Python Linked List

Question

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!

Answer 1

Here's a rather Scheme way to do it:

class cons:
        def __init__(self, f, r):
                self.__f = f
                self.__r = r
        def __str__(self):
                return "(%s, %s)" % (str(self.__f), str(self.__r))
        __repr__ = __str__
        class empty:
                def __init__(self): pass
                __repr__ = lambda self: "empty"
                __str__ = __repr__
        empty = empty()
        def first(self): return self.__f
        def rest(self): return self.__r

I'm looking for a more python way, though, and ideally one that has easier to work with syntax than this:

>>> cons(12, cons(4, cons.empty))
(12, (4, empty))
>>> cons(12, cons(4, cons.empty)).first()
12
>>> cons(12, cons(4, cons.empty)).rest()
(4, empty)

Answer 2

The How to Think Like a Computer Scientist book covers this well in Chapter 17: Linked lists.

Answer 3

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.

Answer 4

I wrote this up the other day

#! /usr/bin/env python

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


class linked_list:
    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 = ll.cur_node
        while node:
            print node.data
            node = node.next



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

ll.list_print()

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

So you have profiled you working code and determined list operations to be a significant bottleneck that you need to replace them?

Answer 9

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

Linked lists have no practical value in Python. I've never used a 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")
  

Answer 10

I want to implement it using an underlying c linked list class and import is as a module.Should i use swig or read the docs and do it?

Answer 11

Here 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__()

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