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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

Implemeting custom iterable object in python

This is my implementation of a custom singly linked list in Python.
class SList:
def __init__(self):
self.root = None
self.size = 0
def insert(self, item):
if not item:
raise ValueError('Cannot add None item to a list')
self.size += 1
if self.root is None:
self.root = Node(item)
else:
p = Node(item)
p.next = self.root
self.root = p
"""Remove the element at the specific index"""
def remove(self, index):
if index < 0 or index >= self.size:
raise ValueError('Index cannot be negative or greater than the size of the list')
current = self.root
if index == 0:
self.root = self.root.next
else:
for _ in range(index -1):
current = current.next
p = current.next.next
if p is not None:
current.next = p
else:
current.next = None
self.size -= 1
def __len__(self):
return self.size
def __repr__(self):
res = '[ '
current = self.root
while current is not None:
res += str(current.data)
res += ' '
current = current.next
res += ']'
return res
def __iter__(self):
return self
def next(self):
........
This is the Node object
class Node:
def __init__(self, data):
try:
if not data:
raise ValueError
self.data = data
self.next = None
except ValueError:
raise ValueError('Node cannot be instantiated without an item')
I'm at a little loss at implementing the iter method. I see there are multiple ways to implement it and yield seems to be the common way forward. Would appreciate some help in implmenting it with yield

You can make your class iterable by making its __iter__ method a generator.
Here's some code that runs correctly on Python 2 or Python 3.
from __future__ import print_function
class Node(object):
def __init__(self, data):
if data is None:
raise ValueError('Node cannot be instantiated without an item')
self.data = data
self.nextnode = None
def __repr__(self):
return 'Node({})'.format(self.data)
class SList(object):
def __init__(self):
self.root = None
self.size = 0
def insert(self, item):
if item is None:
raise ValueError('Cannot add None item to a list')
self.size += 1
if self.root is None:
self.root = Node(item)
else:
p = Node(item)
p.nextnode = self.root
self.root = p
def remove(self, index):
""" Remove the element at the specific index """
if index < 0 or index >= self.size:
raise ValueError('Index cannot be negative or greater than the size of the list')
current = self.root
if index == 0:
self.root = self.root.nextnode
else:
for _ in range(index - 1):
current = current.nextnode
current.nextnode = current.nextnode.nextnode
self.size -= 1
def __len__(self):
return self.size
def __repr__(self):
res = []
current = self.root
while current is not None:
res.append(current.data)
current = current.nextnode
return str(res)
def __iter__(self):
current = self.root
while current is not None:
yield current
current = current.nextnode
# test
a = SList()
for c in 'ABCDE':
a.insert(c)
print(a)
gen = iter(a)
print('root', next(gen))
for node in gen:
print(node)
a.remove(2)
print(list(a))
for node in a:
print(node)
output
['E', 'D', 'C', 'B', 'A']
root Node(E)
Node(D)
Node(C)
Node(B)
Node(A)
[Node(E), Node(D), Node(B), Node(A)]
Node(E)
Node(D)
Node(B)
Node(A)

Try to follow this explanation:
pythontips.com/2013/09/29/the-python-yield-keyword-explained
In short and simplified form - in order to create a generator, you should create a function that contains the yield keyword once or more. Anytime yield is reached in the function execution, it is put on hold, and passes the value after the keyword to the caller.
def my_gen(arg):
yield arg * 10
for i in xrange(5):
if i / 2 == 0:
yield i
for x in my_gen(3):
print(x)
Will print:
30
0
2
4
The constructor of the Node class is also faulty, in addition to catching your own exception, you also have an error with your if condition not data will be True not only if data is False or None, but also if data is 0,an empty list, empty string and many more. Use data is None instead.

How to use Iterator that traverses the doubly linked list and skips null nodes

Since the doubly linked list has two dummy nodes, one is the head and the other one is the tail. I can skip the dummy tail node by self.__current == None: raise StopIteration, but I don't know how to pass the dummy head nodes and continue traverse the following node.
class LinkedListDLL:
def __init__(self):
self.__head = NodeDLL(None)
self.__head.set_next(self.__head)
self.__head.set_prev(self.__head)
self.__count = 0
def size(self):
return self.__count;
def is_empty(self):
return self.__count == 0
def add_before(self, item, curr):
new_node = NodeDLL(item)
new_node.set_next(curr)
new_node.set_prev(curr.get_prev())
curr.set_prev(new_node)
new_node.get_prev().set_next(new_node)
self.__count += 1
def add_to_head(self, item):
self.add_before(item, self.__head.get_next())
def add_to_tail(self, item):
self.add_before(item, self.__head)
def remove_from_head(self):
self.remove(self.__head.get_next())
def remove_from_tail(self):
self.remove(self.__head.get_prev())
def remove(self, curr):
curr.get_prev().set_next(curr.get_next())
curr.get_next().set_prev(curr.get_prev())
self.__count -= 1
def __iter__(self):
return LinkedListIterator(self.__head)
class LinkedListIterator:
def __init__(self, head):
self.__current = head
def __next__(self):
if self.__current == None :
raise StopIteration
else:
item = self.__current.set_data.get_next()
self.__current = self.__current.get_next()
return item

You should refactor your code keeping in mind that NodeDLL(None) may not be the same as None. With that done:
def __next__(self):
if not self.__current.get_next():
raise StopIteration
#Skips first value in list, i.e. head
self.__current = self.__current.get_next()
return self.__current

Implementing Doubly Linked List in Python

I was trying to implement a doubly linked list with the following functions:
_insertItemAfterNode(self,index)
_nodeBeforeIndex(self, index)
insertItemAtTheFront(self,item)
insertItemAtTheEnd(self,item)
insertItemAtIndex(self, index, item)
The last three functions should use the first two private methods.
This is what I have so far but I can't seem to work it. Any help is appreciated!!!
# DoublyLinkedLists
class Node(object):
def __init__(self, prev = None, data=None, next = None):
self._prev = prev
self._data = data
self._next = next
def __str__(self):
return str(self._data)
def __repr__(self):
return "Node(%s,%s,%s)" % (repr(self._prev), repr(self._data), repr (self._next))
def __eq__(self, other):
if other == None:
return False
else:
return self._prev == other._prev and self._data == other._data and self._next == other._next`enter code here`
class DoublyLinkedList(object):
def __init__(self):
self._first = Node(None, None, None)
self._length = 0
self._last = Node(None, None, None)
def __len__(self):
return self._length
def _insertItemAfterNode(self,item,aNode):
newNode = Node(aNode._prev, item, aNode._next)
aNode._next= newNode._prev
aNode._prev=newNode._next
self._length += 1
def _nodeBeforeIndex(self, index):
if 0 <= index <= self._length:
aNode = self._first
for i in range(index):
aNode = aNode._next
return aNode
else:
raise IndexError
def _nodeAfterIndex(self, index):
if 0<= index <= self.length:
aNode = self._last
for i in range(index):
aNode = aNode._prev
return aNode
else:
raise IndexError
def __getitem__(self, index):
return self._nodeBeforeIndex(index)._next._data
def insertAtEnd(self, item):
self._nodeAfterIndex(item)
def __iter__(self):
return DoublyLinkedListIterator(self)
def __setitem__(self, index, item):
self._nodeBeforeIndex(index)._next._data = item
def insertItemAtTheFront(self, item):
self._insertItemAfterNode(item, self._nodeBeforeIndex(0))
def insertItemAtTheEnd(self, item):
self._insertItemAfterNode(item, self._nodeBeforeIndex(self._length))
def insertItemAtIndex(self, index, item):
'''Valid range 0 <= index <= length.'''
self._insertItemAfterNode(item, self._nodeBeforeIndex(index))
def __iter__(self):
return DoublyLinkedListIterator(self)
def __repr__(self):
#used by print in abscence of __str__ .
plist = []
for item in self:
plist.append(item)
return "DoublyLinkedList(%s)" % str(plist)
class DoublyLinkedListIterator(object):
def __init__(self, aList):
self._list = aList
self._currentIndex = -1
self._currentNode = self._list._first
def __iter__(self):
return self
def __next__(self):
if self._currentIndex == self._list._length - 1:
raise StopIteration
else:
self._currentIndex += 1
self._currentNode = self._currentNode._next
return self._currentNode._data
def main():
x = DoublyLinkedList()
x.insertItemAtTheEnd(45)
print(x)
main()

I see two immediate issues with your code. There may be more issues that I haven't noticed, but these will probably prevent you from seeing any others until they are fixed.
The first issue is very simple. You have two sentinel nodes, _first and _last in your DoublyLinkedList class. However, they're not linked to each other, so when more nodes are added later, they won't be connected to both sentinels like they should be.
You can fix this by changing the __init__ method:
def __init__(self):
self._first = Node(None, None, None)
self._length = 0
self._last = Node(self._first, None, None) # pass first node as an arg here
self._first._next = self._last # make a link back too
The second issue is how you're adding nodes. There are a couple of places where you're assigning values from the _prev or _next attributes of a node, rather than referring to the node itself.
Lets think of a list containing A and C nodes, and you want to insert B in between them. You need to make B's node link to both A and C, and then modify C and A to link back to it. In your code, A is aNode, B is newNode and C is (at the start) aNode._next.
def _insertItemAfterNode(self, item, aNode):
newNode = Node(aNode, item, aNode._next) # link B to A and C
aNode._next._prev = newNode # link C back to B
aNode._next = newNode # and A to B too!
self._length += 1

There are a few things strange with your code. Let's dig into the first problem that I found:
def _insertItemAfterNode(self,item,aNode):
newNode = Node(aNode._prev, item, aNode._next)
aNode._next= newNode._prev
aNode._prev=newNode._next
self._length += 1
Assume that you have:
prevNode <-> aNode <-> nextNode
You now create a newNode which points to prevNode as its previous and nextNode as its follow-up. Then you reverse the pointers of aNode. prevNode still thinks its following node is aNode, but aNode now thinks that its following node is prevNode again.
/---------\
/----> aNode < v
prevNode <--------/ \-- nextNode
^ ^
\---- newNode --/
Which is not nearly what you want:
prevNode <-> aNode <-> newNode <-> nextNode
Here's a better version:
def _insertItemAfterNode(self,item,aNode):
# If the new Node follows aNode, than its preceeding node is aNode,
# not aNode._prev!
newNode = Node(aNode, item, aNode._next)
# Now we need to make aNode point at its new following Node, newNode
aNode._next = newNode
# And finally, we need to make the following Node notice that
# its preceeding Node has changed.
newNode._next._prev = newNode
self._length += 1
This doesn't care about any edge cases, for example what to do if aNode is the last node in the linked list.
What I recommend is that you take out a piece of paper and draw how the different links need to be changed for each of the operations that you need to make, because it is hard to keep all this in your head.
Make sure that you look at all the edge cases, ie. putting into an empty list, putting in item into the first and last positions.
And finally, while shuffling links around, always double check where each link is pointing at each instruction, and not to drop any nodes accidentally.

Single Linked List search in Python

I want to search a value/character in a linked list and return the number of times the value/character is in the linked list. Also would it be easier if I just used recursion instead of tail recursion?
class MyList():
__slots__=('head','size')
class Empty():
__slots__=()
class NonEmpty():
__slots__=('data','next')
def mkMyList():
lst = MyList()
lst.head = mkEmpty()
lst.size = 0
return lst
def mkEmpty():
return Empty()
def mkNonEmpty(data,lst):
node = NonEmpty()
node.data = data
node.next = lst
return node
def count(l, value, c = 0):
l = mkMyList()
if l.head != value:
l.head = l.head.next
if l.head == value:
return count(l.head.next, value, c + 1)
if l.size == 0:
return c
When I try to test it, I get this:
count(s,'s',c= 0)
Traceback (most recent call last):
File "<pyshell#2>", line 1, in <module>
count(s,'s',c= 0)
File "C:\Users\Qasim\Desktop\Linked Lists.py", line 30, in count
l.head = l.head.next
AttributeError: 'Empty' object has no attribute 'next'
\

Rather than use recursion, I would use the iterator pattern. Here is one way to do it in the context of your problem:
class LinkedList(object):
class Node(object):
__slots__ = ('prev', 'next', 'value')
def __init__(self, prev=None, next=None, value=None):
self.prev = prev
self.next = next
self.value = value
def __init__(self, iterable=[]):
self.head = LinkedList.Node() # dummy node
self.tail = self.head
self.size = 0
for item in iterable:
self.append(item)
def __iter__(self):
current = self.head
while True:
if current.next is not None:
current = current.next
yield current.value
else:
raise StopIteration
def append(self, value):
self.tail.next = LinkedList.Node(prev=self.tail, value=value)
self.tail = self.tail.next
self.size += 1
def pop(self):
if self.size > 0:
value = self.tail.value
self.tail = self.tail.prev
self.tail.next = None
self.size -= 1
return value
else:
raise IndexError('pop from empty list')
def count(self, value):
cumsum = 0
for item in self:
if item == value:
cumsum += 1
return cumsum
By my defining a Python special method __iter__, one can sequentially access the elements of a LinkedList in the following manner:
l = LinkedList([1, 2, 3, 3, 3, 4, 5])
for value in l:
print(value)
which then makes the desired method count straight-forward to implement.
Note that I have used the Python generator syntax to implement __iter__, you can read about generators and the yield statement here.

Tracing your code:
l = mkMyList() # => head = Empty()
if l.head != value: # True since head is Empty()
l.head = l.head.next # Empty does not have a ".next" attribute
This is what the Traceback is telling you.
EDIT: Two more things: (1) I'm not sure why count is even calling mkMyList when it seems your intent is to pass it the list, l, in the function args. (2) I'm guessing you want to put the size-check if statement at the top of this function:
if l.size == 0:
return c

The issue I see is that the list in count is never initialized properly. In mkMyList(), the head element is set to and Empty, which has no next attribute. In count(), you only use mkMyList(). This means that l.head is an Empty, and there's no way it could have a next attribute. To fix this, I would recommend instantiating the list l using the given input.
With regards to the question about recursion: no, there is very little difference in terms of composing a tail recursive function versus a regularly recursive function.