Comparing two objects properties but not working propery

Comparing two objects properties but not working propery - javascript

Here is my code. I know it's not entirely strict but please shed some light on why the let...in does not work properly here.
const object1 = {here: 1, object: 3};
const obj = {here: 1, object: 2};
function comp(a, b) {
if (typeof a == typeof b) {
let arra = Object.keys(a);
let arrb = Object.keys(b);
for (let key in arra){
if (a[key] == b[key]) return true
}
return false
}
}
console.log(comp(obj, object1))
The above prints true but it is supposed to print false

You're getting true because you return true in your for loop whenever a key value from one object equals the key-value pair of another object. So when your code sees the here property, it will return true, thus stopping your function from running any further code.
You need to remove this check, and only return false in your for loop, such that your for loop will only complete if it never returns (ie: all key-value pairs are equal).
Moreover, the for..in loop will loop over the keys in an object, so, there is no need to get the keys of your objects (using Object.keys) into an array (as then you'll be looping over the keys of an array (ie: the indexes)).
However, with that being said, you can use Object.keys to help with another issue. You can use it to get the number of properties in both objects, as you know the two objects are not the same if they don't have the same number of properties in them
See example below:
const object1 = {
here: 1,
object: 3
};
const obj = {
here: 1,
object: 3
};
function comp(a, b) {
if (typeof a == typeof b) {
if(Object.keys(a).length !== Object.keys(b).length) {
return false; // return false (stop fruther code execution)
}
for (let key in a) { // loop through the properties of larger object (here I've chosen 'a') - no need for Object.keys
if (a[key] != b[key])
return false; // return false (stops any further code executing)
}
return true; // we only reach this point if the for loop never returned false
}
return false; // we reach this point when the two types don't match, and so we can say they're not equal
}
console.log(comp(obj, object1))

You should use for..of (or just a plain old for) instead of for..in which is only used on Objects. You're reading array indices right now, not actual key names. Object.keys returns an Array of key names, not an Object.
Also stop returning early; Right now you return immediately after the first key check.

You need to check for actual values and not property names.
Also, you need to check if b has more properties than a, because if it is the same, but has one property more, it will still output true.
const tester = {here: 1, object: 3};
const obj1 = {here: 1, object: 2};
const obj2 = {here: 1, object: 3};
const obj3 = {here: 1, object: 3, test: 1};
const obj4 = {here: 1, test: 1};
function comp(a, b) {
if (typeof a == typeof b && Object.keys(a).length == Object.keys(b).length) { // Check for the length of the keys array, because if the length is different, they might have different properties
// Dont use Object.keys here. You don't need the keys, you need the objects
for (let key in a){
if (a[key] != b[key])
return false; // If one key property of a does not match b, return false
}
return true; // If nothing returns false, return true
}
return false; // If they are not the same type, return false
}
console.log(comp(tester, obj1))
console.log(comp(tester, obj2))
console.log(comp(tester, obj3))
console.log(comp(tester, obj4))

Here is your problem:
for (let key in arra){
if (a[key] == b[key]) return true
}
return false
}
You should perform the exact opposite while iterating through the object:
for (let key in a){ // a not arra
if (a[key] !== b[key]) return false
}
return true
}
And ommit these lines:
let arra = Object.keys(a);
let arrb = Object.keys(b);

Array.prototype.some() function can be used as below:
const object1 = {here: 1, object: 3, r:4};
const obj = {here: 1, r:4, object: 3};
function comp(a, b) {
if (typeof a == typeof b) {
let arra = Object.keys(a);
return !arra.some(key => { return a[key] != b[key] })
}
}
console.log(comp(obj, object1))

Related

How to compare JavaScript objects obtained by parsing JSON?

wtf1 = JSON.parse('{"asdf": "jkl"}');
wtf2 = JSON.parse('{"asdf": "jkl"}');
wtf1 == wtf2; // false
wtf1 === wtf2; // false
I’ve recently stumbled upon the above problem. This counter-intuitive situation makes it hard to, for example, find a specific object in an array deep in a JSON hierarchy.
Any way to somehow compare such objects?

For simple objects you can stringify them again (with ordered keys) and compare strings. For example:
var wtf1 = JSON.parse('{"a": "a", "b": "b"}');
var wtf2 = JSON.parse('{"b": "b", "a": "a"}');
var s1 = JSON.stringify(wtf1, Object.keys(wtf1).sort());
var s2 = JSON.stringify(wtf2, Object.keys(wtf2).sort());
console.log('wtf1 is equal to wtf2: ', (s1 == s2));
For nested objects with prototypes etc you should probably use _.isEqual or any other lib that provides deep equality test.
Note, that in general it's not trivial to correctly implement deep equality test for objects, it's not as simple as iterating and comparing keys/values. However, since the "objects were obtained by parsing JSON" you can skip most of complications and recursively stringify nested values.

I think it's better to just compare them before you parse into objects. But this only works if they are exactly the same, including the order of the properties
var obj1 = {name: "potato", age: 10}
var obj2 = {name: "potato", age: 10}
console.log(obj1 == obj2) //false
console.log(JSON.stringify(obj1) == JSON.stringify(obj2)) //true
var obj1 = {name: "potato", age: 10}
var obj2 = {age: 10, name: "potato"}
console.log(obj1 == obj2) //false
console.log(JSON.stringify(obj1) == JSON.stringify(obj2)) //also false

You cannot compare objects ( as theyre differnet ), but you can iterate over each property, and compare them (and recursively check if theyre objects again):
function compare(obj1,obj2){
//check for obj2 overlapping props
if(!Object.keys(obj2).every(key=>obj1.hasOwnProperty(key))){
return false;
}
//check every key for being same
return Object.keys(obj1).every(function(key){
//if object
if((typeof obj1[key]=="object" )&&( typeof obj2[key]=="object")){
//recursively check
return compare(obj1[key],obj2[key]);
}else{
//do the normal compare
return obj1[key]===obj2[key];
}
});
}
http://jsbin.com/zagecigawi/edit?js

you can use loadash for the same.
using _.isEqual("object1", "object2");
var obj1 = {"prop1" : 2, "prop2" : 3 };
var obj2 = {"prop1" : 2, "prop2" : 3};
console.dir(_.isEqual(obj1, obj2))
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>
Second approach would be
JSON.stringify(obj1) === JSON.stringify(obj2)

This will compare JSON structure of response and model(It will compare keys only and not values & will work for any hierarchy)
function compareObjects(response, model) {
if (response == "" && model == "") {
return false;
}
if (typeof (response) != "object" && typeof (model) != "object") {
var response = JSON.parse(response);
var model = JSON.parse(model);
}
if (typeof (response) != typeof (model)) {
return false;
} else {
switch (Object.prototype.toString.call(model)) {
case '[object]':
var x;
var mKeys = Object.keys(model);
for (x in mKeys) {
return compareObjects(Object.keys(model)[x], Object.keys(response)[x]);
}
break;
case '[object Array]':
return compareObjects(model[0], response[0]);
case "[object String]":
return model == response;
default:
return true;
}
}
}
var response = '[{"educationId":5,"degreeName":"Bacheltecture - B.Arch"},{"educationId":2,"degreeName":"Bachelor of Arts - B.A. "},{"educationId":3,"degreeName":"Bachelor of Ayurvedic Medicine Surgery - B.A.M.S. "}]';
var model = '[{"degreeName":null},{"educationId":null,"degreeName":null},{"educationId":null,"degreeName":null}]';
var output = compareObjects(response, model);
console.log(output);

Two objects in JS even with same properties are never equal. You can stringify objects and compare those strings or iterate over all properties (but it's hard if you need deep compare)

I recommend using a library implementation for Object equality checking. Try the lodash version.
var object = { 'a': 1 };
var other = { 'a': 1 };
_.isEqual(object, other);
// => true

For in loop in recursive function not completing

I cannot get my for in loop to keep working after the first property in my object. This is a question from Eloquent JavaScript in Chapter 4:
Write a function, deepEqual, that takes two values and returns true
only if they are the same value or are objects with the same
properties whose values are also equal when compared with a recursive
call to deepEqual.
To find out whether to compare two things by identity (use the ===
operator for that) or by looking at their properties, you can use the
typeof operator. If it produces "object" for both values, you should
do a deep comparison. But you have to take one silly exception into
account: by a historical accident, typeof null also produces "object".
Here is my code:
function deepEqual(obj1, obj2) {
if ((typeof obj1 === 'object' && obj1 != null) && (typeof obj2 === 'object' && obj2 != null)) {
for (var property in obj1) {
if (property in obj2) {
return deepEqual(obj1[property], obj2[property])
} else {
return false;
}
}
} else if (obj1 !== obj2) {
return false;
} else {
return true;
}
}
var obj = {object: 3, here: 1};
var obj2 = {object: 3, here: 2};
console.log(deepEqual(obj, obj2));
The console returns true, when it should say false because the 'here' properties are not equal. When looking into the output, it's because the 'for in loop' in the function quits after the first property. Please help me as to why it's not continuing to loop.

your for loop can't move beyond the first property because you return out of the function when you call deepEqual
for (var property in obj1) {
if (property in obj2) {
// returning means no more looping....
return deepEqual(obj1[property], obj2[property])
}
you want to carry on looping if deepEqual returns that its equal, or return if its false.

Check if Javascript Object only contains falsy values

I got an object like this :
var obj1 = {foo: false, bar: ''};
var obj2 = {foo: false, bar: '1'};
var obj3 = {foo: true, bar: ''};
var obj4 = {foo: true, bar: '1'};
I want a simple function to check those objects if all of their values are false or not. In this given example only obj1 should trigger a false - cause all of their values are false. obj2, obj3 and obj4 own at least one value which is true.
Is there a simple solution to do this?

As a one-liner:
!Object.keys(obj1).some(function(k) {return obj1[k];});
// Array.some returns true if any callback returns true
// You want true if all return false, aka if none return true
// So just negate Array.some
As a more readable method:
var ok = true, k;
for( k in obj1) if( obj1.hasOwnProperty(k)) {
if( obj1[k]) {
ok = false;
break;
}
}

You can use the some function:
[ob1, ob2, ob3, obj4].some(function(obj) { return obj.foo })

For each object: if every property value is false return false, otherwise true:
var out = [obj1, obj2, obj3, obj4].map(function (obj) {
return !Object.keys(obj).every(function (p) {
return !obj[p];
})
}); // [false, true, true, true]
DEMO

There is a flaw with the other implementations of .some() you want to return values where all properties are a falsy value, not just where any property is a falsy value as per the OP requirements.
So this will do it
function isFalse(obj) {
return Object
.keys(obj)
.every(function (k) {
return !obj[k];
});
}
See jsFiddle for proof https://jsfiddle.net/0se3yh62/
The other benefit of Object.keys it will only return own properties so you don't have to check in a manual loop.

const isObjectFilled = obj => Object.values(obj).filter(value => !!value).length > 0
isObjectFilled(obj1) // false
isObjectFilled(obj2) // true
isObjectFilled(obj3) // true
isObjectFilled(obj4) // true
"The Object.values() method returns an array of a given object's own enumerable property values" (MDN)
Once i've this array of values, i'm going to filter it by removing each falsy values.
.filter(value => !!value) will filter the array by returning only truthy values.
If the length of the filtered array is greater than 0, it means that at least one of the object's values is truthy. Then the function will return true
If the filtered array is empty, it means that all of the object's values are falsy. Then the function will return false

I assume you want a dynamic function that would work on any object, so how about just iterating all of the object's indexes?
function objIsFalsy(obj) {
for(var i in obj) {
if(obj[i]) return false;
}
return true;
}

Follow these steps :-
Put all your object variables in an array.
Make a foreach loop for the above array.
Inside the loop check for object key's value.

Loop on your object:
function isFalsyObj(obj) {
for(var elt in obj) if (obj[elt]!=false) return false;
return true;
}
var obj1 = {foo: false, bar: ''};
var obj2 = {foo: false, bar: '1'};
var obj3 = {foo: true, bar: ''};
var obj4 = {foo: true, bar: '1'};
console.log(isFalsyObj(obj1));//true
console.log(isFalsyObj(obj2));//false
console.log(isFalsyObj(obj3));//false
console.log(isFalsyObj(obj4));//false

Check this
isvalidObj(obj1); // false
isvalidObj(obj2); // true
function isvalidObj(obj){
for(var i in obj){
if(obj[i]) return true;
}
return false;
}

Check if two objects have common sub-objects

I need to check if two object have common sub-objects.
By common I mean that it exact same value, and not just equal values.
Something like:
function haveCommonObects(value1, value2) {
...
}
var common = {};
haveCommonObjects({a: common}, {b: {c: common}}) // true
haveCommonObjects({a: 1}, {b: 1}) // false
I need to check large objects, so function should be reasonable efficient.
Also I can't change objects, so I can't flag sub-objects with special property. Objects create in 3rd-party library so I can't alter Object.prototype.
Ideal solution would be to get some kind of ID for every object and save it in collection that support fast lookup.
Can I make such function in JS?

Here's how I would do this:
function haveCommonObjects(a,b) {
// check if a and b have any object in common, at any depth level
if (typeof(a) !== 'object' || typeof(b) !== 'object') return false;
for (var key in a) {
var o = a[key];
if (typeof(o) === 'object' && (hasObject(b,o) || haveCommonObjects(o,b)))
return true;
}
return false;
}
function hasObject(x,t) {
// check if x has a reference to object t, at any depth level
for (var key in x) {
var o = x[key];
if (typeof(o) === 'object' && (o === t || hasObject(o,t)))
return true;
}
return false;
}
function log(msg) { document.getElementById('log').innerHTML += msg+'<br/>'; }
var common = {};
log(haveCommonObjects({a: common}, {b: {c: common}})); // true
log(haveCommonObjects({a: 1}, {b: 1})); // false
<div id="log"></div>
Note: You should add a hasOwnProperty() filter in every for..in loop if you want to exclude inherited properties; see for..in and hasOwnProperty.

My == isn't working [duplicate]

This question already has answers here:
How to compare arrays in JavaScript?
(55 answers)
Closed 6 years ago.
var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);
alert(a == b + "|" + b == c);
demo
How can I check these array for equality and get a method which returns true if they are equal?
Does jQuery offer any method for this?

This is what you should do. Please do not use stringify nor < >.
function arraysEqual(a, b) {
if (a === b) return true;
if (a == null || b == null) return false;
if (a.length !== b.length) return false;
// If you don't care about the order of the elements inside
// the array, you should sort both arrays here.
// Please note that calling sort on an array will modify that array.
// you might want to clone your array first.
for (var i = 0; i < a.length; ++i) {
if (a[i] !== b[i]) return false;
}
return true;
}

[2021 changelog: bugfix for option4: no total ordering on js objects (even excluding NaN!=NaN and '5'==5 ('5'===5, '2'<3, etc.)), so cannot use .sort(cmpFunc) on Map.keys() (though you can on Object.keys(obj), since even 'numerical' keys are strings).]
Option 1
Easiest option, works in almost all cases, except that null!==undefined but they both are converted to JSON representation null and considered equal:
function arraysEqual(a1,a2) {
/* WARNING: arrays must not contain {objects} or behavior may be undefined */
return JSON.stringify(a1)==JSON.stringify(a2);
}
(This might not work if your array contains objects. Whether this still works with objects depends on whether the JSON implementation sorts keys. For example, the JSON of {1:2,3:4} may or may not be equal to {3:4,1:2}; this depends on the implementation, and the spec makes no guarantee whatsoever. [2017 update: Actually the ES6 specification now guarantees object keys will be iterated in order of 1) integer properties, 2) properties in the order they were defined, then 3) symbol properties in the order they were defined. Thus IF the JSON.stringify implementation follows this, equal objects (in the === sense but NOT NECESSARILY in the == sense) will stringify to equal values. More research needed. So I guess you could make an evil clone of an object with properties in the reverse order, but I cannot imagine it ever happening by accident...] At least on Chrome, the JSON.stringify function tends to return keys in the order they were defined (at least that I've noticed), but this behavior is very much subject to change at any point and should not be relied upon. If you choose not to use objects in your lists, this should work fine. If you do have objects in your list that all have a unique id, you can do a1.map(function(x)}{return {id:x.uniqueId}}). If you have arbitrary objects in your list, you can read on for option #2.)
This works for nested arrays as well.
It is, however, slightly inefficient because of the overhead of creating these strings and garbage-collecting them.
Option 2
Historical, version 1 solution:
// generally useful functions
function type(x) { // does not work in general, but works on JSONable objects we care about... modify as you see fit
// e.g. type(/asdf/g) --> "[object RegExp]"
return Object.prototype.toString.call(x);
}
function zip(arrays) {
// e.g. zip([[1,2,3],[4,5,6]]) --> [[1,4],[2,5],[3,6]]
return arrays[0].map(function(_,i){
return arrays.map(function(array){return array[i]})
});
}
// helper functions
function allCompareEqual(array) {
// e.g. allCompareEqual([2,2,2,2]) --> true
// does not work with nested arrays or objects
return array.every(function(x){return x==array[0]});
}
function isArray(x){ return type(x)==type([]) }
function getLength(x){ return x.length }
function allTrue(array){ return array.reduce(function(a,b){return a&&b},true) }
// e.g. allTrue([true,true,true,true]) --> true
// or just array.every(function(x){return x});
function allDeepEqual(things) {
// works with nested arrays
if( things.every(isArray) )
return allCompareEqual(things.map(getLength)) // all arrays of same length
&& allTrue(zip(things).map(allDeepEqual)); // elements recursively equal
//else if( this.every(isObject) )
// return {all have exactly same keys, and for
// each key k, allDeepEqual([o1[k],o2[k],...])}
// e.g. ... && allTrue(objectZip(objects).map(allDeepEqual))
//else if( ... )
// extend some more
else
return allCompareEqual(things);
}
// Demo:
allDeepEqual([ [], [], [] ])
true
allDeepEqual([ [1], [1], [1] ])
true
allDeepEqual([ [1,2], [1,2] ])
true
allDeepEqual([ [[1,2],[3]], [[1,2],[3]] ])
true
allDeepEqual([ [1,2,3], [1,2,3,4] ])
false
allDeepEqual([ [[1,2],[3]], [[1,2],[],3] ])
false
allDeepEqual([ [[1,2],[3]], [[1],[2,3]] ])
false
allDeepEqual([ [[1,2],3], [1,[2,3]] ])
false
<!--
More "proper" option, which you can override to deal with special cases (like regular objects and null/undefined and custom objects, if you so desire):
To use this like a regular function, do:
function allDeepEqual2() {
return allDeepEqual([].slice.call(arguments));
}
Demo:
allDeepEqual2([[1,2],3], [[1,2],3])
true
-->
Option 3
function arraysEqual(a,b) {
/*
Array-aware equality checker:
Returns whether arguments a and b are == to each other;
however if they are equal-lengthed arrays, returns whether their
elements are pairwise == to each other recursively under this
definition.
*/
if (a instanceof Array && b instanceof Array) {
if (a.length!=b.length) // assert same length
return false;
for(var i=0; i<a.length; i++) // assert each element equal
if (!arraysEqual(a[i],b[i]))
return false;
return true;
} else {
return a==b; // if not both arrays, should be the same
}
}
//Examples:
arraysEqual([[1,2],3], [[1,2],3])
true
arraysEqual([1,2,3], [1,2,3,4])
false
arraysEqual([[1,2],[3]], [[1,2],[],3])
false
arraysEqual([[1,2],[3]], [[1],[2,3]])
false
arraysEqual([[1,2],3], undefined)
false
arraysEqual(undefined, undefined)
true
arraysEqual(1, 2)
false
arraysEqual(null, null)
true
arraysEqual(1, 1)
true
arraysEqual([], 1)
false
arraysEqual([], undefined)
false
arraysEqual([], [])
true
/*
If you wanted to apply this to JSON-like data structures with js Objects, you could do so. Fortunately we're guaranteed that all objects keys are unique, so iterate over the objects OwnProperties and sort them by key, then assert that both the sorted key-array is equal and the value-array are equal, and just recurse. We CANNOT extend the sort-then-compare method with Maps as well; even though Map keys are unique, there is no total ordering in ecmascript, so you can't sort them... but you CAN query them individually (see the next section Option 4). (Also if we extend this to Sets, we run into the tree isomorphism problem http://logic.pdmi.ras.ru/~smal/files/smal_jass08_slides.pdf - fortunately it's not as hard as general graph isomorphism; there is in fact an O(#vertices) algorithm to solve it, but it can get very complicated to do it efficiently. The pathological case is if you have a set made up of lots of seemingly-indistinguishable objects, but upon further inspection some of those objects may differ as you delve deeper into them. You can also work around this by using hashing to reject almost all cases.)
*/
<!--
**edit**: It's 2016 and my previous overcomplicated answer was bugging me. This recursive, imperative "recursive programming 101" implementation keeps the code really simple, and furthermore fails at the earliest possible point (giving us efficiency). It also doesn't generate superfluous ephemeral datastructures (not that there's anything wrong with functional programming in general, but just keeping it clean here).
If we wanted to apply this to a non-empty arrays of arrays, we could do seriesOfArrays.reduce(arraysEqual).
This is its own function, as opposed to using Object.defineProperties to attach to Array.prototype, since that would fail with a key error if we passed in an undefined value (that is however a fine design decision if you want to do so).
This only answers OPs original question.
-->
Option 4:
(continuation of 2016 edit)
This should work with most objects:
const STRICT_EQUALITY_BROKEN = (a,b)=> a===b;
const STRICT_EQUALITY_NO_NAN = (a,b)=> {
if (typeof a=='number' && typeof b=='number' && ''+a=='NaN' && ''+b=='NaN')
// isNaN does not do what you think; see +/-Infinity
return true;
else
return a===b;
};
function deepEquals(a,b, areEqual=STRICT_EQUALITY_NO_NAN, setElementsAreEqual=STRICT_EQUALITY_NO_NAN) {
/* compares objects hierarchically using the provided
notion of equality (defaulting to ===);
supports Arrays, Objects, Maps, ArrayBuffers */
if (a instanceof Array && b instanceof Array)
return arraysEqual(a,b, areEqual);
if (Object.getPrototypeOf(a)===Object.prototype && Object.getPrototypeOf(b)===Object.prototype)
return objectsEqual(a,b, areEqual);
if (a instanceof Map && b instanceof Map)
return mapsEqual(a,b, areEqual);
if (a instanceof Set && b instanceof Set) {
if (setElementsAreEqual===STRICT_EQUALITY_NO_NAN)
return setsEqual(a,b);
else
throw "Error: set equality by hashing not implemented because cannot guarantee custom notion of equality is transitive without programmer intervention."
}
if ((a instanceof ArrayBuffer || ArrayBuffer.isView(a)) && (b instanceof ArrayBuffer || ArrayBuffer.isView(b)))
return typedArraysEqual(a,b);
return areEqual(a,b); // see note[1] -- IMPORTANT
}
function arraysEqual(a,b, areEqual) {
if (a.length!=b.length)
return false;
for(var i=0; i<a.length; i++)
if (!deepEquals(a[i],b[i], areEqual))
return false;
return true;
}
function objectsEqual(a,b, areEqual) {
var aKeys = Object.getOwnPropertyNames(a);
var bKeys = Object.getOwnPropertyNames(b);
if (aKeys.length!=bKeys.length)
return false;
aKeys.sort();
bKeys.sort();
for(var i=0; i<aKeys.length; i++)
if (!areEqual(aKeys[i],bKeys[i])) // keys must be strings
return false;
return deepEquals(aKeys.map(k=>a[k]), aKeys.map(k=>b[k]), areEqual);
}
function mapsEqual(a,b, areEqual) { // assumes Map's keys use the '===' notion of equality, which is also the assumption of .has and .get methods in the spec; however, Map's values use our notion of the areEqual parameter
if (a.size!=b.size)
return false;
return [...a.keys()].every(k=>
b.has(k) && deepEquals(a.get(k), b.get(k), areEqual)
);
}
function setsEqual(a,b) {
// see discussion in below rest of StackOverflow answer
return a.size==b.size && [...a.keys()].every(k=>
b.has(k)
);
}
function typedArraysEqual(a,b) {
// we use the obvious notion of equality for binary data
a = new Uint8Array(a);
b = new Uint8Array(b);
if (a.length != b.length)
return false;
for(var i=0; i<a.length; i++)
if (a[i]!=b[i])
return false;
return true;
}
Demo (not extensively tested):
var nineTen = new Float32Array(2);
nineTen[0]=9; nineTen[1]=10;
> deepEquals(
[[1,[2,3]], 4, {a:5,'111':6}, new Map([['c',7],['d',8]]), nineTen],
[[1,[2,3]], 4, {111:6,a:5}, new Map([['d',8],['c',7]]), nineTen]
)
true
> deepEquals(
[[1,[2,3]], 4, {a:'5','111':6}, new Map([['c',7],['d',8]]), nineTen],
[[1,[2,3]], 4, {111:6,a:5}, new Map([['d',8],['c',7]]), nineTen],
(a,b)=>a==b
)
true
Note that if one is using the == notion of equality, then know that falsey values and coercion means that == equality is NOT TRANSITIVE. For example ''==0 and 0=='0' but ''!='0'. This is relevant for Sets: I do not think one can override the notion of Set equality in a meaningful way. If one is using the built-in notion of Set equality (that is, ===), then the above should work. However if one uses a non-transitive notion of equality like ==, you open a can of worms: Even if you forced the user to define a hash function on the domain (hash(a)!=hash(b) implies a!=b) I'm not sure that would help... Certainly one could do the O(N^2) performance thing and remove pairs of == items one by one like a bubble sort, and then do a second O(N^2) pass to confirm things in equivalence classes are actually == to each other, and also != to everything not thus paired, but you'd STILL have to throw a runtime error if you have some coercion going on... You'd also maybe get weird (but potentially not that weird) edge cases with https://developer.mozilla.org/en-US/docs/Glossary/Falsy and Truthy values (with the exception that NaN==NaN... but just for Sets!). This is not an issue usually with most Sets of homogenous datatype.
To summarize the complexity of recursive equality on Sets:
Set equality is the tree isomorphism problem http://logic.pdmi.ras.ru/~smal/files/smal_jass08_slides.pdf but a bit simpler
set A =? set B being synonymous with B.has(k) for every k in A implicitly uses ===-equality ([1,2,3]!==[1,2,3]), not recursive equality (deepEquals([1,2,3],[1,2,3]) == true), so two new Set([[1,2,3]]) would not be equal because we don't recurse
trying to get recursive equality to work is kind of meaningless if the recursive notion of equality you use is not 1) reflexive (a=b implies b=a) and 2) symmetric (a=a) and 3) transitive (a=b and b=c implies a=c); this is the definition of an equivalence class
the equality == operator obviously does not obey many of these properties
even the strict equality === operator in ecmascript
does not obey these properties, because the strict equality comparison algorithm of ecmascript has NaN!=NaN; this is why many native datatypes like Set and Map 'equate' NaNs to consider them the same values when they appear as keys
As long as we force and ensure recursive set equality is indeed transitive and reflexive and symmetric, we can make sure nothing horribly wrong happens.
Then, we can do O(N^2) comparisons by recursively comparing everything randomly, which is incredibly inefficient. There is no magical algorithm that lets us do setKeys.sort((a,b)=> /*some comparison function*/) because there is no total ordering in ecmascript (''==0 and 0=='0', but ''!='0'... though I believe you might be able to define one yourself which would certainly be a lofty goal).
We can however .toStringify or JSON.stringify all elements to assist us. We will then sort them, which gives us equivalence classes (two same things won't not have the same string JSON representation) of potentially-false-positives (two different things may have the same string or JSON representation).
However, this introduces its own performance issues because serializing the same thing, then serializing subsets of that thing, over and over, is incredibly inefficient. Imagine a tree of nested Sets; every node would belong to O(depth) different serializations!
Even if that was not an issue, the worst-case performance would still be O(N!) if all the serializations 'hints' were the same
Thus, the above implementation declares that Sets are equal if the items are just plain === (not recursively ===). This will mean that it will return false for new Set([1,2,3]) and new Set([1,2,3]). With a bit of effort, you may rewrite that part of the code if you know what you're doing.
(sidenote: Maps are es6 dictionaries. I can't tell if they have O(1) or O(log(N)) lookup performance, but in any case they are 'ordered' in the sense that they keep track of the order in which key-value pairs were inserted into them. However, the semantic of whether two Maps should be equal if elements were inserted in a different order into them is ambiguous. I give a sample implementation below of a deepEquals that considers two maps equal even if elements were inserted into them in a different order.)
(note [1]: IMPORTANT: NOTION OF EQUALITY: You may want to override the noted line with a custom notion of equality, which you'll also have to change in the other functions anywhere it appears. For example, do you or don't you want NaN==NaN? By default this is not the case. There are even more weird things like 0=='0'. Do you consider two objects to be the same if and only if they are the same object in memory? See https://stackoverflow.com/a/5447170/711085 . You should document the notion of equality you use.) Also note that other answers which naively use .toString and .sort may sometimes fall pray to the fact that 0!=-0 but are considered equal and canonicalizable to 0 for almost all datatypes and JSON serialization; whether -0==0 should also be documented in your notion of equality, as well as most other things in that table like NaN, etc.
You should be able to extend the above to WeakMaps, WeakSets. Not sure if it makes sense to extend to DataViews. Should also be able to extend to RegExps probably, etc.
As you extend it, you realize you do lots of unnecessary comparisons. This is where the type function that I defined way earlier (solution #2) can come in handy; then you can dispatch instantly. Whether that is worth the overhead of (possibly? not sure how it works under the hood) string representing the type is up to you. You can just then rewrite the dispatcher, i.e. the function deepEquals, to be something like:
var dispatchTypeEquals = {
number: function(a,b) {...a==b...},
array: function(a,b) {...deepEquals(x,y)...},
...
}
function deepEquals(a,b) {
var typeA = extractType(a);
var typeB = extractType(a);
return typeA==typeB && dispatchTypeEquals[typeA](a,b);
}

jQuery does not have a method for comparing arrays. However the Underscore library (or the comparable Lodash library) does have such a method: isEqual, and it can handle a variety of other cases (like object literals) as well. To stick to the provided example:
var a=[1,2,3];
var b=[3,2,1];
var c=new Array(1,2,3);
alert(_.isEqual(a, b) + "|" + _.isEqual(b, c));
By the way: Underscore has lots of other methods that jQuery is missing as well, so it's a great complement to jQuery.
EDIT: As has been pointed out in the comments, the above now only works if both arrays have their elements in the same order, ie.:
_.isEqual([1,2,3], [1,2,3]); // true
_.isEqual([1,2,3], [3,2,1]); // false
Fortunately Javascript has a built in method for for solving this exact problem, sort:
_.isEqual([1,2,3].sort(), [3,2,1].sort()); // true

For primitive values like numbers and strings this is an easy solution:
a = [1,2,3]
b = [3,2,1]
a.sort().toString() == b.sort().toString()
The call to sort() will ensure that the order of the elements does not matter. The toString() call will create a string with the values comma separated so both strings can be tested for equality.

With JavaScript version 1.6 it's as easy as this:
Array.prototype.equals = function( array ) {
return this.length == array.length &&
this.every( function(this_i,i) { return this_i == array[i] } )
}
For example, [].equals([]) gives true, while [1,2,3].equals( [1,3,2] ) yields false.

Even if this would seem super simple, sometimes it's really useful. If all you need is to see if two arrays have the same items and they are in the same order, try this:
[1, 2, 3].toString() == [1, 2, 3].toString()
true
[1, 2, 3,].toString() == [1, 2, 3].toString()
true
[1,2,3].toString() == [1, 2, 3].toString()
true
However, this doesn't work for mode advanced cases such as:
[[1,2],[3]].toString() == [[1],[2,3]].toString()
true
It depends what you need.

Based on Tim James answer and Fox32's comment, the following should check for nulls, with the assumption that two nulls are not equal.
function arrays_equal(a,b) { return !!a && !!b && !(a<b || b<a); }
> arrays_equal([1,2,3], [1,3,4])
false
> arrays_equal([1,2,3], [1,2,3])
true
> arrays_equal([1,3,4], [1,2,3])
false
> arrays_equal(null, [1,2,3])
false
> arrays_equal(null, null)
false

jQuery has such method for deep recursive comparison.
A homegrown general purpose strict equality check could look as follows:
function deepEquals(obj1, obj2, parents1, parents2) {
"use strict";
var i;
// compare null and undefined
if (obj1 === undefined || obj2 === undefined ||
obj1 === null || obj2 === null) {
return obj1 === obj2;
}
// compare primitives
if (typeof (obj1) !== 'object' || typeof (obj2) !== 'object') {
return obj1.valueOf() === obj2.valueOf();
}
// if objects are of different types or lengths they can't be equal
if (obj1.constructor !== obj2.constructor || (obj1.length !== undefined && obj1.length !== obj2.length)) {
return false;
}
// iterate the objects
for (i in obj1) {
// build the parents list for object on the left (obj1)
if (parents1 === undefined) parents1 = [];
if (obj1.constructor === Object) parents1.push(obj1);
// build the parents list for object on the right (obj2)
if (parents2 === undefined) parents2 = [];
if (obj2.constructor === Object) parents2.push(obj2);
// walk through object properties
if (obj1.propertyIsEnumerable(i)) {
if (obj2.propertyIsEnumerable(i)) {
// if object at i was met while going down here
// it's a self reference
if ((obj1[i].constructor === Object && parents1.indexOf(obj1[i]) >= 0) || (obj2[i].constructor === Object && parents2.indexOf(obj2[i]) >= 0)) {
if (obj1[i] !== obj2[i]) {
return false;
}
continue;
}
// it's not a self reference so we are here
if (!deepEquals(obj1[i], obj2[i], parents1, parents2)) {
return false;
}
} else {
// obj2[i] does not exist
return false;
}
}
}
return true;
};
Tests:
// message is displayed on failure
// clean console === all tests passed
function assertTrue(cond, msg) {
if (!cond) {
console.log(msg);
}
}
var a = 'sdf',
b = 'sdf';
assertTrue(deepEquals(b, a), 'Strings are equal.');
b = 'dfs';
assertTrue(!deepEquals(b, a), 'Strings are not equal.');
a = 9;
b = 9;
assertTrue(deepEquals(b, a), 'Numbers are equal.');
b = 3;
assertTrue(!deepEquals(b, a), 'Numbers are not equal.');
a = false;
b = false;
assertTrue(deepEquals(b, a), 'Booleans are equal.');
b = true;
assertTrue(!deepEquals(b, a), 'Booleans are not equal.');
a = null;
assertTrue(!deepEquals(b, a), 'Boolean is not equal to null.');
a = function () {
return true;
};
assertTrue(deepEquals(
[
[1, 1, 1],
[2, 'asdf', [1, a]],
[3, {
'a': 1.0
},
true]
],
[
[1, 1, 1],
[2, 'asdf', [1, a]],
[3, {
'a': 1.0
},
true]
]), 'Arrays are equal.');
assertTrue(!deepEquals(
[
[1, 1, 1],
[2, 'asdf', [1, a]],
[3, {
'a': 1.0
},
true]
],
[
[1, 1, 1],
[2, 'asdf', [1, a]],
[3, {
'a': '1'
},
true]
]), 'Arrays are not equal.');
a = {
prop: 'val'
};
a.self = a;
b = {
prop: 'val'
};
b.self = a;
assertTrue(deepEquals(b, a), 'Immediate self referencing objects are equal.');
a.prop = 'shmal';
assertTrue(!deepEquals(b, a), 'Immediate self referencing objects are not equal.');
a = {
prop: 'val',
inside: {}
};
a.inside.self = a;
b = {
prop: 'val',
inside: {}
};
b.inside.self = a;
assertTrue(deepEquals(b, a), 'Deep self referencing objects are equal.');
b.inside.self = b;
assertTrue(!deepEquals(b, a), 'Deep self referencing objects are not equeal. Not the same instance.');
b.inside.self = {foo: 'bar'};
assertTrue(!deepEquals(b, a), 'Deep self referencing objects are not equal. Completely different object.');
a = {};
b = {};
a.self = a;
b.self = {};
assertTrue(!deepEquals(b, a), 'Empty object and self reference of an empty object.');

If you are using lodash and don't want to modify either array, you can use the function _.xor(). It compares the two arrays as sets and returns the set that contains their difference. If the length of this difference is zero, the two arrays are essentially equal:
var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);
_.xor(a, b).length === 0
true
_.xor(b, c).length === 0
true

Check every each value by a for loop once you checked the size of the array.
function equalArray(a, b) {
if (a.length === b.length) {
for (var i = 0; i < a.length; i++) {
if (a[i] !== b[i]) {
return false;
}
}
return true;
} else {
return false;
}
}

Using map() and reduce():
function arraysEqual (a1, a2) {
return a1 === a2 || (
a1 !== null && a2 !== null &&
a1.length === a2.length &&
a1
.map(function (val, idx) { return val === a2[idx]; })
.reduce(function (prev, cur) { return prev && cur; }, true)
);
}

If you wish to check arrays of objects for equality and order does NOT matter, i.e.
areEqual([{id: "0"}, {id: "1"}], [{id: "1"}, {id: "0"}]) // true
you'll want to sort the arrays first. lodash has all the tools you'll need, by combining sortBy and isEqual:
// arr1 & arr2: Arrays of objects
// sortProperty: the property of the object with which you want to sort
// Note: ensure every object in both arrays has your chosen sortProperty
// For example, arr1 = [{id: "v-test_id0"}, {id: "v-test_id1"}]
// and arr2 = [{id: "v-test_id1"}, {id: "v-test_id0"}]
// sortProperty should be 'id'
function areEqual (arr1, arr2, sortProperty) {
return _.areEqual(_.sortBy(arr1, sortProperty), _.sortBy(arr2, sortProperty))
}
EDIT: Since sortBy returns a new array, there is no need to clone your arrays before sorting. The original arrays will not be mutated.
Note that for lodash's isEqual, order does matter. The above example will return false if sortBy is not applied to each array first.

This method sucks, but I've left it here for reference so others avoid this path:
Using Option 1 from #ninjagecko worked best for me:
Array.prototype.equals = function(array) {
return array instanceof Array && JSON.stringify(this) === JSON.stringify(array) ;
}
a = [1, [2, 3]]
a.equals([[1, 2], 3]) // false
a.equals([1, [2, 3]]) // true
It will also handle the null and undefined case, since we're adding this to the prototype of array and checking that the other argument is also an array.

There is no easy way to do this. I needed this as well, but wanted a function that can take any two variables and test for equality. That includes non-object values, objects, arrays and any level of nesting.
In your question, you mention wanting to ignore the order of the values in an array. My solution doesn't inherently do that, but you can achieve it by sorting the arrays before comparing for equality
I also wanted the option of casting non-objects to strings so that [1,2]===["1",2]
Since my project uses UnderscoreJs, I decided to make it a mixin rather than a standalone function.
You can test it out on http://jsfiddle.net/nemesarial/T44W4/
Here is my mxin:
_.mixin({
/**
Tests for the equality of two variables
valA: first variable
valB: second variable
stringifyStatics: cast non-objects to string so that "1"===1
**/
equal:function(valA,valB,stringifyStatics){
stringifyStatics=!!stringifyStatics;
//check for same type
if(typeof(valA)!==typeof(valB)){
if((_.isObject(valA) || _.isObject(valB))){
return false;
}
}
//test non-objects for equality
if(!_.isObject(valA)){
if(stringifyStatics){
var valAs=''+valA;
var valBs=''+valB;
ret=(''+valA)===(''+valB);
}else{
ret=valA===valB;
}
return ret;
}
//test for length
if(_.size(valA)!=_.size(valB)){
return false;
}
//test for arrays first
var isArr=_.isArray(valA);
//test whether both are array or both object
if(isArr!==_.isArray(valB)){
return false;
}
var ret=true;
if(isArr){
//do test for arrays
_.each(valA,function(val,idx,lst){
if(!ret){return;}
ret=ret && _.equal(val,valB[idx],stringifyStatics);
});
}else{
//do test for objects
_.each(valA,function(val,idx,lst){
if(!ret){return;}
//test for object member exists
if(!_.has(valB,idx)){
ret=false;
return;
}
// test for member equality
ret=ret && _.equal(val,valB[idx],stringifyStatics);
});
}
return ret;
}
});
This is how you use it:
_.equal([1,2,3],[1,2,"3"],true)
To demonstrate nesting, you can do this:
_.equal(
['a',{b:'b',c:[{'someId':1},2]},[1,2,3]],
['a',{b:'b',c:[{'someId':"1"},2]},["1",'2',3]]
,true);

It handle all possible stuff and even reference itself in structure of object. You can see the example at the end of code.
var deepCompare = (function() {
function internalDeepCompare (obj1, obj2, objects) {
var i, objPair;
if (obj1 === obj2) {
return true;
}
i = objects.length;
while (i--) {
objPair = objects[i];
if ( (objPair.obj1 === obj1 && objPair.obj2 === obj2) ||
(objPair.obj1 === obj2 && objPair.obj2 === obj1) ) {
return true;
}
}
objects.push({obj1: obj1, obj2: obj2});
if (obj1 instanceof Array) {
if (!(obj2 instanceof Array)) {
return false;
}
i = obj1.length;
if (i !== obj2.length) {
return false;
}
while (i--) {
if (!internalDeepCompare(obj1[i], obj2[i], objects)) {
return false;
}
}
}
else {
switch (typeof obj1) {
case "object":
// deal with null
if (!(obj2 && obj1.constructor === obj2.constructor)) {
return false;
}
if (obj1 instanceof RegExp) {
if (!(obj2 instanceof RegExp && obj1.source === obj2.source)) {
return false;
}
}
else if (obj1 instanceof Date) {
if (!(obj2 instanceof Date && obj1.getTime() === obj2.getTime())) {
return false;
}
}
else {
for (i in obj1) {
if (obj1.hasOwnProperty(i)) {
if (!(obj2.hasOwnProperty(i) && internalDeepCompare(obj1[i], obj2[i], objects))) {
return false;
}
}
}
}
break;
case "function":
if (!(typeof obj2 === "function" && obj1+"" === obj2+"")) {
return false;
}
break;
default: //deal with NaN
if (obj1 !== obj2 && obj1 === obj1 && obj2 === obj2) {
return false;
}
}
}
return true;
}
return function (obj1, obj2) {
return internalDeepCompare(obj1, obj2, []);
};
}());
/*
var a = [a, undefined, new Date(10), /.+/, {a:2}, function(){}, Infinity, -Infinity, NaN, 0, -0, 1, [4,5], "1", "-1", "a", null],
b = [b, undefined, new Date(10), /.+/, {a:2}, function(){}, Infinity, -Infinity, NaN, 0, -0, 1, [4,5], "1", "-1", "a", null];
deepCompare(a, b);
*/

var a= [1, 2, 3, '3'];
var b = [1, 2, 3];
var c = a.filter(function (i) { return ! ~b.indexOf(i); });
alert(c.length);

Develop Reference

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