Stack implementation in Python 3

Question 1

I'm learning Python by implementing commonly used data structures. My primary language is Java, thus would like to know if I'm being Pythonic enough. Any means to better up will be appreciated.

class Node:    def __init__(self, value=None, next=None):        self.value = value        self.next = nextfrom node import Nodeclass Stack:    def __init__(self):        self.top = None        self.size = 0    def push(self, item):        curr = Node(item)        self.size += 1        if self.top is None:            self.top = curr        else:            curr.next = self.top            self.top = curr    def peek(self):        return self.top.value    def pop(self):        if self.top is None:            raise Exception("Nothing to pop.")        curr = self.top        self.top = self.top.next        self.size -= 1        return curr    def __sizeof__(self):        return self.size    def is_empty(self):        return self.size == 0;    def __str__(self):        curr = self.top        while curr:            print(curr.value, end=' ')            curr = curr.next

Question 2

The easiest way to create Stack that I found was this:

class Stack:    def __init__(self):        self.stack = []    def push(self,data):        self.stack.append(data)    def pop(self):        if self.isEmpty():            raise Exception("nothing to pop")        return self.stack.pop(len(self.stack)-1)    def peek(self):        if self.isEmpty():            raise Exception("Nothing to peek")        return self.stack[len(self.stack)-1]    def __sizeOf__(self):        return len(self.stack)    def isEmpty(self):        return len(self.stack) == 0    def __str__(self):        for x in self.stack.reverse(): #Reverse list, so when printing last pushed value is first and the oldest is printed last.            print(x)

What it does is when creating Stack Variable:

it makes new list calledstack.
When you push something to Stack, it is added at the end of list.
When you pop something from Stack, it is taken away form list (methodpop() will not only return value from list, but also delete it)
When you peek something from Stack, it returns last value on list without deleting it.
When you want to check size of stack, it measures list size.
When you want to use__str__ you print all variables from list.

I have reversed the list usingfor x in stack:. This was done because otherwisex would be equivalent tostack[0] in first run,stack[1] in second etc. However we must not forget thatstack[0] is the oldest value placed in stack, and the newest is the value with the highest index in list. Reversing the list allows to print from the newest value to the oldest.

A little bit more of explanation:

self.stack.pop(len(self.stack)-1) - This code will pop last value from list. It contains-1 because length of stack is started from 1 while indexing is started from 0. Functionlen() returns (in this case) length of list
if self.isEmpty(): raise Exception("Nothing to peek") - This code will run its functionisEmpty() from its own class, which returns Boolean value (which is perfect for our if statement).
self.stack[len(self.stack)-1] while using list you will use list name (in this case stack) and then index you want to access, e.g. stack[1] (will return second value in list, dont forget about 0-indexing).

Question 3

__sizeOf__ should be__sizeof__, unless you're implementing your own magic methods on purpose (in which case, don't). But really, it should returnsys.sizeof(self.stack); In most cases,len(obj) != sys.sizeof(obj)!

vakus 2761 gold badge3 silver badges10 bronze badges · Accepted Answer · 2018-04-08 20:02:49Z

The easiest way to create Stack that I found was this:

class Stack:    def __init__(self):        self.stack = []    def push(self,data):        self.stack.append(data)    def pop(self):        if self.isEmpty():            raise Exception("nothing to pop")        return self.stack.pop(len(self.stack)-1)    def peek(self):        if self.isEmpty():            raise Exception("Nothing to peek")        return self.stack[len(self.stack)-1]    def __sizeOf__(self):        return len(self.stack)    def isEmpty(self):        return len(self.stack) == 0    def __str__(self):        for x in self.stack.reverse(): #Reverse list, so when printing last pushed value is first and the oldest is printed last.            print(x)

What it does is when creating Stack Variable:

it makes new list calledstack.
When you push something to Stack, it is added at the end of list.
When you pop something from Stack, it is taken away form list (methodpop() will not only return value from list, but also delete it)
When you peek something from Stack, it returns last value on list without deleting it.
When you want to check size of stack, it measures list size.
When you want to use__str__ you print all variables from list.

I have reversed the list usingfor x in stack:. This was done because otherwisex would be equivalent tostack[0] in first run,stack[1] in second etc. However we must not forget thatstack[0] is the oldest value placed in stack, and the newest is the value with the highest index in list. Reversing the list allows to print from the newest value to the oldest.

A little bit more of explanation:

self.stack.pop(len(self.stack)-1) - This code will pop last value from list. It contains-1 because length of stack is started from 1 while indexing is started from 0. Functionlen() returns (in this case) length of list
if self.isEmpty(): raise Exception("Nothing to peek") - This code will run its functionisEmpty() from its own class, which returns Boolean value (which is perfect for our if statement).
self.stack[len(self.stack)-1] while using list you will use list name (in this case stack) and then index you want to access, e.g. stack[1] (will return second value in list, dont forget about 0-indexing).

__sizeOf__ should be__sizeof__, unless you're implementing your own magic methods on purpose (in which case, don't). But really, it should returnsys.sizeof(self.stack); In most cases,len(obj) != sys.sizeof(obj)!

Movatterモバイル変換

Stack Exchange Network

Stack implementation in Python 3

1 Answer1

A little bit more of explanation:

You mustlog in to answer this question.

Related

Hot Network Questions

Subscribe to RSS