I am learning the linked list implemented in Python:
$ cat linked_stack.py
class LinkedStack:
"""LIFO Stack implementation using a singly linked list for storage"""
#-------------nested Node class ------------------------------------
class _Node:
"""Lightweight, nonpublic class for storing a singly linked node."""
__slots__ = "_element", "_next"
def __init__(self, element, next): #initialize node's fields
self._element = element
self._next = next
#--------------stack methods---------------------------------------
def __init__(self):
"""Create an empty stack"""
self._head = None
self._size = 0
def top(self):
"""Return (but do not remove) the element at the top of the stack.
Raise Empty exception if the stack is empty."""
if self.is_empty():
raise Exception("Stack is empty")
return self._head._element
def __len__(self)->int:
return self._size
def is_empty(self)->bool:
return len(self._size) == 0
def push(self, e) -> None:
self._head = self._Node(e, self._head) #create and link a new node
self._size += 1
def pop(self)
Reference to the top method is:
def top(self):
"""Return (but do not remove) the element at the top of the stack.
Raise Empty exception if the stack is empty."""
if self.is_empty():
raise Exception("Stack is empty")
return self._head._element
I want to annotate it as:
def top(self)-> _Node._element:
How to get it done in a decent way?
Related
Intuitively, using __iter__ and __next__ seems likely to be significantly for efficient, but I curious of the technical specifics of why, and how one maybe go about benchmarking to measure and compare the space and runtime of these?
A common case I can think of is a Linked List,
The following is an implementation illustration the use of __iter__ and __next__:
I have tried various benchmarking tools without much success being able to generate data to more closely analyze they're internals; specifically, how exactly the magic methods function, and if they're truly as significantly more efficient as I surmise they are.
class Node(object):
def __init__(self, value: int, next_node):
self.value = value
self.next = next_node
def __repr__(self):
return str(self.value)
class LinkedList(object):
def __init__(self, head=None):
self.head = None
self._current = None
def __repr__(self):
return '<{} {}>'.format(type(self).__name__, self.head)
def __iter__(self):
self._current = self.head
return self
def __next__(self):
if self._current is None:
raise StopIteration
current, self._current = self._current, self._current.next
return current
#property
def is_empty(self):
return self.head is None
def prepend_node(self, valuetoadd: int):
self.head = IntNode(valuetoadd, self.head)
Compared to the following:
class ListNode:
"""
A node in a singly-linked list.
"""
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def __repr__(self):
return repr(self.data)
class SinglyLinkedList:
def __init__(self):
"""
Create a new singly-linked list.
Takes O(1) time.
"""
self.head = None
def prepend(self, data):
"""
Insert a new element at the beginning of the list.
Takes O(1) time.
"""
self.head = ListNode(data=data, next=self.head)
def find(self, key):
"""
Search for the first element with `data` matching
`key`. Return the element or `None` if not found.
Takes O(n) time.
"""
curr = self.head
while curr and curr.data != key:
curr = curr.next
return curr # Will be None if not found
def remove(self, key):
"""
Remove the first occurrence of `key` in the list.
Takes O(n) time.
"""
# Find the element and keep a
# reference to the element preceding it
curr = self.head
prev = None
while curr and curr.data != key:
prev = curr
curr = curr.next
# Unlink it from the list
if prev is None:
self.head = curr.next
elif curr:
prev.next = curr.next
curr.next = None
Here is my code. I created a linked list manually to check if my method works or not. But the output I get is nothing. There is no output
class Node:
def __init__(self, data=None):
self.data = data
self.next = None
node1=Node(2)
node2=Node(4)
node3=Node(5)
node1.next=node2
node2.next=node3
a=node1
class MyList():
def __init__(self):
self.head=Node()
def isEmpty(self,a):
return self.head.next is None
hello=MyList()
print(hello.isEmpty(a))
In case you want to add data to LinkedList, you need to set the head of the list manually.
This code is probably what you want:
class Node:
def __init__(self, data=None):
self.data = data
self.next = None
node1=Node(2)
node2=Node(4)
node3=Node(5)
node1.next=node2
node2.next=node3
class MyList():
def __init__(self):
self.head=Node()
def isEmpty(self):
return self.head.data is None
hello=MyList()
print(hello.isEmpty())
new_hello = MyList()
new_hello.head=node1
print(new_hello.isEmpty())
Output
True
False
You never set the head node to point to another node in the way you're currently doing it (your head and "a" aren't actually the same node here). Pass in another variable as a node object to change this.
a = node1
class MyList():
def __init__(self, head):
self.head = head
def isEmpty(self):
return self.head is None # a linked list with a head is technically not empty
new_hello = MyList(a)
print (new_hello.isEmpty())
personally I would add an add_node(self, value) method and keep track of the end node as well, instead of doing it the way you are
I'm reading a book on data structures and algorithms and understand the overall goal of the code but for top() and pop() there are strange member variables. Is this some sort of python abstraction that I am running into (ex. self._head._element) or a feature of not using pythons dictionary variable assignment?
'''
class LinkedStack:
class _Node:
__slots__ = '_element','_next' #treats elements not like a dictionary
def __init__(self, element, next):
self._element = element
self._next = next
def __init__(self):
self._head = None
self._size = 0
def __len__(self):
return self._size
def is_empty(self):
return self._size == 0
def push(self, e):
self._head = self._Node(e, self._head)
self._size += 1
def top(self):
if self.is_empty():
raise Empty('Stack is empty')
else:
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
'''
I have a LinkedList class that has close to 200 lines of code. I would like to make a new class LLCircular(LinkedList) by always making sure that any myLL.tail.next is myLL.head. I believe I would need to update append(), push(), remove(), etc. accordingly. Is there a way I can do this to keep the original LinkedList class intact? Maybe a decorator or some dunder method?
For brevity's sake, if reading the code, my push() method is just the inverse of append(). I also have a pop() and remove() method, which would need to be updated, if I just rewrite those methods. As I am trying to avoid that approach, I am not posting that part of the code.
class LinkedListNode:
def __init__(self, value, nextNode=None, prevNode=None):
self.value = value
self.next = nextNode
self.prev = prevNode
def __str__(self):
return str(self.value)
class LinkedList:
def __init__(self, values=None):
self.head = None
self.tail = None
if values is not None:
self.append(values)
def __str__(self):
values = [str(x) for x in self]
return ' -> '.join(values)
def append(self, value=None):
if value is None:
raise ValueError('ERROR: LinkedList.py: append() `value` PARAMETER MISSING')
if isinstance(value, list):
for v in value:
self.append(v)
return
elif self.head is None:
self.head = LinkedListNode(value)
self.tail = self.head
else:
''' We have existing nodes '''
''' Head.next is same '''
''' Tail is new node '''
self.tail.next = LinkedListNode(value, None, self.tail)
self.tail = self.tail.next
if self.head.next is None:
self.head.next = self.tail.prev
return self.tail
'''class LLCircular(LinkedList):'''
''' ??? '''
Test Code:
foo = LinkedList([1,2,3])
foo.tail.next = foo.head #My LL is now circular
cur = foo.head
i = 0
while cur:
print(cur)
cur = cur.next
i +=1
if i>9:
break
What you want is to call your LinkedList base class functions using the super keyword, and then add the slight modifications to the LLCircular class function, i.e:
class LLCircular(LinkedList):
def append(self, value=None):
super(LLCircular, self).append(value)
# In addition to having called the LinkedList append, now you want
# to make sure the tail is pointing at the head
self.tail.next = self.head
self.head.prev = self.tail
If it is "circular" it won't need a tail or head, would it?
Nor "append" makes sense - insert_after and insert_before methods should be enough - also, any node is a reference to the complete circular list, no need for different objects:
class Circular:
def __init__(self, value=None):
self.value = value
self.next = self
self.previous = self
def insert_after(self, value):
node = Circular(value)
node.next = self.next
node.previous = self
self.next.previous = node
self.next = node
def insert_before(self, value):
node = Circular(value)
node.next = self
node.previous = self.previous
self.previous.next = node
self.previous = node
def del_next(self):
self.next = self.next.next
self.next.previous = self
def __iter__(self):
cursor = self.next
yield self
while cursor != self:
yield cursor
cursor = cursor.next
def __len__(self):
return sum(1 for _ in self)
I have the following Linked List implementation. There is a problem with the printlist() function. The while loop is turning an error that there is no attribute next for self. Is there a better way to write this function? Thank you!!!
class Node:
def __init__(self, data, next=None):
self.data=data
def _insert(self, data):
self.next=Node(data)
def _find(self, data):
if self.data==data:
return self
if self.next is None:
return None
while self.next is not None:
if self.next.data==data:
return self.next
return None
def _delete(self, data):
if self.next.data == data:
temp=self.next
self.next =self.next.next
temp=None
def _printtree(self):
while self:
print self.data,
self=self.next
class LinkedList:
def __init__(self):
self.head=None
def insert(self, data):
if self.head:
self.head._insert(data)
else:
self.head=Node(data)
def find(self, data):
if self.head.data==data:
return self.head
return self.head._find(data)
def delete(self, data):
if self.head.data==data:
head=None
return
self.head._delete(data)
def printtree(self):
self.head._printtree()
add next attribute to Node's ini method
you should define printtree of LinkedList this way:
def printree(self):
current_node = self.head
print current_node.data
while current_node.next is not None:
print current_node.next.data
current_node = current_node.next
adding a repr method will make your code nicer