I am brushing up on some data structures and algorithms with Python, so I am implementing an Unordered Linked list. Within the same file I first wrote a Node class followed by a List class. What I don't get is how the "current" variable in my search_item() method seems to be a node object or at least able to access the Node class methods and attributes. I noticed that if I comment out my add_node() method then "current" no longer has access to Node's methods. Now I am not explicitly using neither inheritance nor composition, so I am having a hard time seeing how current just gets to call get_next() the way the code is written below. I would think I'd have to declare current as: current = Node(self.head) but just current = self.head seems to work?
Your help would be greatly appreciated.
class Node:
def __init__(self, data):
self.data = data
self.next = None
def get_data(self):
return self.data
def set_data(self, d):
self.data = d
def get_next(self):
return self .next
def set_next(self, n):
self.next = n
class UnorderedList:
def __init__(self):
self.head = None
def add_node(self, item):
tmp = Node(item)
tmp.set_next(self.head)
self.head = tmp
def search_item(self, item):
current = self.head
# current = Node(self.head)
found = False
while current != None and not found:
if current.get_data() == item:
found = True
else:
current = current.get_next()
return found
Well if you comment out add_node then you do not add nodes to your linked list any longer therefore search_item will always see the initial value of self.head which is None.
Calling current.get_next() just works because via add_node you always ensure that self.head points either to None or to an instance of Node as tmp is created by tmp = Node(item) and then assigned to self.head = tmp. Therefore when setting current = self.head it will already refer to an instance of Node (or None) so you do not need to call current = Node(self.head).
I just recently came across the concept of duck typing and this old post of mine came to mind. It seems that this is what's at play and just didn't understand it at the time. 'current' is set to None by default, by when invoking any of the methods defined in the Node class, it is automatically defined as Node object.
Related
I tried program to delete node from linked list recursively. My program is given below
class node:
def __init__(self, data=None):
self.data = data
self.next = None
class linkedList:
def __init__(self):
self.head=None
def printList(self):
cur = self.head
while(cur != None):
print(cur.data,end="->")
cur=cur.next
print("null")
def push(self,dat):
newNode = node(dat)
temp = self.head
self.head = newNode
newNode.next = temp
#staticmethod
def dnr(head, key):
if head is not None:
if head.data == key:
head=head.next
return head
if head is not None:
head.next = linkedList.dnr(head.next, key)
return head
return head
if __name__ == '__main__':
ll=linkedList()
ll.push(3)
ll.push(6)
ll.push(9)
ll.push(12)
ll.push(15)
ll.printList()
print("*******")
# ll.head = linkedList.dnr(ll.head,2)
linkedList.dnr(ll.head,9)
ll.printList()
The problem with this is that this does not work for first element.To make it work for first element I have to call the function like this
ll.head = linkedList.dnr(ll.head,2)
second thing is that I wanted my function to call this way
ll.dnr(2)
please tell me how to create a recursive function to delete node in linked list in python
I rewrote your code:
class node:
def __init__(self, data=None):
self.data = data
self.next = None
class linkedList:
def __init__(self):
self.__head=None
def printList(self):
cur = self.__head
while(cur != None):
print(cur.data,end="->")
cur=cur.next
print("null")
def push(self,dat):
newNode = node(dat)
temp = self.__head
self.__head = newNode
newNode.next = temp
#staticmethod
def __dnr(head, key):
if head is None:
return head
if head.data == key:
head = head.next
return head
head.next = linkedList.__dnr(head.next, key)
return head
#staticmethod
def dnr(listObj, key):
if listObj is None or listObj.__head is None:
return listObj
if listObj.__head.data == key:
listObj.__head = listObj.__head
listObj.__head = linkedList.__dnr(listObj.__head, key)
def deleteKey(self, key):
linkedList.dnr(self, key)
if __name__ == '__main__':
ll=linkedList()
ll.push(3)
ll.push(6)
ll.push(9)
ll.push(12)
ll.push(15)
ll.printList()
print("*******")
linkedList.dnr(ll, 9)
ll.deleteKey(12)
ll.printList()
I made head variable inside linkedList class private, it's not smart to give access outer world to class core components. And class core components should never leak to the outer world because if it's that not used properly that can cause errors. So I rewrote your dnr function and now it's more clear and it doesn't return head object which is core component of linkedList class. Now dnr function just check if passed listObj is valid and check if head is that node that should be deleted. After that it calls private static __dnr function to delete node with given key. Function deleteKey can be called like this ll.deleteKey(12), that is what you wanted.
Giving core component accessible through outer world is like giving bank customer access to a bank vault. Not everybody will try to steal money from it, but there will be someone who will try.
If you don't understand private variables follow this link.
I wanted my function to call this way ll.dnr(2)
Then you need to define an instance method. Your static function can serve a purpose, but as you noted, you really need to assign its return value back to your list's head attribute to be sure it also works when the original head node is removed. With your static method you cannot avoid this overhead, since that method has no knowledge about your linkedList instance.
You can achieve what you want simply by adding an instance method, that will rely on the existing static method and will deal with this assignment back to the instance's head attribute.
Add this to your linkedList class:
def remove(self, key):
self.head = linkedList.dnr(self.head, key)
Now in your main program you can do:
ll.remove(15)
Side note: you don't need the second if head is not None: check in your static method, as this condition will always be true when the execution reaches that point in your code. Just do the assignment to head.next unconditionally.
Addendum
If you want dnr itself to become an instance method (without addition of a remove method), then you need to temporarily cut off the head node from the list (even if you want to keep it), recur, and then conditionally add that cut-off node again (if it's key is not the one to delete).
It would look like this:
def dnr(self, key):
head = self.head
if head:
self.head = head.next # Skip
if head.data != key: # Need to restore it
self.dnr(key) # But first look further...
head.next = self.head # Prefix the removed node
self.head = head # ...and make it the head again
You would call like:
ll.dnr(15)
I'm writing a linked list in Python and I've come across an issue which is really troublesome and terrible for debugging and I feel like I'm missing something about Python. I'm supposed to create a singly linked list with some basic functionalities. One of them is the take() function, which is meant to create a new list of n first elements of the original list.
However, creating a new instance of the LinkedList class seems to change the .self parameter and the variable node is modified, as the attribute .next is turned to None. In result, when creating a list and then trying to make a new one out of the n elements of it, the program runs indefinitely, but no matter which part I look at, I cannot find the loop or the reason behind it.
class LinkedList:
def __init__(self, head=None):
self.head = head
def is_empty(self):
if self.head == None:
return True
else:
return False
def add_last(self, node):
if self.is_empty():
self.head = node
return
nextEl = self.head
while True:
if nextEl.next is None:
nextEl.next = node
return
nextEl = nextEl.next
def take(self, n):
node = self.head
newHead = self.head
newHead.next = None
newList = LinkedList(newHead)
count = 0
while count < n:
newList.add_last(node.next)
node = node.next
count += 1
return newList
class Node:
def __init__(self, data, next=None):
self.data = data
self.next = next
Thank you for all your help.
In the take() function the line
newHead.next = None
modifies a node of the linked list, breaking this list. You can fix this as follows:
def take(self, n):
node = self.head
newHead = Node(self.head.data)
newList = LinkedList(newHead)
count = 0
while count < n:
newList.add_last(node.next)
node = node.next
count += 1
return newList
This, however, still will not work correctly since there is a problem with the add_last() function too. This function, I think, is supposed to add a node as the last element of a linked list, but since you do not modify the next attribute of the node, you actually append a whole linked list starting with that node. This can be fixed in the following way:
def add_last(self, node):
if self.is_empty():
self.head = node
return
nextEl = self.head
while True:
if nextEl.next is None:
nextEl.next = Node(node.data)
return
nextEl = nextEl.next
There are more issues. For example, take(sefl, n) will actually create a list of n+1 elements and will throw an exception if it is applied to a linked list that does not have that many elements.
I have a Linked Lists assignment for school although I am just getting the hang of class constructors. I am trying to simply get the basics of the linked list data structure down, and I understand the basic concept. I have watched lots of Youtube tutorials and the like, but where I am failing to understand is how to print out the cargo or data in my nodes using a loop.
I have written something along these lines:
class Node:
def __init__(self, value, pointer):
self.value = value
self.pointer = pointer
node4 = Node(31, None)
node3 = Node(37, None)
node2 = Node(62, None)
node1 = Node(23, None)
Now...I understand that each node declaration is a call to the class constructor of Node and that the list is linked because each node contains a pointer to the next node, but I simply don't understand how to print them out using a loop. I've seen examples using global variables for the "head" and I've seen subclasses created to accomplish the task. I'm old and dumb. I was wondering if someone could take it slow and explain it to me like I'm 5. If anyone out there has the compassion and willingness to hold my hand through the explanation, I would be greatly obliged. Thank you in advance, kind sirs.
First of all, your nodes should be created something like this :
node4 = Node(31, node3)
node3 = Node(37, node2)
node2 = Node(62, node1)
node1 = Node(23, None)
Now, i am sure you can see that the last node in the list would point to None. So, therefore, you can loop through the list until you encounter None. Something like this should work :
printhead = node4
while True:
print(printhead.value)
if printhead.pointer is None:
break;
else :
printhead = printhead.pointer
This is a very basic linked list implementation for educational purposes only.
from __future__ import print_function
"""The above is needed for Python 2.x unless you change
`print(node.value)` into `print node.value`"""
class Node(object):
"""This class represents list item (node)"""
def __init__(self, value, next_node):
"""Store item value and pointer to the next node"""
self.value = value
self.next_node = next_node
class LinkedList(object):
"""This class represents linked list"""
def __init__(self, *values):
"""Create nodes and store reference to the first node"""
node = None
# Create nodes in reversed order as each node needs to store reference to next node
for value in reversed(values):
node = Node(value, node)
self.first_node = node
# Initialize current_node for iterator
self.current_node = self.first_node
def __iter__(self):
"""Tell Python that this class is iterable"""
return self
def __next__(self):
"""Return next node from the linked list"""
# If previous call marked iteration as done, let's really finish it
if isinstance(self.current_node, StopIteration):
stop_iteration = self.current_node
# Reset current_node back to reference first_node
self.current_node = self.first_node
# Raise StopIteration to exit for loop
raise stop_iteration
# Take the current_node into local variable
node = self.current_node
# If next_node is None, then the current_node is the last one, let's mark this with StopIteration instance
if node.next_node is None:
self.current_node = StopIteration()
else:
# Put next_node reference into current_node
self.current_node = self.current_node.next_node
return node
linked_list = LinkedList(31, 37, 62, 23)
for node in linked_list:
print(node.value)
This doesn't handle many cases properly (including break statement in the loop body) but the goal is to show minimum requirements for linked list implementation in Python.
Hey everyone I am trying to print (2->None, 3) in the linked list below, but am getting a syntax error. can someone please help me with this syntax cannot find on google for some reason. Code below:
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
def Insert(head, data):
if head == None:
head = Node(data)
print(head)
else:
current = head
while current.next != None:
current = current.next
current.next = Node(data)
return head
print Insert(2->None, 3) # -> is bringing a syntax error, how do I write this in python 2.7?
Honestly, I think your confused about how to implement a linked list in the first place. If you want a linked list in the form of:
[2] -> [3] -> [None]
Then you need to insert each element backwards. First None then 3 then 2. You also need to make your insert method inside of a class because you need to save state. Here is what I'd recommend instead:
class Node(object):
def __init__(self, data, next_node=None):
self.data = data
self.next_node = next_node
# Create a class not a function because we need to save state
# More specficly, we need to create a "global" variable which
# keeps track of the head of the linked list.
class LinkedList(object):
def __init__(self):
self.head = None
# Put the insert function inside of of the class.
# That way, we can save and load the state of self.head.
def insert(self, data):
new_node = Node(data, self.head)
self.head = new_node
# demo
ll = LinkedList()
# insert the elements in the reverse order you want
# them to appear.
ll.insert(None)
ll.insert(3)
ll.insert(2)
print("Head:", ll.head.data) # Head: 2
print("Middle:", ll.head.next_node.data) # Middle: 3
print("Tail:", ll.head.next_node.next_node.data) # Tail: None
Also, I recommend to a little research as well. Such as this article. Or just google Linked List in Python and browse over some of the results.
I have a feeling you just want to insert None as the first argument as what #leaf said. In that case, just directly use None:
print Insert(None, 3)
-> doesn't exist in Python as of now. This is #leaf's comment:
If you want to pass in none, simply pass in None: Insert(None, 3). Python doesn't have pointer in the sense your thinking about.
Below is a simple linked list program, I know how a linked list works conceptually ( adding, removing, etc) but I am finding it hard to understand how it works from an object oriented design perspective.
Code:
class Node():
def __init__(self,d,n=None):
self.data = d
self.next_node = n
def get_next(self):
return self.next_node
def set_next(self,n):
self.next_node = n
def get_data(self):
return self.data
def set_data(self,d):
self.data = d
class LinkedList():
def __init__(self,r = None):
self.root = r
self.size = 0
def get_size(self):
return self.size
def add(self,d):
new_node = Node(d,self.root)
self.root = new_node
self.size += 1
def get_list(self):
new_pointer = self.root
while new_pointer:
print new_pointer.get_data()
new_pointer = new_pointer.get_next()
def remove(self,d):
this_node = self.root
prev_node = None
while this_node:
if this_node.get_data() == d:
if prev_node:
prev_node.set_next(this_node.get_next())
else:
self.root = this_node
self.size -= 1
return True
else:
prev_node = this_node
this_node = this_node.get_next()
return False
def find(self,d):
this_node = self.root
while this_node:
if this_node.get_data() == d:
return d
else:
this_node = this_node.get_next()
return None
myList = LinkedList()
myList.add(5)
myList.add(8)
myList.add(12)
myList.get_list()
I have couple questions here..
How is it storing the values. As far as I understand each variable can hold one value. So how does data / next_node hold multiple values. And does next_node hold the memory location of the next node?
new_pointer.get_data() How is new_pointer able to access get_data()? Don't we need to have an instance to access methods of Node?
This question may be silly, but I am quiet new to object oriented programming. If someone can answer these questions or post an external link addressing these questions it would be really helpful.
Thanks in advance.
next_node is an instance of Node and so it has its own data field. next_node is a reference to the node object, which is some memory address (however it is not a C-like pointer, as you don't need to dereference it or anything).
I'm assuming you are talking about get_list(). new_pointer is an instance of Node. (unless it is None, in which case you would never get into the get_data() call). When you do an add, you create this instance of Node and set root to it. Then in get_list you set new_pointer to root.
myList.root is storing one value only that is the root of the list. See initially when you do:
myList = LinkedList()
in memory myList.root = None (according to __init__ of LinkedList). Now:
myList.add(1)
Then this statement is called:
new_node = Node(d,self.root) #Note here self.root = None
and then:
def init(self,d,n=None):
self.data = d
self.next_node = n
So our list is : 1--> None.Now:
myList.add(2)
then again this statement is called:
new_node = Node(d,self.root) #Note here self.root has some value
now a new node object is created and its next is assigned to myList.root.
So our list becomes : 2-->1--> None
Going in similar fashion whole list is assigned.
Key thing to note here is that myList.root is always storing the top most node which in turn holds the next node and so on.
For your second question, it is quite clear from above explaination that always the object of class node is available to myList.root which in turn has next_node which is again an object of 'node'. So they all have access to 'get_data()' method.