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remove kth node from end of linked list

I am trying to remove the kth element from the END of a linked list. Here is my code
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def removeKthNodeFromEnd(head, k):
if k == 0 or head is None:
return
temp = head
while temp is not None and k > 1:
temp = temp.next
k -= 1
if temp is None or k > 1:
return head
trailing = None
leading = head
while temp.next is not None:
trailing = leading
leading = leading.next
temp = temp.next
if trailing is None:
head = head.next
else:
trailing.next = leading.next
head = LinkedList(0)
head.next = LinkedList(1)
head.next.next = LinkedList(2)
head.next.next.next = LinkedList(3)
head.next.next.next.next = LinkedList(4)
head.next.next.next.next.next = LinkedList(5)
head.next.next.next.next.next.next = LinkedList(6)
head.next.next.next.next.next.next.next = LinkedList(7)
head.next.next.next.next.next.next.next.next = LinkedList(8)
head.next.next.next.next.next.next.next.next.next = LinkedList(9)
removeKthNodeFromEnd(head, 10)
while head is not None:
print(head.value)
head = head.next
But this does not work and prints out all the values in the linked list from 0 to 9. Why is this the case? If the node to delete is the head node which I check by checking trailing is None, then I update head = head.next. I am changing head to be the next node. I can return head after this update and the result of head = removeKthNodeFromEnd(head, k) will give me the desired output but why can't I do it the other way without returning anything? For instance in my solution I can get rid of any element in between the first and last node including the last and works just fine. The original head gets updated and node gets removed. But when trying to update the head node head = head.next it does not work.
One way I achieved this is by doing the following...
if trailing is None:
head.value = head.next.value
head.next = head.next.next
But why must I use the values? To me this seems the same as
if trailing is None:
head = head.next
head.next = head.next.next
but does not work

It doesn't work because head in that function is a local name. Assigning to a local variable never does anything to other variables, even not when they happen to have the same name (like the global head).
The "trick" you have by moving a linked list value with head.value = head.next.value will not work when the list only has one node.
One way to do this, is to return the (potentially modified) value of head, and expect the caller to assign this back to their own head variable.
So:
in removeKthNodeFromEnd make sure all return statements are like return head, and that you add a return head at the very end. In short, removeKthNodeFromEnd should always return head.
in the main program, change the call to head = removeKthNodeFromEnd(head, 10)
That will solve your issue.
Here are some ideas to make your code more elegant:
Create a separate class for maintaining the head of the linked list
Improve the constructor, so that it is easier to initialise a linked list
Add an iterator so it is easier to print a linked list
class Node: # renamed
def __init__(self, value, nxt=None):
self.value = value
self.next = nxt # is now an argument
class LinkedList: # added
def __init__(self, *values):
self.head = None
if values: # populate the list with values
for value in reversed(values):
self.push_front(value)
def push_front(self, value):
self.head = Node(value, self.head)
def __iter__(self):
node = self.head
while node:
yield node.value
node = node.next
def removeKthNodeFromEnd(self, k):
if k <= 0:
return
temp = self.head
for _ in range(k):
if not temp: # k is out of range
return
temp = temp.next
if temp:
trailing = self.head
while temp.next:
trailing = trailing.next
temp = temp.next
trailing.next = trailing.next.next
else:
self.head = self.head.next
lst = LinkedList(*range(10))
print(*lst)
lst.removeKthNodeFromEnd(0)
print(*lst)
And if you maintain a size attribute, you can do:
class LinkedList: # added
def __init__(self, *values):
self.head = None
self.size = 0
if values:
for value in reversed(values):
self.push_front(value)
def push_front(self, value):
self.head = Node(value, self.head)
self.size += 1
def __iter__(self):
node = self.head
while node:
yield node.value
node = node.next
def removeAt(self, k):
if 1 < k <= self.size:
trailing = self.head
for _ in range(k - 2):
trailing = trailing.next
trailing.next = trailing.next.next
elif 1 == k <= self.size:
self.head = self.head.next
def removeKthNodeFromEnd(self, k):
self.removeAt(self.size + 1 - k)

How to directly reference an object through a variable in Python

Basically, I'm implementing a LinkedList class and then implementing various methods to use it. Below is the code for the LinkedList class(and its dependent Node class)
# A simple linked list implementation
class Node:
# attributes:
# data (can be anything)
# next (another Node)
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# attributes:
# head (a Node)
# ****************
# methods:
# insert
# find
# delete
def __init__(self):
self.head = None
def __str__(self):
output = []
current_node = self.head
while current_node:
output.append(str(current_node.data))
current_node = current_node.next
return(", ".join(output))
# Creates a Node with the given data and inserts into the front of the list.
def insert(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node
# Finds the first node with the given data. Returns None if there is no such node.
def find(self, data):
current_node = self.head
while(current_node):
if (current_node.data == data):
return current_node
current_node = current_node.next
return None # We reached the end of the list without finding anything.
# Deletes given node. Can be used in conjunction with find.
def delete(self, deleted_node):
if deleted_node == self.head:
self.head = self.head.next
return
current_node = self.head
while(current_node.next != deleted_node):
current_node = current_node.next
current_node.next = deleted_node.next
Then, I'm trying to implement a rotate(my_list, N) function, which will, well, rotate my_list by N. The following is my code:
import linkedlists as ll
from ErrorHandler import sanitize
import random, math, time, copy
def length(my_list):
sanitize(my_list, ll.LinkedList)
if my_list.head == None: return 0
count = 1 #First item! Ah, ah, ah
current_node = my_list.head
while current_node.next != None:
current_node = current_node.next
count += 1 #One more item! Ah, ah, ah
return count
def get_Nth(my_list, N):
sanitize(my_list, ll.LinkedList)
if my_list.head == None: return None
current_node = my_list.head
count = 0
while current_node.next != None:
if count == N:
return current_node
count +=1
current_node = current_node.next
return current_node
def rotate(my_list, N):
sanitize(my_list, ll.LinkedList)
if my_list.head == None: return None
N = N % length(my_list)
lent = length(my_list)
get_Nth(my_list, lent-1).next = my_list.head
my_list.head = get_Nth(my_list, lent-N)
get_Nth(my_list, lent-N-1).next = None
However, calling rotate() on a LinkedList containing the numbers from 0 to 9 in ascending order returns 8,9,0,1,2,3,4,5. Why? I'm pretty sure it has to do with the third- and second-to-last lines, because when assigning get_Nth(my_list, lent-1).next to my_list.head, it only points to my_list.head, instead of the Node object my_list.head points to at the time.
How can I fix this?

Your error is right here: get_Nth(my_list, lent-N-1).next = None
I'm assuming you called rotate(my_list, 2) so at this point, the list looks like this [8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, ...]. So when you call get_Nth(my_list, lent-N-1), lent-N-1 is 7, and well, the element at index 7 is actually 5.
You should just use lent-1.

Doubly linked lists in Python

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!

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

Python lists and computational complexity [duplicate]

What's the easiest way to use a linked list in Python? In Scheme, a linked list is defined simply by '(1 2 3 4 5).
Python's lists, [1, 2, 3, 4, 5], and tuples, (1, 2, 3, 4, 5), are not, in fact, linked lists, and linked lists have some nice properties such as constant-time concatenation, and being able to reference separate parts of them. Make them immutable and they are really easy to work with!

For some needs, a deque may also be useful. You can add and remove items on both ends of a deque at O(1) cost.
from collections import deque
d = deque([1,2,3,4])
print d
for x in d:
print x
print d.pop(), d

I wrote this up the other day
#! /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()

Here is some list functions based on Martin v. Löwis's representation:
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")
where w = sys.stdout.write
Although doubly linked lists are famously used in Raymond Hettinger's ordered set recipe, singly linked lists have no practical value in Python.
I've never used a singly linked list in Python for any problem except educational.
Thomas Watnedal suggested a good educational resource How to Think Like a Computer Scientist, Chapter 17: Linked lists:
A linked list is either:
the empty list, represented by None, or
a node that contains a cargo object and a reference to a linked list.
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")

The accepted answer is rather complicated. Here is a more standard design:
L = LinkedList()
L.insert(1)
L.insert(1)
L.insert(2)
L.insert(4)
print L
L.clear()
print L
It is a simple LinkedList class based on the straightforward C++ design and Chapter 17: Linked lists, as recommended by 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__()

Immutable lists are best represented through two-tuples, with None representing NIL. To allow simple formulation of such lists, you can use this function:
def mklist(*args):
result = None
for element in reversed(args):
result = (element, result)
return result
To work with such lists, I'd rather provide the whole collection of LISP functions (i.e. first, second, nth, etc), than introducing methods.

Here's a slightly more complex version of a linked list class, with a similar interface to python's sequence types (ie. supports indexing, slicing, concatenation with arbitrary sequences etc). It should have O(1) prepend, doesn't copy data unless it needs to and can be used pretty interchangably with tuples.
It won't be as space or time efficient as lisp cons cells, as python classes are obviously a bit more heavyweight (You could improve things slightly with "__slots__ = '_head','_tail'" to reduce memory usage). It will have the desired big O performance characteristics however.
Example of usage:
>>> 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])
Implementation:
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 — Linked list datatypes for Python
llist module implements linked list data structures. It supports a doubly linked list, i.e. dllist and a singly linked data structure sllist.
dllist objects
This object represents a doubly linked list data structure.
first
First dllistnode object in the list. None if list is empty.
last
Last dllistnode object in the list. None if list is empty.
dllist objects also support the following methods:
append(x)
Add x to the right side of the list and return inserted dllistnode.
appendleft(x)
Add x to the left side of the list and return inserted dllistnode.
appendright(x)
Add x to the right side of the list and return inserted dllistnode.
clear()
Remove all nodes from the list.
extend(iterable)
Append elements from iterable to the right side of the list.
extendleft(iterable)
Append elements from iterable to the left side of the list.
extendright(iterable)
Append elements from iterable to the right side of the list.
insert(x[, before])
Add x to the right side of the list if before is not specified, or insert x to the left side of dllistnode before. Return inserted dllistnode.
nodeat(index)
Return node (of type dllistnode) at index.
pop()
Remove and return an element’s value from the right side of the list.
popleft()
Remove and return an element’s value from the left side of the list.
popright()
Remove and return an element’s value from the right side of the list
remove(node)
Remove node from the list and return the element which was stored in it.
dllistnode objects
class llist.dllistnode([value])
Return a new doubly linked list node, initialized (optionally) with value.
dllistnode objects provide the following attributes:
next
Next node in the list. This attribute is read-only.
prev
Previous node in the list. This attribute is read-only.
value
Value stored in this node.
Compiled from this reference
sllist
class llist.sllist([iterable])
Return a new singly linked list initialized with elements from iterable. If iterable is not specified, the new sllist is empty.
A similar set of attributes and operations are defined for this sllist object. See this reference for more information.

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())

I based this additional function on 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
The method he has adds the new node at the beginning while I have seen a lot of implementations which usually add a new node at the end but whatever, it is fun to do.

The following is what I came up with. It's similer to Riccardo C.'s, in this thread, except it prints the numbers in order instead of in reverse. I also made the LinkedList object a Python Iterator in order to print the list out like you would a normal Python list.
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
"""

I just did this as a fun toy. It should be immutable as long as you don't touch the underscore-prefixed methods, and it implements a bunch of Python magic like indexing and len.

Here is my solution:
Implementation
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)
Original Implementation Idea
http://interactivepython.org/runestone/static/pythonds/BasicDS/ImplementinganUnorderedListLinkedLists.html

When using immutable linked lists, consider using Python's tuple directly.
ls = (1, 2, 3, 4, 5)
def first(ls): return ls[0]
def rest(ls): return ls[1:]
Its really that ease, and you get to keep the additional funcitons like len(ls), x in 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)

I've put a Python 2.x and 3.x singly-linked list class at https://pypi.python.org/pypi/linked_list_mod/
It's tested with CPython 2.7, CPython 3.4, Pypy 2.3.1, Pypy3 2.3.1, and Jython 2.7b2, and comes with a nice automated test suite.
It also includes LIFO and FIFO classes.
They aren't immutable though.

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

Here is my simple implementation:
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)
Output:
Data 1: Next -> Data 2: Next -> Data 34: Next -> Data 4: Next -> None

I did also write a Single Linked List based on some tutorial, which has the basic two Node and Linked List classes, and some additional methods for insertion, delete, reverse, sorting, and such.
It's not the best or easiest, works OK though.
"""
🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏🍎🍏
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()

I think the implementation below fill the bill quite gracefully.
'''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)

First of all, I assume you want linked lists. In practice, you can use collections.deque, whose current CPython implementation is a doubly linked list of blocks (each block contains an array of 62 cargo objects). It subsumes linked list's functionality. You can also search for a C extension called llist on pypi. If you want a pure-Python and easy-to-follow implementation of the linked list ADT, you can take a look at my following minimal implementation.
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)

Linked List Class
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(), '-|----> ',
Output
| 55 | 25 -|----> | 25 | 10 -|----> | 10 | --> None

Sample of a doubly linked list (save as 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
Testing (save as 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()
Output:
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

Expanding Nick Stinemates's answer
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()

My 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()

my double Linked List might be understandable to noobies.
If you are familiar with DS in C, this is quite readable.
# 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()
Though many more simplifications can be added to this code, I thought a raw implementation would me more grabbable.

Current Implementation of Linked List in Python requires for creation of a separate class, called Node, so that they can be connected using a main Linked List class. In the provided implementation, the Linked List is created without defining a separate class for a node. Using the proposed implementation, Linked Lists are easier to understand and can be simply visualized using the print function.
class Linkedlist:
def __init__(self):
self.outer = None
def add_outermost(self, dt):
self.outer = [dt, self.outer]
def add_innermost(self, dt):
p = self.outer
if not p:
self.outer = [dt, None]
return
while p[1]:
p = p[1]
p[1] = [dt, None]
def visualize(self):
p = self.outer
l = 'Linkedlist: '
while p:
l += (str(p[0])+'->')
p = p[1]
print(l + 'None')
ll = Linkedlist()
ll.add_innermost(8)
ll.add_outermost(3)
ll.add_outermost(5)
ll.add_outermost(2)
ll.add_innermost(7)
print(ll.outer)
ll.visualize()

If you want to just create a simple liked list then refer this code
l=[1,[2,[3,[4,[5,[6,[7,[8,[9,[10]]]]]]]]]]
for visualize execution for this cod Visit http://www.pythontutor.com/visualize.html#mode=edit

single linked list reverse in python

i'm trying to create a simple single linked list in python.
(i know there is no need to implement list in python, but that's not the point)
here is my code:
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
list1.inserthead(nodeC)
list1.inserthead(nodeD)
list1.inserthead(nodeE)
print "list"
list1.printlist()
print "list reverse"
list1.reverte_list1()
list1.printlist()
list1.reverte_iterative()
print "list reverse reverse"
list1.printlist()
and there are the result:
None
E
D
C
B
list reverse
A
B
C
D
E
list reverse reverse
E
D
C
B
for some reason i can't print all the list and in the first case doesn't print the "A" Node
but print the first node(but i checkd and the B node points to A)
the first reverse is OK
but the third again don't print the A node even if it's pointed by the B Node.
probably the problem of the printing is in the __show function.
but i suppose i a conceptual bug.
thanks

def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
that only prints the current node if it has a next node, thats why the last node is never printed.
should be:
def __show(self,start):
if start:
print start.data
self.__show(start.next)
you make similar mistakes of checking/assigning a node instead of it's next and vice versa throughout the code (in inserthead() for example - which is causing the None being printed)