Kotlin Koans—Part 14

I have always had a huge complaint about Java’s collection classes. The complaint involves switching between types of collections. For example, let’s suppose you have an ArrayList and you want to switch it to a set. Here is how you would do it in Java.

public class CollectionConvert {
    public static void main(String [] args){
        List lst = Arrays.asList(1, 2, 3);
        Set set = new HashSet(lst);
    }
}

Ok so let’s be honest here. It’s not as if it’s really that difficult to convert a List to a Set in Java. All of the constructors of the Java collection class accept another collection so we can pass an existing collection into a new collection to handle switching between collection types.

But let’s face the fact. It can be better than this. We are using two lines of code that could be done with one piece of code instead. Here is how it’s done in Kotlin.

fun Shop.getSetOfCustomers(): Set {
    // Return a set containing all the customers of this shop
    return customers.toSet()
}

Kotlin extends Java’s collection classes with these sort of conversion method. It may not seems like a big deal until you want to do something like this.

val lst = arrayListOf(1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9)

//Filter out duplicates and maintain a list
val noDuplicates = lst.toSet().toList()
//continue working with list instead of set...

The Set interface lacks method such as get(index). Sometimes it’s helpful to remove all duplicates in a list prior before working on the list. Java 8 does a lot of help with this problem, but I still think Kotlin is more consice.

List lst = Arrays.asList(1, 1, 1, 2, 2, 2, 3, 3, 3);
List noDuplicates = lst.stream().distinct().collect(Collectors.toList());

You can click here to see Part 13

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Kotlin Koan—Part 13

Kotlin has nice extensions on Collection classes. JDK has provided developers a way to sort collections for a long time now, but often times, it’s done with external classes such as Collections. This puts us in a odd position of using procedural programming when performing operations such as sorting a collection.

A more OOP approach would be to have a sort method directly on a Collection as opposed to passing a collection to an external class with a static method. Kotlin extends Java collections by using its extension feature to complement the collection classes found in JDK.

Here is an example that demonstrates how to sort an ArrayList in Java.

fun task12(): List {
    val lst =  arrayListOf(1, 5, 2)
    lst.sortDescending()
    return lst
}

If all fairness to Java, JDK 8 did address this issue. Here is the equivalent Java code that does the same thing.

public class CollectionSort {
    public List sortList(){
        List lst = Arrays.asList(1, 5, 2);
        lst.sort(Comparator.reverseOrder());
        return lst;
    }
}

I’m not going to complain too much about the Java 8 syntax, but I think sortDescending() and sortAcending() is more readable than Comparator.natualOrder() and Comparator.reverseOrder(). However, the Kotlin approach is much superior to JDK 7 and earlier.

You can read the previous post here.

Lists—Python

Lists are a sequence type object that Python provides to us for grouping data together into a single variable. Let’s consider a common application of a list before we go into details.

names = ['Bob Belcher',
         'Linda Belcher',
         'Tina Belcher',
         'Gene Belcher',
         'Louise Belcher']
for name in names:
    print (name)

This code creates a list called names and populates it with five names. Then it prints the names to the console. We could accomplish the same output by using this code.

bob = 'Bob Belcher'
linda = 'Linda Belcher'
tina = 'Tina Belcher'
gene = 'Gene Belcher'
louise = 'Louise Belcher'

print(bob)
print(linda)
print(tina)
print(gene)
print(louise)

A quick comparison shows that the first example is not only much easier to read, but it is also more maintainable. If we want to print additional names, we only need to add them to the names list in the first example. However, in the second example, we need to add new name variable and another print statement. This may not seem like a big deal with six names, but obviously 6,000 is a much different case.

It’s for this reason that almost every programming language provides some sort of a collection object. List is one such data type that Python provides out of the box. Let’s look at common list operations.

Check if item is in the list

We may want to check if a list has a certain item.

names = ['Bob Belcher',
         'Linda Belcher',
         'Tina Belcher',
         'Gene Belcher',
         'Louise Belcher']
if 'Bob Belcher' in names:
    print('Found Bob')

This code prints ‘Found Bob’ because 'Bob Belcher' in names returns True.

Check if item is not in list

We can also check if a list does not have an item.

names = ['Bob Belcher',
         'Linda Belcher',
         'Tina Belcher',
         'Gene Belcher',
         'Louise Belcher']
if 'Teddy' not in names:
    print('No Teddy here!')

This code would print ‘No Teddy here!’ because 'Teddy' not in names is True. Our names list does not have ‘Teddy’

Combine lists

We can add two lists together (called concatenation) using the + operator.

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']

pestos = ['Jimmy Pesto',
          'Jimmy Pesto Jr.',
          'Andy Pesto',
          'Ollie Pesto']

family_frackus = belchers + pestos
print(family_frackus)

This code will print all of the names to standard out when run.

Accessing items

Lists use the index operator access items

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']
print(belchers[0])
print(belchers[3])

Keep in mind that lists are 0 based, so belchers[0] is ‘Bob Belcher’ while belchers[3] is ‘Gene Belcher’

Add item to list

We can add an item to a list using the append() method.

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']
belchers.append('Teddy')
print(belchers[5])

Remove item from a list

We use the del operator to remove an item from a list

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']
del belchers[0]

Using del belchers[0] removes ‘Bob Belcher’ from the list.

Replace item in a list

We can replace an item in a list by specifying the index and assigning a new value to it.

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']
belchers[0] = 'Mort'

The code belchers[0] = 'Mort' replaces ‘Bob Belcher’ with ‘Mort’

Length of the list

We get the length of the list using len.

belchers = ['Bob Belcher',
            'Linda Belcher',
            'Tina Belcher',
            'Gene Belcher',
            'Louise Belcher']
print(str(len(belchers))) 

The print(str(len(belchers))) prints 5 to the console.

Conclusion

We can do a lot more with lists than what was discussed in this post. Make sure you check out the Python documentation for a complete list of features!

Stack—Python

Implementing a Stack is a common problem that computer science students are asked to do while learning about data structures. Stacks are a data structure that implemenet the LIFO (last in, first out) principle. We can implement stacks using a linked list methodology or an array backed list.

In the real world, Python’s built in list object is one of many production tested objects that can and should be used for stacks. Our stack is a linked list implementation of a simple stack.

Here is the example code that we will work with followed by an explanation about how it works.

import unittest

class Node:
    def __init__(self, element=None, next_node=None):
        self.element = element
        self.next_node = next_node

    def __str__(self):
        if self.element:
            return self.element.__str__()
        else:
            return 'Empty Node'

    def __repr__(self):
        return self.__str__()

class EmptyStackException(Exception):
    pass

class Stack:
    def __init__(self):
        self.top = None
        self.size = 0

    def push(self, data):
        self.top = Node(data, self.top)
        self.size += 1

    def pop(self):
        if self.top:
            val = self.top.element
            self.top = self.top.next_node
            self.size -= 1
            return val
        else:
            raise EmptyStackException

    def peek(self):
        if self.top:
            return self.top.element
        else:
            raise EmptyStackException

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

    def __str__(self):
        elements = []
        curr = self.top
        while curr is not None:
            elements.append(curr.element)
            curr = curr.next_node
        return elements.__str__()

    def __repr__(self):
        return self.__str__()

class TestStack(unittest.TestCase):
    def __init__(self, method_name='runTest'):
        super().__init__(method_name)
        self.names = ['Bob Belcher',
                      'Linda Belcher',
                      'Tina Belcher',
                      'Gene Belcher',
                      'Louise Belcher']

    def test_init(self):
        stack = Stack()
        self.assertIsNone(stack.top)
        self.assertEqual(stack.size, 0)

    def test_push(self):
        stack = Stack()
        for name in self.names:
            stack.push(name)

        names = list(self.names)
        names.reverse()

        self.assertEqual(names.__str__(), stack.__str__())
        self.assertEqual(len(names), stack.size)

    def test_pop(self):
        stack = Stack()
        for name in self.names:
            stack.push(name)

        self.assertEqual(stack.pop(), 'Louise Belcher')
        self.assertEqual(stack.size, 4)

        self.assertEqual(stack.pop(), 'Gene Belcher')
        self.assertEqual(stack.size, 3)

        self.assertEqual(stack.pop(), 'Tina Belcher')
        self.assertEqual(stack.size, 2)

        self.assertEqual(stack.pop(), 'Linda Belcher')
        self.assertEqual(stack.size, 1)

        self.assertEqual(stack.pop(), 'Bob Belcher')
        self.assertEqual(stack.size, 0)

        with self.assertRaises(EmptyStackException):
            stack.pop()

    def test_peek(self):
        stack = Stack()
        for name in self.names:
            stack.push(name)

        self.assertEqual(stack.peek(), 'Louise Belcher')
        self.assertEqual(stack.size, 5)

        stack = Stack()
        with self.assertRaises(EmptyStackException):
            stack.peek()

    def test_empty(self):
        stack = Stack()
        self.assertTrue(stack.empty())

        for name in self.names:
            stack.push(name)

        self.assertFalse(stack.empty())

if __name__ == '__main__':
    unittest.main()

Node

Since this is a linked list implementation of a stack, we are going to begin with the Node class. The Node does the work of holding the data in the stack and contains a reference to the next item in the stack.

class Node:
    def __init__(self, element=None, next_node=None):
        self.element = element
        self.next_node = next_node

    def __str__(self):
        if self.element:
            return self.element.__str__()
        else:
            return 'Empty Node'

    def __repr__(self):
        return self.__str__()

There isn’t a lot going on in this class in terms of methods. It has an __init__ method which takes optional element and next_node parameters that default to None. This class also implements __str__ and __repr__ so that debuggers such as PyCharm print out user readable information.

EmptyStackException

The EmptyStackException class is a custom exception that gets raised when client code attempts to remove items from an empty stack.

class EmptyStackException(Exception):
    pass

This class doesn’t do anything other than subclass the Exception base class. For this reason, it’s implemented with a pass statement.

Stack

Stack is our main class the implements a Stack. We begin with the code for this class followed by an explanation about how it works.

class Stack:
    def __init__(self):
        self.top = None
        self.size = 0

    def push(self, data):
        self.top = Node(data, self.top)
        self.size += 1

    def pop(self):
        if self.top:
            val = self.top.element
            self.top = self.top.next_node
            self.size -= 1
            return val
        else:
            raise EmptyStackException

    def peek(self):
        if self.top:
            return self.top.element
        else:
            raise EmptyStackException

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

    def __str__(self):
        elements = []
        curr = self.top
        while curr is not None:
            elements.append(curr.element)
            curr = curr.next_node
        return elements.__str__()

    def __repr__(self):
        return self.__str__()

__init__

The __init__ method intializes the data structure to an empty stack.

def __init__(self):
    self.top = None
    self.size = 0

We have two class variables in our stack. The first variable, self.top, is a Node that represents the last item inserted into the Stack. It’s next_node property points to the next item in the stack, and so one. The other class variable is self.size, which represents the number of items that are placed in the stack.

push

The push method is used to add items to the stack.

def push(self, data):
    self.top = Node(data, self.top)
    self.size += 1

Whenever we add items to a Stack, the newest Node becomes the top of the stack. If you remember back from our Node class, it’s __init__ method has an optional next_node argument we can use to initialize the new Node with it’s next_node. This works well for us here because we can pass self.top as the next_node of the new Node and then immediatly assign the new Node to self.top. In this way, the last item in the Stack get’s shifted right and the new item becomes the top of the Stack. The next line of code simply increments the size of the stack by one, thus completing the operation of adding (or pushing) an item onto the Stack.

pop

The pop method is opposite operation to the push method. This method removes an item from the Stack.

def pop(self):
    if self.top:
        val = self.top.element
        self.top = self.top.next_node
        self.size -= 1
        return val
    else:
        raise EmptyStackException

The pop method needs to consider two possible cases. The fist case involves a Stack that has items on it and needs an item removed. The other case involves an empty stack. Let’s begin with the first case.

We start by checking if the stack has items. If the stack is empty, self.top is None. Assuming that self.top is not None, we begin by storing the data contained in self.top.element in a variable called val. Our next job is to delete that Node from the Stack by changing self.top to point at self.top.next_node. The original Node will get garbage collected by the Python runtime. Now that we have removed the Node, we need to shrink the size of the stack by one. After we decrease self.size, we can return val to the caller thus completing the pop operation.

If it turns our that the stack is empty, we need to notify the caller. I think it would be a very bad practice to simply return None, since calling pop() on an empty Stack would indicate a programming error. Thus, we raise EmptyStackException. The client code will either need to handle the Exception or allow the program to crash.

peek

The peek operation lets client code take a look at the first element in the Stack without removing the item.

def peek(self):
    if self.top:
        return self.top.element
    else:
        raise EmptyStackException

In this case, we just just need to return self.top.element if the stack has an item, or raise an EmptyStackException if the stack is empty.

empty

Clients of this class need to check if the Stack is empty prior to calling pop() or peek()

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

This method simply returns True if self.size == 0 or False if self.size is non-zero.

__str__ and __repr__

We don’t need these methods to represent a Stack, but if you choose to implement them, debuggers such as the on found in PyCharm will use these methods to print more user friendly data to the debugger window.

def __str__(self):
    elements = []
    curr = self.top
    while curr is not None:
        elements.append(curr.element)
        curr = curr.next_node
    return elements.__str__()

def __repr__(self):
    return self.__str__()

We are going to leverage Python’s list object to get a reader friendly String representation of our Stack. We start by making an empty list called elements. Then we create a curr variable that points at self.stop. Now, we are going to loop until the end of the Stack by checking if curr is None.

During each iteration of the while loop, we add curr.element to elements. Then we advance to the next node by assigning curr = curr.next_node. When we are done, we will return elements.__str__() which creates a String for us. The __repr__ method works by calling self.__str__().

Conclusion

This article is a simple linked list implementation of a Stack. Stacks are useful in cases where you need a LIFO data structure to track the order of which items are added to the stack. As you can see, a Stack is a much simplier data structure to implement than a linked list.

Doubly Linked List—Python

We saw in Singly Linked List the benefits of using a linked list over an array. Linked lists are data structures that grow and expand as needed. Not only is it easy to add an element at the beginning or end of a linked list, but it is also easy to insert elements at arbitrary positions within the list. We can also easily remove elements from the linked list when they are no longer needed.

The doubly linked list is a varient of the singly linked list. The main complaint about a singly linked list is that it can only traverse the list in one direction starting at the head and working until it reaches the end of the list. Clients may not notice a performance hit when operating on small lists, but large lists will almost certainly have a performance hit.

Doubly linked lists work by tracking both the next node and previous nodes in the list. Tracking both nodes creates more overhead when inserting or removing from the list, but it allows the list to work in a bi-directional fashion. That can allow for a major performance boost when operating on large lists.

Here is the entire code followed by an explanation as to how it works. Note that the topic is the Doubly Linked List so I won’t be covering the unit testing code in this module, but I did include for those who are interested.

class Node:
    def __init__(self, element=None, next_node=None, prev_node=None):
        self.element = element
        self.next_node = next_node
        self.prev_node = prev_node

    def __str__(self):
        if self.element:
            return self.element.__str__()
        else:
            return 'Empty Node'

    def __repr__(self):
        return self.__str__()


class DoublyLinkedList:
    def __init__(self):
        self.head = Node(element='Head')
        self.tail = Node(element='Tail')

        self.head.next_node = self.tail
        self.tail.prev_node = self.head

    def size(self):
        count = 0
        current = self.head.next_node

        while current is not None and current != self.tail:
            count += 1
            current = current.next_node

        return count

    def insert_front(self, data):
        node = Node(element=data, next_node=self.head.next_node, prev_node=self.head)
        self.head.next_node.prev_node = node
        self.head.next_node = node

    def insert_last(self, data):
        node = Node(element=data, next_node=self.tail, prev_node=self.tail.prev_node)
        self.tail.prev_node.next_node = node
        self.tail.prev_node = node

    def insert(self, data, position):
        if position == 0:
            self.insert_front(data)
        elif position == self.size():
            self.insert_last(data)
        else:
            if 0 < position < self.size():
                current_node = self.head.next_node
                count = 0
                while count < (position - 1):
                    current_node = current_node.next_node
                    count += 1

                node = Node(element=data, next_node=current_node.next_node, prev_node=current_node)
                current_node.next_node.prev_node = node
                current_node.next_node = node
            else:
                raise IndexError

    def remove_first(self):
        self.head = self.head.next_node
        self.head.prev_node = None

    def remove_last(self):
        self.tail = self.tail.prev_node
        self.tail.next_node = None

    def remove(self, position):
        if position == 0:
            self.remove_first()
        elif position == self.size():
            self.remove_last()
        else:
            if 0 < position < self.size():
                current_node = self.head.next_node
                current_pos = 0

                while current_pos < position:
                    current_node = current_node.next_node
                    current_pos += 1

                next_node = current_node.next_node
                prev_node = current_node.prev_node

                next_node.prev_node = prev_node
                prev_node.next_node = next_node
            else:
                raise IndexError

    def fetch(self, position):
        if 0 <= position < self.size():
            current_node = self.head.next_node
            current_pos = 0

            while current_pos < position:
                current_node = current_node.next_node
                current_pos += 1

            return current_node.element
        else:
            raise IndexError


import unittest
from random import randint


class TestDoublyLinkedList(unittest.TestCase):
    names = ['Bob Belcher',
             'Linda Belcher',
             'Tina Belcher',
             'Gene Belcher',
             'Louise Belcher']

    def test_init(self):
        dll = DoublyLinkedList()
        self.assertIsNotNone(dll.head)
        self.assertIsNotNone(dll.tail)
        self.assertEqual(dll.size(), 0)

    def test_insert_front(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_front(name)

        self.assertEqual(dll.fetch(0), TestDoublyLinkedList.names[4])
        self.assertEqual(dll.fetch(1), TestDoublyLinkedList.names[3])
        self.assertEqual(dll.fetch(2), TestDoublyLinkedList.names[2])
        self.assertEqual(dll.fetch(3), TestDoublyLinkedList.names[1])
        self.assertEqual(dll.fetch(4), TestDoublyLinkedList.names[0])

    def test_insert_last(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_last(name)

        for i in range(len(TestDoublyLinkedList.names) - 1):
            self.assertEqual(dll.fetch(i), TestDoublyLinkedList.names[i])

    def test_insert(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_last(name)

        pos = randint(0, len(TestDoublyLinkedList.names) - 1)

        dll.insert('Teddy', pos)
        self.assertEqual(dll.fetch(pos), 'Teddy')

    def test_remove_first(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_last(name)

        for i in range(dll.size(), 0, -1):
            self.assertEqual(dll.size(), i)
            dll.remove_first()

    def test_remove_last(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_last(name)

        for i in range(dll.size(), 0, -1):
            self.assertEqual(dll.size(), i)
            dll.remove_last()

    def test_remove(self):
        dll = DoublyLinkedList()
        for name in TestDoublyLinkedList.names:
            dll.insert_last(name)

        dll.remove(1)

        self.assertEqual(dll.fetch(0), 'Bob Belcher')
        self.assertEqual(dll.fetch(1), 'Tina Belcher')
        self.assertEqual(dll.fetch(2), 'Gene Belcher')
        self.assertEqual(dll.fetch(3), 'Louise Belcher')


if __name__ == '__main__':
    unittest.main()

Node

Like the singly linked list, the doubly linked list begins with a Node class that holds the data contained in the list and references to the previous and next nodes. Here is the code for the Node.

class Node:
    def __init__(self, element=None, next_node=None, prev_node=None):
        self.element = element
        self.next_node = next_node
        self.prev_node = prev_node

    def __str__(self):
        if self.element:
            return self.element.__str__()
        else:
            return 'Empty Node'

    def __repr__(self):
        return self.__str__()

Here is a break down of each methods in the Node class.

__init__

The __init__ method creates the node. All three of its parameters are optional, but basically we are setting the data contained in the node, and building refrences to next_node and previous_node.

def __init__(self, element=None, next_node=None, prev_node=None):
    self.element = element
    self.next_node = next_node
    self.prev_node = prev_node

__str__ and __repr__

I found that there was more work invloved with debugging doubly linked list, so in this version of the Node class, I chose to implement __str__ and __repr__ so that I could easily identify each Node in my debugger.

def __str__(self):
    # Check if self.element is null
    if self.element:
        # Just return the string representation
        # of self.element
        return self.element.__str__()
    else:
        # Otherwise return Empty Node
        # to indicate the node does not
        # hold data
        return 'Empty Node'

# We are just going to reuse the
# code in __str__ here
def __repr__(self):
    return self.__str__()

Doubly Linked List

This is an example of a doubly linked list. It’s not opitmized for the simply fact that we are learning. If you need an optimized list structure, you Python’s list object.

class DoublyLinkedList:
    def __init__(self):
        self.head = Node(element='Head')
        self.tail = Node(element='Tail')

        self.head.next_node = self.tail
        self.tail.prev_node = self.head

    def size(self):
        count = 0
        current = self.head.next_node

        while current is not None and current != self.tail:
            count += 1
            current = current.next_node

        return count

    def insert_front(self, data):
        node = Node(element=data, next_node=self.head.next_node, prev_node=self.head)
        self.head.next_node.prev_node = node
        self.head.next_node = node

    def insert_last(self, data):
        node = Node(element=data, next_node=self.tail, prev_node=self.tail.prev_node)
        self.tail.prev_node.next_node = node
        self.tail.prev_node = node

    def insert(self, data, position):
        if position == 0:
            self.insert_front(data)
        elif position == self.size():
            self.insert_last(data)
        else:
            if 0 < position < self.size():
                current_node = self.head.next_node
                count = 0
                while count < (position - 1):
                    current_node = current_node.next_node
                    count += 1

                node = Node(element=data, next_node=current_node.next_node, prev_node=current_node)
                current_node.next_node.prev_node = node
                current_node.next_node = node
            else:
                raise IndexError

    def remove_first(self):
        self.head = self.head.next_node
        self.head.prev_node = None

    def remove_last(self):
        self.tail = self.tail.prev_node
        self.tail.next_node = None

    def remove(self, position):
        if position == 0:
            self.remove_first()
        elif position == self.size():
            self.remove_last()
        else:
            if 0 < position < self.size():
                current_node = self.head.next_node
                current_pos = 0

                while current_pos < position:
                    current_node = current_node.next_node
                    current_pos += 1

                next_node = current_node.next_node
                prev_node = current_node.prev_node

                next_node.prev_node = prev_node
                prev_node.next_node = next_node
            else:
                raise IndexError

    def fetch(self, position):
        if 0 <= position < self.size():
            current_node = self.head.next_node
            current_pos = 0

            while current_pos < position:
                current_node = current_node.next_node
                current_pos += 1

            return current_node.element
        else:
            raise IndexError

__init__

This doubly linked list implementation makes use of a head and a tail node. We traversing the list, the code will look for either head or tail as a means to detect if we are at the end of the list.

def __init__(self):
    self.head = Node(element='Head')
    self.tail = Node(element='Tail')

    self.head.next_node = self.tail
    self.tail.prev_node = self.head

We pass the strings “Head” and “Tail” as the element data for each of these nodes for debugging purpose. Since Node implements __str__ and __repr__, we will see Heads and Tail in the debugger (at least when using PyCharm).

The next two lines do the work of pointing head and tail at each other. When the list is created, head.next_node is tail. Conversly, tail.prev_node is head. This indicates an empty list.

size

This code inefficient on purpose. A better implementation would use a length varaible and increment or decrment it as Nodes are added and removed. However, in this case, I wanted to show how to traverse a list without the clutter of adding or removing Nodes.

def size(self):
    count = 0
    current = self.head.next_node

    while current is not None and current != self.tail:
        count += 1
        current = current.next_node

    return count

We start by making a count variable and a current variable that points at self.head.next_node. It’s really important that we use self.head.next_node instead of self.head because self.head’s purpose is to mark the beginning of the list, not contain data. If we failed to make this distinction, the size of the list will be off by one.

Now, we are going to traverse the list increment count by one an each iteration of the loop. We should check for None for defensive programming purposes, but the critical part of the loop condition is if current != self.tail. The self.tail Node marks the end of the list and so we do not wish to include it in the size of our list.

insert_front

This method inserts a new Node at the front of the Linked List.

def insert_front(self, data):
    node = Node(element=data, next_node=self.head.next_node, prev_node=self.head)
    self.head.next_node.prev_node = node
    self.head.next_node = node

Nodes inserted at the front of the list need their prev_node to point at self.head and their next_node to point at the Node located at self.head.next_node.

To complete the insertion, we now need to update self.head.next_node.prev_node to point back at the new Node that we created (otherwise, it would still point at self.head). Likewise, we need to update self.head.next_node to point at our new Node.

insert_last

The code for this is almost identical to insert_head with the main difference being that we are working on the tail Node rather than the head Node.

def insert_last(self, data):
    node = Node(element=data, next_node=self.tail, prev_node=self.tail.prev_node)
    self.tail.prev_node.next_node = node
    self.tail.prev_node = node

Once again, we create a new Node. It’s next_node has to point at self.tail and it’s prev_node needs to point at self.tail.prev_node.

To complete the insertion, we need to update self.tail.prev_node.next_node to point at our new Node (otherwise it will continue to point at self.tail). We also need to point self.tail.prev_node at our new Node.

insert

This method handles the insertion of a Node into the middle of the list. Keep in mind that it has to handle four possible cases.

  • Position could be 0 (front of list)
  • Position could be equal to size() (end of list)
  • Position is in the middle of the list
  • Position is size() (out of bounds)
def insert(self, data, position):
    if position == 0:
        # First case, we insert at front
        self.insert_front(data)
    elif position == self.size():
        # Second case, we insert at end
        self.insert_last(data)
    else:
        if 0 < position < self.size():
            # Third case, insert in middle
            current_node = self.head.next_node
            count = 0
            while count < (position - 1):
                current_node = current_node.next_node
                count += 1

            node = Node(element=data, next_node=current_node.next_node, prev_node=current_node)
            current_node.next_node.prev_node = node
            current_node.next_node = node
        else:
            # Fourth case, index out of bounds
            raise IndexError

The comments point out how this code handles each of the possible cases. Let's focus on the insertion. We begin by creating a count variable and current_node variable. Now, we need to traverse the list until we arrive at the node right before the desired position. Each interation of the loop requires us to point current_node at current_node.next_node.

Once we arrive at our destination, we create a new Node. We will point it's next_node at current_node.next_node and its prev_node at current_node. Doing the insertion at this point in the list causes the new node to appear at the position index specified in the method parameters.

To complete the insertion, we point current_node.next_node.prev_node at the new Node (otherwise it would still point at current_node). Likewise, we point current_node.next_node at our new Node.

Again, I should mention that this code is inefficient again. Normally we would make use of the bidirectional capabilities of the list and decide if we want to traverse the list in forward (like we do here) or in reverse. For example, if we are trying insert into a list at size() – 2, it doesn't make sense to start at the head node and traverse forward to find our insertion point when we could start at tail and move backwards.

remove_first

It’s really simple to remove nodes from the head of a doubly linked list.

def remove_first(self):
     self.head = self.head.next_node
     self.head.prev_node = None

You will notice that all we need to do is point self.head at self.head.next_node. Then we just set self.head.prev_node to None so that it doesn’t continue to point at the old self.head. The old Node will get garbage collected by the Python runtime.

remove_last

Removing the tail from the list is just as easy as removing the head.

def remove_last(self):
    self.tail = self.tail.prev_node
    self.tail.next_node = None

In this case, we just point self.tail at self.tail.prev_node. To remove the old Node, we next poitn self.tail.next_node to None. The old Node will get garbage collected by the Python runtime.

remove

Remove needs to handle the same cases that insert has to handle. Otherwise it works by traversing the list of the position to remove and deletes the Node.

def remove(self, position):
    if position == 0:
        # First case, remove at front
        self.remove_first()
    elif position == self.size():
        # Second case, remove from end
        self.remove_last()
    else:
        if 0 < position < self.size():
            # 3rd case, remove at middle
            current_node = self.head.next_node
            current_pos = 0

            while current_pos < position:
                current_node = current_node.next_node
                current_pos += 1

            next_node = current_node.next_node
            prev_node = current_node.prev_node

            next_node.prev_node = prev_node
            prev_node.next_node = next_node
        else:
            # 4th case, invalid position
            raise IndexError

Again, we will focus on the removal part of this code. We begin by traversing the list to find our desired position. This time, we stop at the actual position rather than the position before it. Once we arrive at our desired position, we store current_node.next_node and current_node.prev_node into their respective variables. To remove current_node, we simply need to point next_node.prev_node at prev_node and likewise point prev_node.next_node to next_node. Since there are no longer any references to current_node, it is garbage collected by the Python runtime.

fetch

Our final method is the fetch method, which let’s use retreive items from the Linked List.

def fetch(self, position):
    if 0 <= position < self.size():
        current_node = self.head.next_node
        current_pos = 0

        while current_pos < position:
             current_node = current_node.next_node
             current_pos += 1

        return current_node.element
    else:
        raise IndexError

By now readers should be familiar with how the list is traversed. We simply traverse the list to the desired position and return the Node.element to the caller.

Conclusion

Doubly linked list are a more advance form of linked list that supports bi-directional traversals. Bi-directional navigation is possible because each Node in the linked list stores a refrence to both the next and previous Node in the list.

Our linked list implementation only utilizes forward direction traversal to help keep the implementation simple. However, it would only require a small amount of work to make use of reverse traversals.