_OPTION == 12:
                self.reverse_sll()
            elif UI_OPTION == 13:
                self.bubble_sort_sll_nodes_data()
            elif UI_OPTION == 14:
                self.bubble_sort_sll()
            elif UI_OPTION == 15:
                self.merge_sort_sll()
            elif UI_OPTION == 16:
                data = int(input("👉   Enter an integer at which a cycle has to be formed: "))
                self.insert_cycle_in_sll(data)
            elif UI_OPTION == 17:
                if self.sll_has_cycle():
                    print("💛 The linked list has a cycle. ")
                else:
                    print("💚 YAAAY! The linked list does not have a cycle. ")
            elif UI_OPTION == 18:
                self.remove_cycle_from_sll()
            elif UI_OPTION == 19:
                self.test_merge_using_new_list_and_inplace()
            elif UI_OPTION == 20:
                list2 = self.create_single_linked_list()
                self.concat_second_list_to_sll(list2)
            elif UI_OPTION == 21:
                break
            else:
                print("💛 Option must be an integer, between 1 to 21.")

            print()     



if __name__ == '__main__':
    # Instantiates a new SLL object
    SLL_OBJECT = SingleLinkedList()
    SLL_OBJECT.test_all_methods()

Élargir la réponse de Nick Stinemates

class Node(object):
    def __init__(self):
        self.data = None
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def prepend_node(self, data):
        new_node = Node()
        new_node.data = data
        new_node.next = self.head
        self.head = new_node

    def append_node(self, data):
        new_node = Node()
        new_node.data = data
        current = self.head
        while current.next:
            current = current.next
        current.next = new_node

    def reverse(self):
        """ In-place reversal, modifies exiting list"""
        previous = None
        current_node = self.head

        while current_node:
            temp =  current_node.next
            current_node.next = previous
            previous = current_node
            current_node = temp
        self.head = previous

    def search(self, data):
        current_node = self.head
        try:
            while current_node.data != data:
                current_node = current_node.next
            return True
        except:
            return False

    def display(self):
        if self.head is None:
            print("Linked list is empty")
        else:
            current_node = self.head
            while current_node:
                print(current_node.data)
                current_node = current_node.next

    def list_length(self):
        list_length = 0
        current_node = self.head
        while current_node:
            list_length += 1
            current_node = current_node.next
        return list_length


def main():
    linked_list = LinkedList()

    linked_list.prepend_node(1)
    linked_list.prepend_node(2)
    linked_list.prepend_node(3)
    linked_list.append_node(24)
    linked_list.append_node(25)
    linked_list.display()
    linked_list.reverse()
    linked_list.display()
    print(linked_list.search(1))
    linked_list.reverse()
    linked_list.display()
    print("Lenght of singly linked list is: " + str(linked_list.list_length()))


if __name__ == "__main__":
    main()

Mes 2 cents

class Node:
    def __init__(self, value=None, next=None):
        self.value = value
        self.next = next

    def __str__(self):
        return str(self.value)


class LinkedList:
    def __init__(self):
        self.first = None
        self.last = None

    def add(self, x):
        current = Node(x, None)
        try:
            self.last.next = current
        except AttributeError:
            self.first = current
            self.last = current
        else:
            self.last = current

    def print_list(self):
        node = self.first
        while node:
            print node.value
            node = node.next

ll = LinkedList()
ll.add("1st")
ll.add("2nd")
ll.add("3rd")
ll.add("4th")
ll.add("5th")

ll.print_list()

# Result: 
# 1st
# 2nd
# 3rd
# 4th
# 5th

enter code here
enter code here

class node:
    def __init__(self):
        self.data = None
        self.next = None
class linked_list:
    def __init__(self):
        self.cur_node = None
        self.head = None
    def add_node(self,data):
        new_node = node()
        if self.head == None:
            self.head = new_node
            self.cur_node = new_node
        new_node.data = data
        new_node.next = None
        self.cur_node.next = new_node
        self.cur_node = new_node
    def list_print(self):
        node = self.head
        while node:
            print (node.data)
            node = node.next
    def delete(self):
        node = self.head
        next_node = node.next
        del(node)
        self.head = next_node
a = linked_list()
a.add_node(1)
a.add_node(2)
a.add_node(3)
a.add_node(4)
a.delete()
a.list_print()

ma double liste liée pourrait être compréhensible pour les noobies. Si vous êtes familier avec DS en C, c'est assez lisible.

# LinkedList..

class node:
    def __init__(self):           ##Cluster of Nodes' properties 
        self.data=None
        self.next=None
        self.prev=None

class linkedList():
    def __init__(self):
        self.t = node()                    // for future use
        self.cur_node = node()             // current node
        self.start=node()

    def add(self,data):                          // appending the LL

        self.new_node = node()
        self.new_node.data=data
        if self.cur_node.data is None:          
            self.start=self.new_node               //For the 1st node only

        self.cur_node.next=self.new_node
        self.new_node.prev=self.cur_node
        self.cur_node=self.new_node


    def backward_display(self):                  //Displays LL backwards
        self.t=self.cur_node
        while self.t.data is not None:
            print(self.t.data)
            self.t=self.t.prev

    def forward_display(self):                   //Displays LL Forward
        self.t=self.start
        while self.t.data is not None:
            print(self.t.data)
            self.t=self.t.next
            if self.t.next is None:
                print(self.t.data)
                break

    def main(self):                          //This is kind of the main 
                                               function in C
        ch=0
        while ch is not 4:                    //Switch-case in C 
            ch=int(input("Enter your choice:"))
            if ch is 1:
                data=int(input("Enter data to be added:"))
                ll.add(data)
                ll.main()
            elif ch is 2:
                ll.forward_display()
                ll.main()
            elif ch is 3:
                ll.backward_display()
                ll.main()
            else:
                print("Program ends!!")
                return


ll=linkedList()
ll.main()

Bien que de nombreuses simplifications puissent être ajoutées à ce code, j'ai pensé qu'une implémentation brute me serait plus facile à saisir.

Si vous souhaitez simplement créer une simple liste de favoris, reportez-vous à ce code

l = [1, [2, [3, [4, [5, [6, [7, [8, [9, [10]]]]]]]]]]

pour visualiser l'exécution de cette morue Visitez http://www.pythontutor.com/visualize.html#mode=edit