stack implementation in python using class

stackoverflow.com › questions › 18279775 › implementing-stack-with-python

I corrected a few problems below. Also, a 'stack', in abstract programming terms, is usually a collection where you add and remove from the top, but the way you implemented it, you're adding to the top and removing from the bottom, which makes it a queue.

class myStack:
     def __init__(self):
         self.container = []  # You don't want to assign [] to self - when you do that, you're just assigning to a new local variable called `self`.  You want your stack to *have* a list, not *be* a list.

     def isEmpty(self):
         return self.size() == 0   # While there's nothing wrong with self.container == [], there is a builtin function for that purpose, so we may as well use it.  And while we're at it, it's often nice to use your own internal functions, so behavior is more consistent.

     def push(self, item):
         self.container.append(item)  # appending to the *container*, not the instance itself.

     def pop(self):
         return self.container.pop()  # pop from the container, this was fixed from the old version which was wrong

     def peek(self):
         if self.isEmpty():
             raise Exception("Stack empty!")
         return self.container[-1]  # View element at top of the stack

     def size(self):
         return len(self.container)  # length of the container

     def show(self):
         return self.container  # display the entire stack as list


s = myStack()
s.push('1')
s.push('2')
print(s.pop())
print(s.show())

Answer from Brionius on Stack Overflow

W3Schools

w3schools.com › python › python_dsa_stacks.asp

Stacks with Python

To better understand the benefits with using arrays or linked lists to implement stacks, you should check out this page that explains how arrays and linked lists are stored in memory. This is how a stack can be implemented using a linked list. ... class Node: def __init__(self, value): self.value = value self.next = None class Stack: def __init__(self): self.head = None self.size = 0 def push(self, value): new_node = Node(value) if self.head: new_node.next = self.head self.head = new_node self.size += 1 def pop(self): if self.isEmpty(): return "Stack is empty" popped_node = self.head self.head

Stack Overflow

stackoverflow.com › questions › 18279775 › implementing-stack-with-python

algorithm - Implementing Stack with Python - Stack Overflow

Top answer

1 of 12

class myStack:
     def __init__(self):
         self.container = []  # You don't want to assign [] to self - when you do that, you're just assigning to a new local variable called `self`.  You want your stack to *have* a list, not *be* a list.

     def isEmpty(self):
         return self.size() == 0   # While there's nothing wrong with self.container == [], there is a builtin function for that purpose, so we may as well use it.  And while we're at it, it's often nice to use your own internal functions, so behavior is more consistent.

     def push(self, item):
         self.container.append(item)  # appending to the *container*, not the instance itself.

     def pop(self):
         return self.container.pop()  # pop from the container, this was fixed from the old version which was wrong

     def peek(self):
         if self.isEmpty():
             raise Exception("Stack empty!")
         return self.container[-1]  # View element at top of the stack

     def size(self):
         return len(self.container)  # length of the container

     def show(self):
         return self.container  # display the entire stack as list


s = myStack()
s.push('1')
s.push('2')
print(s.pop())
print(s.show())

2 of 12

Assigning to self won't turn your object into a list (and if it did, the object wouldn't have all your stack methods any more). Assigning to self just changes a local variable. Instead, set an attribute:

def __init__(self):
    self.stack = []

and use the attribute instead of just a bare self:

def push(self, item):
    self.stack.append(item)

Also, if you want a stack, you want pop() rather than pop(0). pop(0) would turn your data structure into a(n inefficient) queue.

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