Dictionary assignment in python or strange abstraction - python

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
'''

Related

Add typing hints to linkedlist's top method

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?

What is the easiest way to turn a LinkedList class into a Circular Linked List class?

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)

What method do I need to override for list child to return internal data array instead of itself?

I needed a dynamic size queue so I inherited list and Iterable and I tested all the stuff like appending, indexing and iterating but there is a small problem. Methods like
str(DataSet) return an empty list, and the Python debugger does show an empty list, which is expected, since I don't append to the list parent but the internal data. What do I need to override to not return self but return self.data?
class DataSet(collections.Iterable, list):
data = collections.deque()
index = 0
max = 3
def __add__(self, other):
self.append(other)
def append(self, item):
if len(self.data) >= self.max:
self.data.popleft()
self.data.append(item)
def __next__(self):
if self.index < len(self.data):
ret = self.data[self.index]
self.index += 1
return ret
raise StopIteration
def __getitem__(self, i):
return self.data[i]
def __iter__(self):
self.index = 0
return self
def setMax(self, max):
self.max = max
def __len__(self):
return len(self.data)
Override
def __str__(self):
return str(self.data)

Mirrorring a queue in python

I need to mirror a queue using a function called mirror
I have made the code for the queue class, but i dont know how to create a mirror of it. It needs to print out the original queue and then the same queue reversed
Any help would be appreciated
My Code:
class Queue:
def __init__(self):
self.items = []
def is_empty(self):
return self.items == []
def enqueue(self, item):
self.items.insert(0,item)
def dequeue(self):
return self.items.pop()
def is_empty(self):
return not self.items
def size(self):
return len(self.items)
class Stack:
def __init__(self):
self.items = []
def is_empty(self):
return self.items == []
def push(self, item):
self.items.append(item)
def pop(self):
return self.items.pop()
def peek(self):
return self.items[len(self.items)-1]
def size(self):
return len(self.items)
def mirror(n):
pass
This will work. Your queue is composed of a list, so you can use slice syntax on the list to get a reversed version of the queue.
class Queue:
def __init__(self):
self.items = []
def enqueue(self, item):
self.items.append(item)
def __str__(self):
'''Allow print to be called on the queue object itself'''
return str(self.items)
def __getitem__(self, i):
'''Allow the queue object to be indexable directly'''
return self.items[i]
def mirror(q):
return q[::-1]
q = Queue()
for i in range(10):
q.enqueue(i)
print q
print mirror(q)
Note: A queue appends to the end, not the start. That's the behaviour of a stack.
Maybe you give this one a try: How can I reverse a list in python
You can create a new queue using the reversed self.items list in a member function mirror.
q = ArrayQueue()
def mirror(q):
copy = ArrayQueue()
stack = ArrayStack()
while not q.is_empty():
stack.push(q.first())
copy.enqueue(q.dequeue())
while not stack.is_empty():
copy.enqueue(stack.pop())
for i in range(len(copy)):
print(copy.dequeue(),end=" ")

Printing Linked List

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

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