Comparing 2 JSON objects structure in JavaScript

Comparing 2 JSON objects structure in JavaScript - javascript

I am using angular-translate for a big application. Having several people committing code + translations, many times the translation objects are not in sync.
I am building a Grunt plugin to look at both files' structure and compare it (just the keys and overall structure, not values).
The main goals are:
Look into each file, and check if the structure of the whole object
(or file, in this case) is the exact same as the translated ones;
On error, return the key that doesn't match.
It turns out it was a bit more complicated than I anticipated. So i figured I could do something like:
Sort the object;
Check the type of data the value contains (since they are translations, it will only have strings, or objects for the nestings) and store it in another object, making the key equal to the original key and the value would be a string 'String', or an object in case it's an object. That object contains the children elements;
Recursively repeat steps 1-2 until the whole object is mapped and sorted;
Do the same for all the files
Stringify and compare everything.
A tiny example would be the following object:
{
key1: 'cool',
key2: 'cooler',
keyWhatever: {
anotherObject: {
key1: 'better',
keyX: 'awesome'
},
aObject: 'actually, it\'s a string'
},
aKey: 'more awesomeness'
}
would map to:
{
aKey: 'String',
key1: 'String',
key2: 'String',
keyWhatever: {
aObject: 'String',
anotherObject: {
key1: 'String',
keyX: 'String'
}
}
}
After this, I would stringify all the objects and proceed with a strict comparison.
My question is, is there a better way to perform this? Both in terms of simplicity and performance, since there are many translation files and they are fairly big.
I tried to look for libraries that would already do this, but I couldn't find any.
Thank you
EDIT: Thank you Jared for pointing out objects can't be sorted. I am ashamed for saying something like that :D Another solution could be iterating each of the properties on the main translation file, and in case they are strings, compare the key with the other files. In case they are objects, "enter" them, and do the same. Maybe it is even simpler than my first guess. What should be done?

Lets say you have two JSON objects, jsonA and jsonB.
function compareValues(a, b) {
//if a and b aren't the same type, they can't be equal
if (typeof a !== typeof b) {
return false;
}
// Need the truthy guard because
// typeof null === 'object'
if (a && typeof a === 'object') {
var keysA = Object.keys(a).sort(),
keysB = Object.keys(b).sort();
//if a and b are objects with different no of keys, unequal
if (keysA.length !== keysB.length) {
return false;
}
//if keys aren't all the same, unequal
if (!keysA.every(function(k, i) { return k === keysB[i];})) {
return false;
}
//recurse on the values for each key
return keysA.every(function(key) {
//if we made it here, they have identical keys
return compareValues(a[key], b[key]);
});
//for primitives just use a straight up check
} else {
return a === b;
}
}
//true if their structure, values, and keys are identical
var passed = compareValues(jsonA, jsonB);
Note that this can overflow the stack for deeply nested JSON objects. Note also that this will work for JSON but not necessarily regular JS objects as special handling is needed for Date Objects, Regexes, etc.

Actually you do need to sort the keys, as they are not required to be spit out in any particular order. Write a function,
function getComparableForObject(obj) {
var keys = Object.keys(obj);
keys.sort(a, b => a > b ? 1 : -1);
var comparable = keys.map(
key => key + ":" + getValueRepresentation(obj[key])
).join(",");
return "{" + comparable + "}";
}
Where getValueRepresentation is a function that either returns "String" or calls getComparableForObject. If you are worried about circular references, add a Symbol to the outer scope, repr, assign obj[repr] = comparable in the function above, and in getValueRepresentation check every object for a defined obj[repr] and return it instead of processing it recursively.

Sorting an array of the keys from the object works. However, sorting has an average time complexity of O(n⋅log(n)). We can do better. A fast general algorithm for ensuring two sets A and B are equivalent is as follows:
for item in B
if item in A
remove item from A
else
sets are not equivalent
sets are equivalent iff A is empty
To address #Katana31, we can detect circular references as we go by maintaining a set of visited objects and ensuring that all descendents of that object are not already in the list:
# poorly written pseudo-code
fn detectCycles(A, found = {})
if A in found
there is a cycle
else
found = clone(found)
add A to found
for child in A
detectCycles(child, found)
Here's a complete implementation (you can find a simplified version that assumes JSON/non-circular input here):
var hasOwn = Object.prototype.hasOwnProperty;
var indexOf = Array.prototype.indexOf;
function isObjectEmpty(obj) {
for (var key in obj) {
return false;
}
return true;
}
function copyKeys(obj) {
var newObj = {};
for (var key in obj) {
newObj[key] = undefined;
}
return newObj;
}
// compares the structure of arbitrary values
function compareObjectStructure(a, b) {
return function innerCompare(a, b, pathA, pathB) {
if (typeof a !== typeof b) {
return false;
}
if (typeof a === 'object') {
// both or neither, but not mismatched
if (Array.isArray(a) !== Array.isArray(b)) {
return false;
}
if (indexOf.call(pathA, a) !== -1 || indexOf.call(pathB, b) !== -1) {
return false;
}
pathA = pathA.slice();
pathA.push(a);
pathB = pathB.slice();
pathB.push(b);
if (Array.isArray(a)) {
// can't compare structure in array if we don't have items in both
if (!a.length || !b.length) {
return true;
}
for (var i = 1; i < a.length; i++) {
if (!innerCompare(a[0], a[i], pathA, pathA)) {
return false;
}
}
for (var i = 0; i < b.length; i++) {
if (!innerCompare(a[0], b[i], pathA, pathB)) {
return false;
}
}
return true;
}
var map = copyKeys(a), keys = Object.keys(b);
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (!hasOwn.call(map, key) || !innerCompare(a[key], b[key], pathA,
pathB)) {
return false;
}
delete map[key];
}
// we should've found all the keys in the map
return isObjectEmpty(map);
}
return true;
}(a, b, [], []);
}
Note that this implementation directly compares two objects for structural equivalency, but doesn't reduce the objects to a directly comparable value (like a string). I haven't done any performance testing, but I suspect that it won't add significant value, though it will remove the need to repeatedly ensure objects are non-cyclic. For that reason, you could easily split compareObjectStructure into two functions - one to compare the structure and one to check for cycles.

Related

Javascript - Determine if any scoped references in object

I execute some function and get a return value. This value could be anything (string, array, object, reference, function). I then pass this result along using JSON.stringify.
Now, the functions and references don't do me much good in the scope they're being delivered to, so the whole "to string, eval" method isn't much use. I'm going to store them in a local array and just pass along an ID to reference them by later. But, I do go ahead and send string data, arrays, and objects (in the "associated array" sense of javascript objects) as those all play very nicely with JSON.stringify.
I'm already using try... JSON.stringify() catch to do this with recursive objects (which JSON.stringify errors on.) But that doesn't account for anything else mentioned above.
What is the most efficient way to check if an value contains a function?
And not
typeof foo === "function"
Because the return might be
["foo", "bar", ["foo", "bar"], function(){...something}]
I don't want to pick apart each individual piece's type either, just return on the whole whether there's ANY functions/objects that cannot be safely stringified. I could probably work out how to loop and check each individual value, but if there's a shortcut or more efficient method someone can think of, I'd like to hear it.
Thanks!

Refining welcome and appreciated!
//your favorite object length checking function could go here
$.objectLength = (function(){
//ie<9
if (typeof Object.keys === "undefined" ){
return function(o){
var count = 0, i;
for (i in o) {
if (o.hasOwnProperty(i)) {
count++;
}
}
return count;
};
//everyone else
} else {
return function(o){
return Object.keys(o).length;
}
}
})();
//comparing our two objects
$.checkMatch = function(a, b){
//if they're not the same length, we're done here. (functions exist, etc)
if (typeof a !== typeof b || typeof a === "object" && typeof b === "object" && a && b && $.objectLength(a) !== $.objectLength(b)){
return false;
//if they are the same length, they may contain deeper objects we need to check.
} else {
var key;
for (key in a){
//make sure it's not prototyped key
if (a.hasOwnProperty(key)){
//if it doesn't exist on the other object
if (!b.hasOwnProperty(key)){
return false;
//if this an object as well
} else if (typeof a[key] === "object"){
//check for a match
if (!$.checkMatch(a[key], b[key])){
return false;
}
//then check if they're not equal (simple values)
} else if (a[key] !== b[key]){
return false
}
}
}
return true;
}
};
//...stuff
//catch recursive objects in parameters
var good = true, sendObject = {"ourobject", "values"}, finalSendObject;
//try to stringify, which rejects infinitely recursive objects
try {
finalSendObject = JSON.stringify(sendObject);
} catch(e){
good = false;
}
//if that passes, try matching the original against the parsed JSON string
if (good && sendObject !== JSON.parse(finalSendObject)){
good = $.checkMatch(sendObject, JSON.parse(finalSendObject));
}

This will not work
But I will leave it up for anyone who thinks of trying it. Working solution coming shortly.
30 seconds later, I figure it out myself.
String it, parse it. If anything changed, it wont be equal to itself.
var checkParse = function(obj){
return obj === JSON.parse(JSON.strigify(obj));
}

How to make JavaScript objects and arrays behave the same way?

I wanted to write one version of a function that iterated over both Array objects and Objects, with minimal duplicate code. Something like:
function (arr_or_obj) {
arr_or_obj.forEach( function(key,value) {
// do stuff with key and value
});
}
This was before I realized that (in Chrome at least), Object.keys returns a list of the keys for an array. So I could use that to treat the Array like an object. Like so:
function (arr_or_obj) {
var keys = Object.keys(arr_or_obj);
keys.forEach( function(key) {
var value = arr_or_obj[key];
// do stuff with key and value
});
}
Problem solved. But this was not before I wrote my own "JavaScript pseudo-Array iterator". The only advantage of this was that instead of getting a list of keys for the Array (which could be very long), we save memory by producing only the return values we need. Like so:
function create_iterator(max) {
var jkeys = {};
jkeys.count_ = 0;
jkeys.length = max;
jkeys.getter_ = function() {
var returnable = jkeys.count_;
jkeys.count_ += 1;
jkeys.__defineGetter__(jkeys.count_,jkeys.getter_);
delete jkeys[jkeys.count_-1];
return returnable;
};
jkeys.__defineGetter__(0, jkeys.getter_);
return jkeys;
}
Which you can then call by going:
var z = create_iterator(100);
z[0];
>> 0
z[0];
>> undefined
z[1];
>> 1;
z[2]
>> 2;
...
This is sort of a question and answer in one, but the obvious question is, is there a better way to do this without using Object.keys?

Object.keys gives you an array of the object's own enumerable properties. That is, properties it has itself, not from its prototype, and that are enumerable (so not including things like length on arrays, which is a non-enumerable property).
You can do the same thing with for-in if you use correct safeguards:
var key;
for (key in arr_or_obj) {
if (arr_or_obj.hasOwnProperty(key)) {
// do something with it
}
}
Now you're doing the same thing you were doing with Object.keys.
But, note that this (and Object.keys) will include properties people put on arrays that aren't array indexes (something which is perfectly valid in JavaScript). E.g.:
var key;
var a = [1, 2, 3];
a.foo = "bar";
for (key in a) {
if (a.hasOwnProperty(key)) {
console.log(key); // Will show "foo" as well as "0", "1", and "2"
}
}
If you want to only process array indexes and not other properties, you'll need to know whether the thing is an array, and then if so use the "is this an index" test:
var key;
var isArray = Array.isArray(arr_or_obj);
// Or (see below):
//var isArray = Object.prototype.toString.call(arr_or_obj) === "[object Array]";
for (key in arr_or_obj) {
if (a.hasOwnProperty(key) &&
(!isArray || isArrayIndex(key))
) {
console.log(key); // Will only show "0", "1", and "2" for `a` above
}
}
...where Array.isArray is a new feature of ES5 which is easily shimmed for older browsers if necessary:
if (!Array.isArray) {
Array.isArray = (function() {
var toString = Object.prototype.toString;
function isArray(arg) {
return toString.call(arg) === "[object Array]";
}
return isArray;
})();
}
...and isArrayIndex is:
function isArrayIndex(key) {
return /^0$|^[1-9]\d*$/.test(key) && key <= 4294967294; // 2^32 - 2
}
See this other answer for details on that (where the magic numbers come from, etc.).
That loop above will loop through an object's own enumerable properties (all of them) if it's not an array, and loop through the indexes (but not other properties) if it's an array.
Taking it a step further, what if you want to include the properties the object gets from its prototype? Object.keys omits those (by design). If you want to include them, just don't use hasOwnProperty in the for-in loop.

You can try something like this...
Live Demo
function forIn(obj) {
var isArray = Array.isArray(obj),
temp = isArray ? obj : Object.keys(obj);
temp.forEach(function (value, index, array) {
console.log(this[isArray ? index : value]);
}, obj);
}

You can write:
for (key in arr_or_obj) {
if (arr_or_obj.hasOwnProperty(key) {
value = arr_or_obj[key];
// do stuff with key and value
}
}

Sort Array B after Array A, such that references and equal Primitives, maintain the exact Position

Update 2
I added a weight lookup to the sort function, which increased the performance by about 100% as well as the stability, as the previous sort function didn't consider all types, and as 1 == "1" the result depends on the initial order of the Array, as #Esailija points out.
The intent of the question is to improve this Answer of mine, I liked the question and since it got accepted and I felt like there is some performance to squeeze out of the sort function. I asked this question here since I hadn't many clues left where to start.
Maybe this makes things clearer as well
Update
I rephrased the complete question, as many people stated I was not specific enough, I did my best to specify what I mean. Also, I rewrote the sort function to better express the intent of the question.
Let arrayPrev be an Array (A) ,where A consists of 0 to n Elements' (E)
Let an Element either be
of a Primitive type
boolean, string, number, undefined, null
a Reference to an Object O, where O.type = [object Object] and O can consist of
0 to n Properties P, where P is defined like Element plus
an circular Reference to any P in O
Where any O can be contained 1 to n times. In the sense of GetReferencedName(E1) === GetReferencedName(E2)...
a Reference to an O, where O.type = [object Array] and O is defined like A
a circular Reference to any E in A
Let arrayCurr be an Array of the same length as arrayPrev
Illustrated in the following example
var x = {
a:"A Property",
x:"24th Property",
obj: {
a: "A Property"
},
prim : 1,
}
x.obj.circ = x;
var y = {
a:"A Property",
x:"24th Property",
obj: {
a: "A Property"
},
prim : 1,
}
y.obj.circ = y;
var z = {};
var a = [x,x,x,null,undefined,1,y,"2",x,z]
var b = [z,x,x,y,undefined,1,null,"2",x,x]
console.log (sort(a),sort(b,a))
The Question is, how can I efficiently sort an Array B, such that any Reference to an Object or value of a Primitive, shares the exact same Position as in a Previously, by the same compareFunction, sorted, Array A.
Like the above example
Where the resulting array shall fall under the rules.
Let arrayPrev contain the Elements' of a and arrayCurr contain the Elements' of b
Let arrayPrev be sorted by a CompareFunction C.
Let arrayCurr be sorted by the same C.
Let the result of sorting arrayCur be such, that when accessing an E in arrayCur at Position n, let n e.g be 5
if type of E is Object GetReferencedName(arrayCurr[n]) === GetReferencedName(arrayPrev[n])
if type of E is Primitive GetValue(arrayCurr[n]) === GetValue(arrayPrev[n])
i.e b[n] === a[n] e.g b[5] === a[5]
Meaning all Elements should be grouped by type, and in this sorted by value.
Where any call to a Function F in C shall be at least implemented before ES5, such that compatibility is given without the need of any shim.
My current approach is to Mark the Objects in arrayPrev to sort them accordingly in arrayCurr and later delete the Marker again. But that seems rather not that efficient.
Heres the current sort function used.
function sort3 (curr,prev) {
var weight = {
"[object Undefined]":6,
"[object Object]":5,
"[object Null]":4,
"[object String]":3,
"[object Number]":2,
"[object Boolean]":1
}
if (prev) { //mark the objects
for (var i = prev.length,j,t;i>0;i--) {
t = typeof (j = prev[i]);
if (j != null && t === "object") {
j._pos = i;
} else if (t !== "object" && t != "undefined" ) break;
}
}
curr.sort (sorter);
if (prev) {
for (var k = prev.length,l,t;k>0;k--) {
t = typeof (l = prev[k]);
if (t === "object" && l != null) {
delete l._pos;
} else if (t !== "object" && t != "undefined" ) break;
}
}
return curr;
function sorter (a,b) {
var tStr = Object.prototype.toString
var types = [tStr.call(a),tStr.call(b)]
var ret = [0,0];
if (types[0] === types[1] && types[0] === "[object Object]") {
if (prev) return a._pos - b._pos
else {
return a === b ? 0 : 1;
}
} else if (types [0] !== types [1]){
return weight[types[0]] - weight[types[1]]
}
return a>b?1:a<b?-1:0;
}
}
Heres a Fiddle as well as a JSPerf (feel free to add your snippet)
And the old Fiddle

If you know the arrays contain the same elements (with the same number of repetitions, possibly in a different order), then you can just copy the old array into the new one, like so:
function strangeSort(curr, prev) {
curr.length = 0; // delete the contents of curr
curr.push.apply(curr, prev); // append the contents of prev to curr
}
If you don't know the arrays contain the same elements, it doesn't make sense to do what you're asking.
Judging by the thing you linked, it's likely that you're trying to determine whether the arrays contain the same elements. In that case, the question you're asking isn't the question you mean to ask, and a sort-based approach may not be what you want at all. Instead, I recommend a count-based algorithm.
Compare the lengths of the arrays. If they're different, the arrays do not contain the same elements; return false. If the lengths are equal, continue.
Iterate through the first array and associate each element with a count of how many times you've seen it. Now that ES6 Maps exist, a Map is probably the best way to track the counts. If you don't use a Map, it may be necessary or convenient to maintain counts for items of different data types differently. (If you maintain counts for objects by giving them a new property, delete the new properties before you return.)
Iterate through the second array. For each element,
If no count is recorded for the element, return false.
If the count for the element is positive, decrease it by 1.
If the count for the element is 0, the element appears more times in the second array than in the first. Return false.
Return true.
If step 4 is reached, every item appears at least as many times in the first array as it does in the second. Otherwise, it would have been detected in step 3.1 or 3.3. If any item appeared more times in the first array than in the second, the first array would be bigger, and the algorithm would have returned in step 1. Thus, the arrays must contain the same elements with the same number of repetitions.

from the description, it appears you can simply use a sort function:
function sortOb(a,b){a=JSON.stringify(a);b=JSON.stringify(b); return a===b?0:(a>b?1:-1);}
var x = {a:1};
var y = {a:2};
var a = [1,x,2,3,y,4,x]
var b = [1,y,3,4,x,x,2]
a.sort(sortOb);
b.sort(sortOb);
console.log (a,b);

Your function always returns a copy of sort.el, but you can have that easier:
var sort = (function () {
var tmp;
function sorter(a, b) {
var types = [typeof a, typeof b];
if (types[0] === "object" && types[1] === "object") {
return tmp.indexOf(a) - tmp.indexOf(b); // sort by position in original
}
if (types[0] == "object") {
return 1;
}
if (types[1] == "object") {
return -1;
}
return a > b ? 1 : a < b ? -1 : 0;
}
return function (el) {
if (tmp) {
for (var i = 0; i < el.length; i++) {
el[i] = tmp[i]; // edit el in-place, same order as tmp
}
return el;
}
tmp = [].slice.call(el); // copy original
return tmp = el.sort(sorter); // cache result
};
})();
Note that I replaced sort.el by tmp and a closure. Fiddle: http://jsfiddle.net/VLSKK/
Edit: This (as well as your original solution)
only works when the arrays contain the same elements.
maintains the order of different objects in a
but
does not mess up when called more than twice

Try this
function ComplexSort (a, b)
{
if(typeof a == "object") {
a = a.a;
}
if(typeof b == "object") {
b = b.a;
}
return a-b;
}
var x = {a:1};
var y = {a:2};
var a = [1,x,2,3,y,4,x]
var b = [1,y,3,4,x,x,2]
a.sort(ComplexSort);
b.sort(ComplexSort);
console.log ("Consistent", a,b);

Pluck specific javascript value from an object based on an array of indexes

Given a nested object like this:
var cars = {
"bentley": {
"suppliers": [
{
"location": "England",
"name": "Sheffield Mines"}
]
// ...
}
};
and an array like this ["bentley", "suppliers", "0", "name"], is there an existing function that will pluck the deepest element, i.e. pluck_innards(cars, ['bentley', "suppliers", "0", "name"]) and that returns "Sheffield Mines".
In other words, is there a function (which I will name deep_pluck) where
deep_pluck(cars, ['bentley', 'suppliers', '0', 'name'])
=== cars['bentley']['suppliers']['0']['name']
It seems to me that this is simple, yet common enough, to have probably been done in one of the Javascript utility libraries such as jQuery or lo-dash/underscore - but I have not seen it.
My thought is something trivial, along the lines of:
function deep_pluck(array, identities) {
var this_id = identities.shift();
if (identities.length > 0) {
return deep_pluck(array[this_id], identities);
}
return array[this_id];
}
Which I have posted on jsFiddle.
It would be helpful of course if the function were smart enough to identify when numerical indexes in arrays are needed. I am not sure offhand what other caveats may be a concern.
This is all a fairly long question for something I imagine has already been cleverly solved, but I thought to post this as I would interested in seeing what solutions are out there.

I don't think you'll have problems with Array indexes if you pass them as number 0.
Here's alternative version of your function without recursion:
function deep_pluck(object, identities) {
var result = object;
for(var i = 0; i < identities.length; i++) {
result = result[identities[i]];
}
return result;
}
Working example here: http://jsfiddle.net/AmH2w/1/

dotty.get(obj, pathspec) does it, accepting either an array or a dotted string as the pathspec.
Dotty is open source, and also has an exists method, and a putter.
The methodology is recursion and very similar to your idea, except that dotty includes a test for null/undefined objects so that it doesn't throw exceptions for trying to access an element of something that doesn't exist.
The dotty.get() source from the docs is posted below:
var get = module.exports.get = function get(object, path) {
if (typeof path === "string") {
path = path.split(".");
}
if (!(path instanceof Array) || path.length === 0) {
return;
}
path = path.slice();
var key = path.shift();
if (typeof object !== "object" || object === null) {
return;
}
if (path.length === 0) {
return object[key];
}
if (path.length) {
return get(object[key], path);
}
};

Although not a generic library, it seems that CasperJS has something of this kind with its utils.getPropertyPath function.
/**
* Retrieves the value of an Object foreign property using a dot-separated
* path string.
*
* Beware, this function doesn't handle object key names containing a dot.
*
* #param Object obj The source object
* #param String path Dot separated path, eg. "x.y.z"
*/
function getPropertyPath(obj, path) {
if (!isObject(obj) || !isString(path)) {
return undefined;
}
var value = obj;
path.split('.').forEach(function(property) {
if (typeof value === "object" && property in value) {
value = value[property];
} else {
value = undefined;
}
});
return value;
}
Edit:
I have come across implementations to solve this a couple times since, including:
the getObject plugin by Ben Alman (on Github).
one I rolled - see gist
Edit (2014)
I would also note the relatively new lodash.deep.

Here's a short ES6 implementation using reduce:
function get(obj, keyPath) {
return keyPath
.split(".")
.reduce((prev, curr) => prev[curr], obj);
}
Usage:
get(cars, "bentley.suppliers.0.name") // -> "Sheffield Mines"

How to check if two arrays are equal with JavaScript? [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);

We Keep Coding

JavaScript is the programming language of the Web.