How to write equals method

Question

Situation: I'm trying to get a good handle on the doubly-linked structure. I've got a decent grip on the methods so far. I want to be able to create two objects for this class and check if every item in it is equal. I don't have any syntax errors, and the error I'm getting is kind of confusing. So here's what I have so far.

class LinkedList:
    class Node:
        def __init__(self, val, prior=None, next=None):
            self.val = val
            self.prior = prior
            self.next  = next

    def __init__(self):
        self.head = LinkedList.Node(None) # sentinel node (never to be removed)
        self.head.prior = self.head.next = self.head # set up "circular" topology
        self.length = 0

    def append(self, value):
        n = LinkedList.Node(value, prior=self.head.prior, next=self.head)
        n.prior.next = n.next.prior = n
        self.length += 1

    def _normalize_idx(self, idx):
        nidx = idx
        if nidx < 0:
            nidx += len(self)
            if nidx < -1:
                raise IndexError  
        return nidx

    def __getitem__(self, idx):
        """Implements `x = self[idx]`"""
        nidx = self._normalize_idx(idx)
        currNode = self.head.next
        for i in range(nidx):
            currNode = currNode.next
        if nidx >= len(self):
            raise IndexError
        return currNode.val


    def __setitem__(self, idx, value):
        """Implements `self[idx] = x`"""
        nidx = self._normalize_idx(idx)
        currNode = self.head.next
        if nidx >= len(self):
            raise IndexError
        for i in range(nidx):
            currNode = currNode.next
        currNode.val = value

    def __iter__(self):
        """Supports iteration (via `iter(self)`)"""
        cursor = self.head.next
        while cursor is not self.head:
            yield cursor.val
            cursor = cursor.next

    def __len__(self):
        """Implements `len(self)`"""
        return self.length

    def __eq__(self, other):
        currNode = self.head.next
        currNode2 = other.head.next
        for currNode, currNode2 in zip(self, other):
            if currNode.val != currNode2.val:
                return False
        return True

Test:

from unittest import TestCase
tc = TestCase()
lst = LinkedList()
lst2 = LinkedList()

tc.assertEqual(lst, lst2)

lst2.append(100)
tc.assertNotEqual(lst, lst2)

When I test this code I get I get an Assertion error saying " [] == [100] " I'm unsure why my code recognizes this as equal, when I want it to actually check specific values in the node.

score 2 · Accepted Answer · answered Jun 14 '16 at 05:45

zip only goes as far as the shortest list. You want itertools.zip_longest, and you don't want .val (your iterator returns the actual values already). Try this:

def __eq__(self, other):
    for val1, val2 in zip_longest(self, other):
        if val1 != val2:
            return False
    return True

or perhaps better?

def __eq__(self, other):
    return all(val1 == val2 for val1, val2 in zip_longest(self, other))

EDIT

I like @BrenBarn's suggestion of checking length first. Here's a more efficient answer:

def __eq__(self, other):
    return len(self) == len(other) and all(
        val1 == val2 for val1, val2 in zip(self, other))

This particular `LinkedList` appears to store its length, so the length check is O(1). — user94559, Jun 14 '16 at 05:54

score 0 · Answer 2 · answered Jun 14 '16 at 05:44

zip(self.other) only gives you as many elements as the shorter of the two lists. It discards the extra part of the longer list. So for [] == [100], zip doesn't give any elements, and your code returns True without checking anything.

You could just do a check at the beginning to see if the lists have different length. If they do, they can't be equal.

How to write equals method

2 Answers2