Object Comparing: check if an object contains the whole other object - javascript

I have two objects. Their structure looks a bit like this:
{
education: ["school", "institute"],
courses: ["HTML", "JS", "CSS"],
Computer: {
"OS":"WXP",
"WS":"NotePad"
}
}
The second:
{
education: ["school", "university", "institute", "collage"],
courses: ["HTML", "CSS", "JS", "Managing", "Directing"],
Computer: {
"OS":"WXP",
"WS":"NotePad",
"AV":"Avast"
},
something: function(){...},
other: "thing"
}
As you may noticed, the second object containes the whole first object, plus some items that the first one doesn't have.
I need to compare these two objects, and get an answer(true-false) if the second objects containes every single item of the first object.
true - if all of the items of the first object are also in the second one
false - if at least one of the items of the first object is not also in the second one, for example: if the second object wouldn't have the "css" course.
(The first one is requirements, the second is what the person has. I need to check if the person has all of the requirements)
Could be plain JS, jQuery, whatever. I prefer not to use server-side languages for that.
is there a way of doing that?
THANKS!

Just recursively check it:
function isContainedIn(a, b) {
if (typeof a != typeof b)
return false;
if (Array.isArray(a) && Array.isArray(b)) {
// assuming same order at least
for (var i=0, j=0, la=a.length, lb=b.length; i<la && j<lb;j++)
if (isContainedIn(a[i], b[j]))
i++;
return i==la;
} else if (Object(a) === a) {
for (var p in a)
if (!(p in b && isContainedIn(a[p], b[p])))
return false;
return true;
} else
return a === b;
}
> isContainedIn(requirements, person)
true
For a more set-logic-like approach to arrays, where order does not matter, add something like
a.sort();
b = b.slice().sort()
(assuming orderable contents) before the array comparison loop or replace that by the quite inefficient
return a.every(function(ael) {
return b.some(function(bel) {
return isContainedIn(ael, bel);
});
});

JavaScript (in ES5) has two composite native types (I'm assuming you don't have any custom collections in your code, if you do - I assume they support the 'old' iteration protocol (having .length)
Here is an annotated sketch of a solution. I did not run this - it's there to get you an idea of how to implement this algorithm. Note that this enters an endless loop for back references (var a = {}; a.a =a}).
function sub(big,small){
if(typeof big === "function") return small === big; // function reference equality.
if(big.length){ // iterable, for example array, nodelist etc. (even string!)
if(small.length > big.length) return false; // small is bigger!
for(var i = 0; i < small.length; i++ ){
if(!sub(big[i],small[i])){ // doesn't have a property
return false;
}
}
return true; // all properties are subproperties recursively
}
if(typeof big === "object" && big !== null){
// I assume null is not a subset of an object, you may change this, it's conceptual
if(typeof small !== "object" || small === null) return false;
for(var key in small){
// I consider the prototype a part of the object, you may filter this with a
// hasOwnProperty check here.
if(!sub(big[key],small[key])){ // doesn't have a property
return false;
}
return true;
}
}
return big === small; // primitive value type equality
// , or ES7 value type equality, future compat ftw :P
}

Edit: didn't notice that merge changes the first argument... changed the code, but it still would cause obj2 to change. You can add _.cloneDeep(obj2) which should take care of that, but by then my solution doesn't seem as elegant. Updated the demo with cloneDeep as well.
Edit2: Since JSON.stringify requires the order of object properties to be the same in the objects you compare, you could instead use something like Object comparison in JavaScript. However, in the demo you can see that it works, so I would say there is a good chance that for your case, using _.merge with JSON.stringify is reliable.
With lo-dash, you can use _.merge and check whether the result is the same as the larger object.
function(obj1, obj2) {
var obj3 =_.merge(_.cloneDeep(obj2), obj1);
return JSON.stringify(obj3) === JSON.stringify(obj1);
}
demo
Of course, another option would be to iterate over the entire object with vanilla JS.

// When order of objects is not same
function isContainedIn(a, b) {
if (typeof a != typeof b)
return false;
if (Array.isArray(a) && Array.isArray(b)) {
if(a.length == 1) {
var j=0;
while (j < b.length) {
if ((isContainedIn( a[0], b[j]))) {
return true;
}
j++;
}
return false;
} else {
var k=0;
while (k < a.length) {
if (!(isContainedIn([a[k]], b))) {
return false;
}
k++;
}
return true;
}
} else if (Object(a) === a) {
for (var p in a)
if (!(p in b && isContainedIn(a[p], b[p])))
return false;
return true;
} else
return a === b;
};
isContainedIn(requirements, person)
true

In addition to Benjamin's answer - you could test this:
const sub = (big, small) => {
if (typeof big === 'function' || typeof small === 'string') return small === big; // function or string reference equality
if (big && big.length) { // iterable, for example array, nodelist etc. (even string!)
if (small.length > big.length) return false; // small is bigger!
for (let i = 0; i < small.length; i++)
if (!sub(big[i], small[i])) // doesn't have a property
return false;
return true; // all properties are subproperties recursively
}
if (typeof big === 'object' && big !== null) {
// I assume null is not a subset of an object, you may change this, it's conceptual
if (typeof small !== 'object' || small === null) return false;
// console.log(Object.keys(small));
for (const key of Object.keys(small)) {
// I consider the prototype a part of the object, you may filter this with a
// hasOwnProperty check here.
if (sub(big[key], small[key]) === false) // doesn't have a property
return false;
continue;
}
return true;
}
return big === small; // primitive value type equality
};
or even use a much cleaner solution:
https://github.com/blackflux/object-deep-contain

Related

React Native Redux: what's the best way to compare prevProps and this.props?

In my React Native app I want to trigger a function if a certain prop has changed. Something like this:
componentDidUpdate(prevProps) {
if (prevProps.a !== this.props.a) {
<trigger event>
}
}
The problem is that I assume prevProps.a !== this.props.a will always be true because it's comparing by reference.
What's the best way to approach this, so that my event will be triggered only if a property of this.props.a has changed?
It's better to compare the properties directly instead of comparing the whole objects like this:
componentDidUpdate(prevProps) {
if (prevProps.a.field1 === this.props.a.field1) {
// Do stuff here
}
}
However, if you still want to compare two objects with deeply nested properties, a simple way to do this is to stringify both objects, and then directly compare the strings:
if (JSON.stringify(prevProps.a) === JSON.stringify(this.props.a) {
// Do stuff here
}
It depends somewhat on the nature of the object a being compared. If it's a matter of specific properties, then you can obviously compare them individually and that is always going to be the most efficient approach. However, if you want to trigger the code on any change to a, including deep changes inside nested objects, then comparing the stringified representations of the objects (JSON.stringify) can do the trick. However it is also inefficient. An intermediate approach is to do a shallow compare, and here is some code (from the React team no less) to do that:
export function shallowEqual(objA: any, objB: any, compare?, compareContext?) {
let ret = compare ? compare.call(compareContext, objA, objB) : void 0;
if (ret !== void 0) {
return !!ret;
}
if (objA === objB) {
return true;
}
if (typeof objA !== 'object' || !objA || typeof objB !== 'object' || !objB) {
return false;
}
const keysA = Object.keys(objA);
const keysB = Object.keys(objB);
if (keysA.length !== keysB.length) {
return false;
}
const bHasOwnProperty = Object.prototype.hasOwnProperty.bind(objB);
// Test for A's keys different from B.
for (let idx = 0; idx < keysA.length; idx++) {
const key = keysA[idx];
if (!bHasOwnProperty(key)) {
return false;
}
const valueA = objA[key];
const valueB = objB[key];
ret = compare ? compare.call(compareContext, valueA, valueB, key) : void 0;
if (ret === false || (ret === void 0 && valueA !== valueB)) {
return false;
}
}
If you need to do an efficient deep compare (more so than JSON.stringify), then using a library such as react-fast-compare (here) is your best bet.
(Note also that JSON.stringify is not perfect for some cases, since some objects like dates and functions are not stringified properly or at all. In practice those limitations usually aren't an issue in React, but you should be aware of them at least.)

Checking if object is already in array not working

I am trying to check if an object is already in an array, following this answer here: How to determine if object is in array
I adjusted the function to suit my needs, and now it looks like this:
var _createDatesArray, _objInArray;
_objInArray = function(array, obj) {
var i;
i = 0;
while (i < array.length) {
console.log("array[i] == obj is ", array[i] === obj, " array[i] is ", array[i], " and obj is ", obj);
if (array[i] === obj) {
return true;
}
i++;
}
};
_createDatesArray = function(val) {
var result;
if (val != null) {
result = {
text: val
};
if (!_objInArray(scope.datesQuestion.dates, result)) {
scope.datesQuestion.dates.push(result);
}
return console.log(scope.datesQuestion.dates);
}
};
What I need to do, is basically see if the object is already in the array, and if is,t return true.
When debugging, the result of the console log is the following:
array[i] == obj is false array[i] is {text: "10/08/17"} and obj is
{text: "10/08/17"}
and the function says they are different (array[i] == obj is false) but they look the same to me.
I also checked the type of both, which is this:
typeof array[i] is "object"
typeof obj is "object"
can you help me with this? Why are they different? what can be different?
_createDatesArray is called when $scope of my angular app changes its value based on a ng-model, but I don't think this is relevant
They're two different objects with the same content. Comparing them with == or === will yield false.
Since you're using AngularJS, you can use angular.equals() instead to perform a deep comparison of the object's properties.
The objects you are comparing don't have the same reference, so == is returning false. See Object comparison in JavaScript for a more detailed explanation.
In this particular case, you could simply compare the text of dates to see if they are equivilant. However this wouldn't work for all objects like the function name suggests.
if (arr[i].text === obj.text)
Alternatively, you could create a method specific for checking if your array includes a given date and simplify it greatly using Array.prototype.some:
dateInArray = function (array, date) {
return array.some(function (arrayDate) {
return arrayDate.text === date.text
})
}
Or, more succinctly using ES6 arrow functions:
dateInArray = (array, date) => array.some(arrayDate => arrayDate.text === date.text)
array[i] === obj will return true ONLY if its the same object. In the link that you have referred the object being checked is the same object that is inserted in the array, thats why it returns true. In your case, you are creating a new object 'result' and adding the value in there. So the array does not contain the exact same object and hence returns false.
If 'text' is the only property in the object, instead of checking for the entire object you could check if the 'text' property in both the objects is same.
_objInArray = function(array, obj) {
var i;
i = 0;
while (i < array.length) {
if (array[i].text === obj.text) {
return true;
}
i++;
}
};
This happens because objects in JS are compared by reference, but not by values they have. But you need to compare objects by their value. So you need to get some third-party function or to write your own. One more option is to use angular built-in equals function.
angular.equals($scope.user1, $scope.user2);
For a better understanding you can read a good article on this subject here.

How to check if values in one JavaScript object are present in another one?

I am trying to compare json_str1 and json_str2, here it should return true as all elements in json_str1 are present in json_str2.
For now I am doing this the long way like this
json_str1 = '{"0":"a","1":"b","2":"c"}';
json_str2 = '{"0":"c","1":"b","2":"a"}';
json_obj1 = $.parseJSON(json_str1);
json_obj2 = $.parseJSON(json_str2);
arr1 = $.map(json_obj1, function(el) { return el });
arr2 = $.map(json_obj2, function(el) { return el });
if($(arr1).not(arr2).length === 0 && $(arr2).not(arr1).length === 0)
alert("equal");
else
alert("not equal");
How could I make it short and simple, without converting the objects into an array ?
https://jsfiddle.net/kq9gtdr0/
Use the following code:
Object.keys(json_obj1) . every(k1 =>
Object.keys(json_obj2) . some(k2 =>
json_obj1[k1] === json_obj2[k2]
)
);
In English:
Every key k1 in json_obj1 satisfies the condition that some key k2 in json_obj2 satisifies the condition that the value of json_obj1 with key k1 is equal to the value of json_obj2 with key k2.
Or in more conversational English:
Every value in the first object matches some value in the second.
Using lodash
var _ = require('lodash');
function compareValues(jstr1, jstr2) {
return _.isEqual(_.valuesIn(JSON.parse(jstr1)).sort(), _.valuesIn(JSON.parse(jstr2)).sort());
}
json_str1 = '{"0":"a","1":"b","2":"c"}';
json_str2 = '{"0":"c","1":"b","2":"a"}';
console.log(compareValues(json_str1, json_str2));
There is short and easy accurate way to this.
You can use a third party but extremely popular utility library called Lodash. Chaining functions you can check for equality.
First parse both JSON into objects
Then use _.values() to extract the values of all keys of each into separate arrays
Find difference of two arrays. If its an empty array then both of them are equal.
You can chain all the steps into one statement like:
_.isEmpty(_.difference(_.values(json_obj1), _.values(json_obj2)))
Example: https://jsfiddle.net/kq9gtdr0/4/
For more information:
https://lodash.com/docs#values
https://lodash.com/docs#difference
https://lodash.com/docs#isEmpty
You can include the library from CDN(https://cdn.jsdelivr.net/lodash/4.5.1/lodash.min.js) or download and use it as normal script. Lodash offers plenty of useful utility functions that makes JS programming a lot easier. You better try it out.
If you prefer using libraries, then you could use underscore isMatch
_.isMatch(object, properties)
Tells you if the keys and values in properties are contained in
object.
Extending the awesome answer by #user663031, in case you need to do deep comparison, here's code that works:
export function objectOneInsideObjectTwo(jsonObj1: any, jsonObj2: any): boolean {
return Object.keys(jsonObj1).every((k1) => {
if (parseType(jsonObj1[k1]) === 'dict') {
return objectOneInsideObjectTwo(jsonObj1[k1], jsonObj2[k1]);
}
if (parseType(jsonObj1[k1]) === 'array') {
const results: boolean[] = [];
jsonObj1[k1].forEach((o: any, i: number) => {
if (parseType(o) === 'dict') {
results.push(objectOneInsideObjectTwo(o, jsonObj2[k1][i]));
} else {
results.push(o === jsonObj2[k1][i]);
}
});
return results.every((r) => r);
}
return Object.keys(jsonObj2).some((k2) => jsonObj1[k1] === jsonObj2[k2]);
});
}
export function parseType<T>(v: T): string {
if (v === null || v === undefined) {
return 'null';
}
if (typeof v === 'object') {
if (v instanceof Array) {
return 'array';
}
if (v instanceof Date) {
return 'date';
}
return 'dict';
}
return typeof v;
}
You can try this
var json_str1 = {"0":"a","1":"b","2":"c"};
var json_str2 = {"0":"c","1":"b","2":"a"};
var flag = 1;
if(Object.keys(json_str1).length == Object.keys(json_str2).length){
Object.keys(json_str1).forEach(function(x){
if(!json_str2.hasOwnProperty(x) || json_str2[x] != json_str1[x]){
flag = 0;
return;
}
});
}
if(flag)
alert('equal');
else
alert('Not Equal');
If you want to find out if both Objects have the same keys there is no way to do this without at least converting the keys of both Objects to an array with Object.keys or looping through both Objects!
The reason is simple: It's clear that you have to compare the number of keys of both Objects and the only way to do this is by looping through all properties or Object.keys.
So I think the shortest way to do this is:
json_obj1 = JSON.parse('{"0":"a","1":"b","2":"c"}');
json_obj2 = JSON.parse('{"0":"c","1":"b","2":"a"}');
keys_1 = Object.keys(json_obj1);
keys_2 = Object.keys(json_obj2);
if(keys_1.length === keys_2.length && keys_1.every(key => keys_2.indexOf(key) >= 0)) {
alert('equal')
} else {
alert('not equal')
}
If you only want to check if all keys from json1 are present in json2 you can do:
json_obj1 = JSON.parse('{"0":"a","1":"b","2":"c"}');
json_obj2 = JSON.parse('{"0":"c","1":"b","2":"a"}');
if(Object.keys(json_obj1).every(key => key in json_obj2)) {
alert('equal');
} else {
alert('not equal');
}
In your question and comments you indicate you are only looking to verify that "all elements in json_str1 are present in json_str2". Your example code doesn't just do that, it checks for the complete equality of keys by testing if all the keys (not values) in the first object are in the second object AND all the keys in the second object are in the first object. By looking at your code, i assume that when you say "elements" you mean keys.
All that aside, this might help:
// Your first few lines of code
json_str1 = '{"0":"a","1":"b","2":"c"}';
json_str2 = '{"0":"c","1":"b","2":"a"}';
json_obj1 = $.parseJSON(json_str1);
json_obj2 = $.parseJSON(json_str2);
// My new code
var values1 = Object.keys(json_obj1).map(key => json_obj1[key]);
var values2 = Object.keys(json_obj2).map(key => json_obj2[key]);
// Check if every key in the first object is in the second object.
values1.every(k1 => values2.indexOf(k1) >= 0);
// OR
// Check for equality of keys by checking both directions.
values1.every(k1 => values2.indexOf(k1) >= 0) && values2.every(k2 => values1.indexOf(k2) >= 0);
That's 2 lines to get the keys, and one line to check. You only need one of those two checks.

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);

is this valid JavaScript arrays comparing algorithm?

Primordial topic, I know, but can anyone please comment on the following algorithm for comparing two ES5 arrrays:
function equal_arrays(a, b) {
"use strict";
var c = 0;
return a.every(function (e) {
return e === b[c++];
});
}
I think this is very clean and obvious use of [].every(), to quickly compare two arrays?
I have that nagging "surely it can not be that simple ?" feeling about it ?
NOTE: two arrays are equal when they contain all elements in the same position strictly equal to each other. So both the element index and element value have to be exactly equal. Values of different types are considered not equal. Sparse arrays are also compared.
TEST CASES:
equal_arrays([],[]) ; // => true
equal_arrays([1,2],[1,2]) ; // => true
equal_arrays([1,,2],[1,,2]) ; // => true
equal_arrays([,],[,]); // => true
equal_arrays([1,,3,,,],[1,,3,,,]); // => true
Uses cases yielding => false, one can imagine herself. Comparing non-arrays is a syntax error.
Many thanks to the well meaning and helpful contributors. It appears that the "best" (there is never such a thing) implementation is this:
function has(element, index)
{
return this[index] === element;
}
function equal_arrays(a, b)
{
return (a.length === b.length) && a.every(has, b) && b.every(has, a);
}
#tom's implementation of the two-way every() which is necessary sp that test cases like this one work:
equal_arrays([1,,3],[1,2,3]); //=> false
Once again thanks ...
No, it is not valid. From the MDN docs:
callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
So if the first array has 'gaps' they will be skipped over.
equal_arrays([1, 2, , 4], [1, 2, 4]); // true
equal_arrays([1, 2, 4], [1, 2, , 4]); // false
Here is a better implementation:
function equal_arrays(a, b) {
if (a.length != b.length) return false;
var i;
for (i = 0; i < a.length; i++) {
if (a[i] !== b[i]) return false;
}
return true;
}
Here is an elegant implementation of #RobG's two-way every() idea:
function has(element, index)
{
return this[index] === element;
}
function equal_arrays(a, b)
{
return (a.length === b.length) && a.every(has, b) && b.every(has, a);
}
I have that nagging "surely it can not be that simple ?" feeling about it ?
No, it isn't. Using every on one array will test it for every member of that array but not the other. You have to test both ways.
Also, it depends on your criteria for "equal". You might want every member of the same index to have the same value, but you might not be concerned about order so is [1,2] equal to [2,1]? and is [,,,2] equal to [2]?
Also, should [1,,2] equal [1,undefined,2], i.e. should a member that doesn't exist be "equal" to one that does but has the value undefined?
An array comparison function
The following may do the job. It compares both ways, checking own properties and values. I think it covers all cases but am willing to be convinced otherwise. Might be better as a separate function, but adding to Array.prototype is convenient.
// Test that own properties of two arrays have the same values
Array.prototype.isEqualTo = function(a) {
var visited = {};
if (this.length != a.length) return false;
// Test every own property of this, remember tested properties
for (var p in this) {
if (this.hasOwnProperty(p)) {
visited[p] = p;
if (!(a.hasOwnProperty(p)) || this[p] !== a[p]) {
return false;
}
}
}
// Reverse test that comparison array only has tested properties
for (var q in a) {
if (a.hasOwnProperty(q) && !(q in this)) {
return false;
}
}
return true;
}
console.log([1,,2].isEqualTo([1,undefined,2])); // false, not defined != undefined
console.log([1,,2].isEqualTo([1,2])); // false, different length
console.log([1,2].isEqualTo([1,2])); // true
Note that inherited properties should be ignored as if arrays from different windows are compared (e.g. one is from a frame) then inherited properties won't be equal.
As an alternative, if you just want to check if both arrays are exactly the same you could just do this:
var a = [1,2,3,4];
var b = [1,2,3,4];
JSON.stringify(a) == JSON.stringify(b); //= true
That should work with arrays of numbers and strings.
As CrazyTrain said, the increment value is useless and you need to check for length first:
function equalArrays( a, b ) {
"use strict";
if (a.length !== b.length) return false;
return a.filter(function(el){ return el;}).every(function(e, i) {
return e === b[i];
});
}
Is a valid algorithm to compare arrays by value.

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