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why is this removeDuplicate not working?

I making a function that simply removes duplicate strings, so there are only unique strings left:
var d = ['a','b','B','C','d','e','f','e'];
d.length; //8
function removeDuplicates(data) {
var i, j, a, b;
for(i = 0; i < data.length - 1; i++) {
a = data[i];
for(j = i; j < data.length; j++) {
b = data[j];
if (a.toLowerCase() == b.toLowerCase()) {
data.splice(j, 1);
}
}
}
}
removeDuplicates(d)
d.length; //4 ????
There were only two duplicates, Only B, e should have gotten removed.
but I get this:
["b", "C", "e", "e"]

Your issue is coming from the fact that any time i === j you'll have a match, and it'll be removed. In your inner loop, just put a check so that you only do your "remove them if they're equal" in situations where i !== j
Also, as nnn noted in the comments, splicing will mess up the index. The cleanest way to do this is with a filter and an every.
Updated solution below:
var d = ['a', 'b', 'B', 'C', 'd', 'e', 'f', 'e'];
var e = ['e','e','e','e'];
// d.length; //8
function removeDuplicates(data) {
const noDuplicates = data.filter((letter, ogIndex) => {
return data.every((innerLetter, innerIndex) => {
if (ogIndex < innerIndex) {
const check = letter.toLowerCase() !== innerLetter.toLowerCase();
return check;
} else {
return true;
}
})
})
return noDuplicates
}
console.log(removeDuplicates(d));
// ["a", "B", "C", "d", "f", "e"]
console.log(removeDuplicates(e));
// ["e"]

With ES6/ES2015, a simple way of getting a list of unique items is to use the new Set data type.
To "unique"-ify an array, just convert it to a set and then back to an array. i.e.
[...new Set(originalArray)]
(Update: My answer originally used Array.from(S_E_T) to convert the set back to an array. I have now changed it to [...S_E_T] to do that conversion.)
To do the same in a case-insensitive manner, follow the same logic, but just convert each original array element to its lowercase equivalent before converting that whole array to the set.
To do keep only the first instance of a string of any case (e.g. to keep 'cAt' from ['cAt', 'CaT'] instead of just the all-lowercased 'cat'), first perform the case-insensitive search as above, then get the index of each unique lowercase element from the lowercased un-uniqified original array, then use that index to retrieve the whichever-cased element from the original-cased un-uniqified original array.
const show = msg => {console.log(JSON.stringify(msg))};
const f = (arr) => {
const arrLC = arr.map(x => x.toLowerCase(x));
return [...new Set(arrLC)].map(x => arr[arrLC.indexOf(x)])
};
const d1 = ['a','b','B','C','d','e','f','e'];
const d2 = ['e', 'e', 'e', 'e', 'e'];
show(d1);
show([...new Set(d1)]);
show([...new Set(d1.map(x => x.toLowerCase(x)))]);
show(f(d1));
show('------');
show(d2);
show([...new Set(d2)]);

Partial string matching in javascript

I have this javascript function:
var y;
y = ['ab', 'f', 'c'].indexOf('b')
This returns -1 because b is not a full match against ab
In javascript, what is the easiest way of looking for a partial match.
So I want :
y = ['ab', 'f', 'c'].indexOf('b')
to return 0(index of first element because it found a partial match while comparing b to ab
I understand i cannot do it using indexOf, what is the best way to do it?

You will have to loop through the Array:
var arr = ['ab', 'f', 'c']
function find(a, c) {
for(var i = 0; i < a.length; ++i) {
if(a[i].indexOf(c) != -1)
return i
}
return -1
}
// Example:
document.body.innerHTML = find(arr, "b")

With ES6 Array.prototype.findIndex and arrow functions:
y = ['ab', 'f', 'c'];
y.findIndex(el => el.indexOf('b') > -1); //0

You have multiple way, you can use the method Array.prototype.filter:
var testRegex = /b/;
var result = ['ab','f','c'].map(function(item, index) {
return testRegex.test(item) ? index : null; //return index in the array if found.
}).filter(function(index) {
return index !== null; // filter the result and remove all results that are null;
});
result should equal [0];
if you try with an array like this:
['ab', 'cgt', 'abcd', 'cd', 'efb'];
It should return:
[0, 2, 4]

How to Compare two Arrays are Equal using Javascript? [duplicate]

This question already has answers here:
How to compare arrays in JavaScript?
(55 answers)
Closed 3 years ago.
I want position of the array is to be also same and value also same.
var array1 = [4,8,9,10];
var array2 = [4,8,9,10];
I tried like this
var array3 = array1 === array2 // returns false

You could use Array.prototype.every().(A polyfill is needed for IE < 9 and other old browsers.)
var array1 = [4,8,9,10];
var array2 = [4,8,9,10];
var is_same = (array1.length == array2.length) && array1.every(function(element, index) {
return element === array2[index];
});
THE WORKING DEMO.

A less robust approach, but it works.
a = [2, 4, 5].toString();
b = [2, 4, 5].toString();
console.log(a===b);

var array3 = array1 === array2
That will compare whether array1 and array2 are the same array object in memory, which is not what you want.
In order to do what you want, you'll need to check whether the two arrays have the same length, and that each member in each index is identical.
Assuming your array is filled with primitives—numbers and or strings—something like this should do
function arraysAreIdentical(arr1, arr2){
if (arr1.length !== arr2.length) return false;
for (var i = 0, len = arr1.length; i < len; i++){
if (arr1[i] !== arr2[i]){
return false;
}
}
return true;
}

A more modern version:
function arraysEqual(a, b) {
a = Array.isArray(a) ? a : [];
b = Array.isArray(b) ? b : [];
return a.length === b.length && a.every((el, ix) => el === b[ix]);
}
Coercing non-array arguments to empty arrays stops a.every() from exploding.
If you just want to see if the arrays have the same set of elements then you can use Array.includes():
function arraysContainSame(a, b) {
a = Array.isArray(a) ? a : [];
b = Array.isArray(b) ? b : [];
return a.length === b.length && a.every(el => b.includes(el));
}

You could try this simple approach
var array1 = [4,8,9,10];
var array2 = [4,8,9,10];
console.log(array1.join('|'));
console.log(array2.join('|'));
if (array1.join('|') === array2.join('|')) {
console.log('The arrays are equal.');
} else {
console.log('The arrays are NOT equal.');
}
array1 = [[1,2],[3,4],[5,6],[7,8]];
array2 = [[1,2],[3,4],[5,6],[7,8]];
console.log(array1.join('|'));
console.log(array2.join('|'));
if (array1.join('|') === array2.join('|')) {
console.log('The arrays are equal.');
} else {
console.log('The arrays are NOT equal.');
}
If the position of the values are not important you could sort the arrays first.
if (array1.sort().join('|') === array2.sort().join('|')) {
console.log('The arrays are equal.');
} else {
console.log('The arrays are NOT equal.');
}

If you comparing 2 arrays but values not in same index, then try this
var array1=[1,2,3,4]
var array2=[1,4,3,2]
var is_equal = array1.length==array2.length && array1.every(function(v,i) { return ($.inArray(v,array2) != -1)})
console.log(is_equal)

Use lodash.
In ES6 syntax:
import isEqual from 'lodash/isEqual';
let equal = isEqual([1,2], [1,2]); // true
Or previous js versions:
var isEqual = require('lodash/isEqual');
var equal = isEqual([1,2], [1,2]); // true

Here goes the code. Which is able to compare arrays by any position.
var array1 = [4,8,10,9];
var array2 = [10,8,9,4];
var is_same = array1.length == array2.length && array1.every(function(element, index) {
//return element === array2[index];
if(array2.indexOf(element)>-1){
return element = array2[array2.indexOf(element)];
}
});
console.log(is_same);

function isEqual(a) {
if (arrayData.length > 0) {
for (var i in arrayData) {
if (JSON.stringify(arrayData[i]) === JSON.stringify(a)) {
alert("Ya existe un registro con esta informacion");
return false;
}
}
}
}
Check this example

Try doing like this: array1.compare(array2)=true
Array.prototype.compare = function (array) {
// if the other array is a falsy value, return
if (!array)
return false;
// compare lengths - can save a lot of time
if (this.length != array.length)
return false;
for (var i = 0, l=this.length; i < l; i++) {
// Check if we have nested arrays
if (this[i] instanceof Array && array[i] instanceof Array) {
// recurse into the nested arrays
if (!this[i].compare(array[i]))
return false;
}
else if (this[i] != array[i]) {
// Warning - two different object instances will never be equal: {x:20} != {x:20}
return false;
}
}
return true;
}

Check if all values of array are equal

I need to find arrays where all values are equal. What's the fastest way to do this? Should I loop through it and just compare values?
['a', 'a', 'a', 'a'] // true
['a', 'a', 'b', 'a'] // false

const allEqual = arr => arr.every( v => v === arr[0] )
allEqual( [1,1,1,1] ) // true
Or one-liner:
[1,1,1,1].every( (val, i, arr) => val === arr[0] ) // true
Array.prototype.every (from MDN) :
The every() method tests whether all elements in the array pass the test implemented by the provided function.

Edit: Be a Red ninja:
!!array.reduce(function(a, b){ return (a === b) ? a : NaN; });
Results:
var array = ["a", "a", "a"] => result: "true"
var array = ["a", "b", "a"] => result: "false"
var array = ["false", ""] => result: "false"
var array = ["false", false] => result: "false"
var array = ["false", "false"] => result: "true"
var array = [NaN, NaN] => result: "false"
Warning:
var array = [] => result: TypeError thrown
This is because we do not pass an initialValue. So, you may wish to check array.length first.

You can turn the Array into a Set. If the size of the Set is equal to 1, then all elements of the Array are equal.
function allEqual(arr) {
return new Set(arr).size == 1;
}
allEqual(['a', 'a', 'a', 'a']); // true
allEqual(['a', 'a', 'b', 'a']); // false

This works. You create a method on Array by using prototype.
if (Array.prototype.allValuesSame === undefined) {
Array.prototype.allValuesSame = function() {
for (let i = 1; i < this.length; i++) {
if (this[i] !== this[0]) {
return false;
}
}
return true;
}
}
Call this in this way:
let a = ['a', 'a', 'a'];
let b = a.allValuesSame(); // true
a = ['a', 'b', 'a'];
b = a.allValuesSame(); // false

In JavaScript 1.6, you can use Array.every:
function AllTheSame(array) {
var first = array[0];
return array.every(function(element) {
return element === first;
});
}
You probably need some sanity checks, e.g. when the array has no elements. (Also, this won't work when all elements are NaN since NaN !== NaN, but that shouldn't be an issue... right?)

And for performance comparison I also did a benchmark:
function allAreEqual(array){
if(!array.length) return true;
// I also made sure it works with [false, false] array
return array.reduce(function(a, b){return (a === b)?a:(!b);}) === array[0];
}
function same(a) {
if (!a.length) return true;
return !a.filter(function (e) {
return e !== a[0];
}).length;
}
function allTheSame(array) {
var first = array[0];
return array.every(function(element) {
return element === first;
});
}
function useSome(array){
return !array.some(function(value, index, array){
return value !== array[0];
});
}
Results:
allAreEqual x 47,565 ops/sec ±0.16% (100 runs sampled)
same x 42,529 ops/sec ±1.74% (92 runs sampled)
allTheSame x 66,437 ops/sec ±0.45% (102 runs sampled)
useSome x 70,102 ops/sec ±0.27% (100 runs sampled)
So apparently using builtin array.some() is the fastest method of the ones sampled.

update 2022 version: use Set()
let a = ['a', 'a', 'b', 'a'];
let b = ['a', 'a', 'a', 'a'];
const check = (list) => {
const setItem = new Set(list);
return setItem.size <= 1;
}
const checkShort = (list) => (new Set(list)).size <= 1
check(a); // false;
check(b); // true;
checkShort(a); // false
checkShort(b); // true
Update new solution: check index
let a = ['a', 'a', 'b', 'a'];
let b = ['a', 'a', 'a', 'a'];
let check = (list) => list.every(item => list.indexOf(item) === 0);
check(a); // false;
check(b); // true;
Updated with ES6:
Use list.every is the fastest way:
let a = ['a', 'a', 'b', 'a'];
let check = (list) => list.every(item => item === list[0]);
old version:
var listTrue = ['a', 'a', 'a', 'a'];
var listFalse = ['a', 'a', 'a', 'ab'];
function areWeTheSame(list) {
var sample = list[0];
return (list.every((item) => item === sample));
}

If you're already using underscore.js, then here's another option using _.uniq:
function allEqual(arr) {
return _.uniq(arr).length === 1;
}
_.uniq returns a duplicate-free version of the array. If all the values are the same, then the length will be 1.
As mentioned in the comments, given that you may expect an empty array to return true, then you should also check for that case:
function allEqual(arr) {
return arr.length === 0 || _.uniq(arr).length === 1;
}

Shortest answer using underscore/lodash
function elementsEqual(arr) {
return !_.without(arr, arr[0]).length
}
spec:
elementsEqual(null) // throws error
elementsEqual([]) // true
elementsEqual({}) // true
elementsEqual([1]) // true
elementsEqual([1,2]) // false
elementsEqual(NaN) // true
edit:
Or even shorter, inspired by Tom's answer:
function elementsEqual2(arr) {
return _.uniq(arr).length <= 1;
}
spec:
elementsEqual2(null) // true (beware, it's different than above)
elementsEqual2([]) // true
elementsEqual2({}) // true
elementsEqual2([1]) // true
elementsEqual2([1,2]) // false
elementsEqual2(NaN) // true

every() function check if all elements of an array
const checkArr = a => a.every( val => val === a[0] )
checkArr(['a','a','a']) // true

You can use Array.every if supported:
var equals = array.every(function(value, index, array){
return value === array[0];
});
Alternatives approach of a loop could be something like sort
var temp = array.slice(0).sort();
var equals = temp[0] === temp[temp.length - 1];
Or, if the items are like the question, something dirty like:
var equals = array.join('').split(array[0]).join('').length === 0;
Also works.

Yes, you can check it also using filter as below, very simple, checking every values are the same as the first one:
//ES6
function sameValues(arr) {
return arr.filter((v,i,a)=>v===a[0]).length === arr.length;
}
also can be done using every method on the array:
//ES6
function sameValues(arr) {
return arr.every((v,i,a)=>v===a[0]);
}
and you can check your arrays like below:
sameValues(['a', 'a', 'a', 'a']); // true
sameValues(['a', 'a', 'b', 'a']); // false
Or you can add it to native Array functionalities in JavaScript if you reuse it a lot:
//ES6
Array.prototype.sameValues = Array.prototype.sameValues || function(){
this.every((v,i,a)=>v===a[0]);
}
and you can check your arrays like below:
['a', 'a', 'a', 'a'].sameValues(); // true
['a', 'a', 'b', 'a'].sameValues(); // false

You can get this one-liner to do what you want using Array.prototype.every, Object.is, and ES6 arrow functions:
const all = arr => arr.every(x => Object.is(arr[0], x));

Now you can make use of sets to do that easily.
let a= ['a', 'a', 'a', 'a']; // true
let b =['a', 'a', 'b', 'a'];// false
console.log(new Set(a).size === 1);
console.log(new Set(b).size === 1);

I think the simplest way to do this is to create a loop to compare the each value to the next. As long as there is a break in the "chain" then it would return false. If the first is equal to the second, the second equal to the third and so on, then we can conclude that all elements of the array are equal to each other.
given an array data[], then you can use:
for(x=0;x<data.length - 1;x++){
if (data[x] != data[x+1]){
isEqual = false;
}
}
alert("All elements are equal is " + isEqual);

You can convert array to a Set and check its size
In case of primitive array entries, i.e. number, string:
const isArrayWithEqualEntries = array => new Set(array).size === 1
In case of array of objects with some field to be tested for equivalence, say id:
const mapper = ({id}) => id
const isArrayWithEqualEntries = array => new Set(array.map(mapper)).size === 1

You can use this:
function same(a) {
if (!a.length) return true;
return !a.filter(function (e) {
return e !== a[0];
}).length;
}
The function first checks whether the array is empty. If it is it's values are equals..
Otherwise it filter the array and takes all elements which are different from the first one. If there are no such values => the array contains only equal elements otherwise it doesn't.

arr.length && arr.reduce(function(a, b){return (a === b)?a:false;}) === arr[0];

Its Simple.
Create a function and pass a parameter.
In that function copy the first index into a new variable.
Then Create a for loop and loop through the array.
Inside a loop create an while loop with a condition checking whether the new created variable is equal to all the elements in the loop.
if its equal return true after the for loop completes else return false inside the while loop.
function isUniform(arra){
var k=arra[0];
for (var i = 0; i < arra.length; i++) {
while(k!==arra[i]){
return false;
}
}
return true;
}

The accepted answer worked great but I wanted to add a tiny bit. It didn't work for me to use === because I was comparing arrays of arrays of objects, however throughout my app I've been using the fast-deep-equal package which I highly recommend. With that, my code looks like this:
let areAllEqual = arrs.every((val, i, arr) => equal(val, arr[0]) );
and my data looks like this:
[
[
{
"ID": 28,
"AuthorID": 121,
"VisitTypeID": 2
},
{
"ID": 115,
"AuthorID": 121,
"VisitTypeID": 1
},
{
"ID": 121,
"AuthorID": 121,
"VisitTypeID": 1
}
],
[
{
"ID": 121,
"AuthorID": 121,
"VisitTypeID": 1
}
],
[
{
"ID": 5,
"AuthorID": 121,
"VisitTypeID": 1
},
{
"ID": 121,
"AuthorID": 121,
"VisitTypeID": 1
}
]
]

You could use a for loop:
function isEqual(arr) {
var first = arr[0];
for (let i = 1; i < arr.length; i++) {
if (first !== arr[i]) {
return false;
}
}
return true;
}

Underscore's _.isEqual(object, other) function seems to work well for arrays. The order of items in the array matter when it checks for equality. See http://underscorejs.org/#isEqual.

var listTrue = ['a', 'a', 'a', 'a'];
var listFalse = ['a', 'a', 'a', 'ab'];
function areWeTheSame(list) {
var sample = list[0];
return !(list.some(function(item) {
return !(item == sample);
}));
}

function isUniform(array) {
for (var i=1; i< array.length; i++) {
if (array[i] !== array[0]) { return false; }
}
for (var i=1; i< array.length; i++) {
if (array[i] === array[0]) { return true; }
}
}
For the first loop; whenever it detects uneven, returns "false"
The first loop runs, and if it returns false, we have "false"
When it's not return false, it means there will be true, so we do the second loop. And of course we will have "true" from the second loop (because the first loop found it's NOT false)

Create a string by joining the array.
Create string by repetition of the first character of the given array
match both strings
function checkArray(array){
return array.join("") == array[0].repeat(array.length);
}
console.log('array: [a,a,a,a]: ' + checkArray(['a', 'a', 'a', 'a']));
console.log('array: [a,a,b,a]: ' + checkArray(['a', 'a', 'b', 'a']));
And you are DONE !

Another interesting way when you use ES6 arrow function syntax:
x = ['a', 'a', 'a', 'a']
!x.filter(e=>e!==x[0])[0] // true
x = ['a', 'a', 'b', 'a']
!x.filter(e=>e!==x[0])[0] // false
x = []
!x.filter(e=>e!==x[0])[0] // true
And when you don't want to reuse the variable for array (x):
!['a', 'a', 'a', 'a'].filter((e,i,a)=>e!==a[0])[0] // true
IMO previous poster who used array.every(...) has the cleanest solution.

this might work , you can use the comment out code as well that also woks well with the given scenerio.
function isUniform(){
var arrayToMatch = [1,1,1,1,1];
var temp = arrayToMatch[0];
console.log(temp);
/* return arrayToMatch.every(function(check){
return check == temp;
});*/
var bool;
arrayToMatch.forEach(function(check){
bool=(check == temp);
})
console.log(bool);
}
isUniform();

Use index of operator for every item of array
to check if it exists or not. If even one item returns -1 (doesn't exist then it will be false)
nst arr1 = [1, 3, 5];
const arr2 = [5, 7, 9];
const arr3 = [1, 3, 5];
arr1.every(item => arr2.indexOf(item) != -1)
// this will return false
arr1.every(item => arr3.indexOf(item) != -1)
// this will return true

Simple one line solution, just compare it to an array filled with the first entry.
if(arr.join('') === Array(arr.length).fill(arr[0]).join(''))

**// Logical Solution:- Declare global array and one variable(To check the condition) whether all element of an array contains same value or not.**
var arr =[];
var isMatching = false;
for(var i=0;i<arr.length;i++){
if(String(arr[i]).toLowerCase()== "Your string to check"){
isMatching=true;
// Array has same value in all index of an array
}
else{
isMatching=false;
// Array Doesn't has same value in all index of an array
break;
}
}
// **Check isMatching variable is true or false**
if(isMatching){ // True
//If Array has same value in all index, then this block will get executed
}
else{ //False
//If Array doesn't has same value in all index, then this block will get executed
}

Using jQuery to compare two arrays of Javascript objects

I have two arrays of JavaScript Objects that I'd like to compare to see if they are the same. The objects may not (and most likely will not) be in the same order in each array. Each array shouldn't have any more than 10 objects. I thought jQuery might have an elegant solution to this problem, but I wasn't able to find much online.
I know that a brute nested $.each(array, function(){}) solution could work, but is there any built in function that I'm not aware of?
Thanks.

There is an easy way...
$(arr1).not(arr2).length === 0 && $(arr2).not(arr1).length === 0
If the above returns true, both the arrays are same even if the elements are in different order.
NOTE: This works only for jquery versions < 3.0.0 when using JSON objects

I was also looking for this today and found:
http://www.breakingpar.com/bkp/home.nsf/0/87256B280015193F87256BFB0077DFFD
Don't know if that's a good solution though they do mention some performance considerations taken into account.
I like the idea of a jQuery helper method.
#David I'd rather see your compare method to work like:
jQuery.compare(a, b)
I doesn't make sense to me to be doing:
$(a).compare(b)
where a and b are arrays. Normally when you $(something) you'd be passing a selector string to work with DOM elements.
Also regarding sorting and 'caching' the sorted arrays:
I don't think sorting once at the start of the method instead of every time through the loop is 'caching'. The sort will still happen every time you call compare(b). That's just semantics, but...
for (var i = 0; t[i]; i++) { ...this loop finishes early if your t array contains a false value in it somewhere, so $([1, 2, 3, 4]).compare([1, false, 2, 3]) returns true!
More importantly the array sort() method sorts the array in place, so doing var b = t.sort() ...doesn't create a sorted copy of the original array, it sorts the original array and also assigns a reference to it in b. I don't think the compare method should have side-effects.
It seems what we need to do is to copy the arrays before working on them. The best answer I could find for how to do that in a jQuery way was by none other than John Resig here on SO! What is the most efficient way to deep clone an object in JavaScript? (see comments on his answer for the array version of the object cloning recipe)
In which case I think the code for it would be:
jQuery.extend({
compare: function (arrayA, arrayB) {
if (arrayA.length != arrayB.length) { return false; }
// sort modifies original array
// (which are passed by reference to our method!)
// so clone the arrays before sorting
var a = jQuery.extend(true, [], arrayA);
var b = jQuery.extend(true, [], arrayB);
a.sort();
b.sort();
for (var i = 0, l = a.length; i < l; i++) {
if (a[i] !== b[i]) {
return false;
}
}
return true;
}
});
var a = [1, 2, 3];
var b = [2, 3, 4];
var c = [3, 4, 2];
jQuery.compare(a, b);
// false
jQuery.compare(b, c);
// true
// c is still unsorted [3, 4, 2]

My approach was quite different - I flattened out both collections using JSON.stringify and used a normal string compare to check for equality.
I.e.
var arr1 = [
{Col: 'a', Val: 1},
{Col: 'b', Val: 2},
{Col: 'c', Val: 3}
];
var arr2 = [
{Col: 'x', Val: 24},
{Col: 'y', Val: 25},
{Col: 'z', Val: 26}
];
if(JSON.stringify(arr1) == JSON.stringify(arr2)){
alert('Collections are equal');
}else{
alert('Collections are not equal');
}
NB: Please note that his method assumes that both Collections are sorted in a similar fashion, if not, it would give you a false result!

Convert both array to string and compare
if (JSON.stringify(array1) == JSON.stringify(array2))
{
// your code here
}

I found this discussion because I needed a way to deep compare arrays and objects. Using the examples here, I came up with the following (broken up into 3 methods for clarity):
jQuery.extend({
compare : function (a,b) {
var obj_str = '[object Object]',
arr_str = '[object Array]',
a_type = Object.prototype.toString.apply(a),
b_type = Object.prototype.toString.apply(b);
if ( a_type !== b_type) { return false; }
else if (a_type === obj_str) {
return $.compareObject(a,b);
}
else if (a_type === arr_str) {
return $.compareArray(a,b);
}
return (a === b);
}
});
jQuery.extend({
compareArray: function (arrayA, arrayB) {
var a,b,i,a_type,b_type;
// References to each other?
if (arrayA === arrayB) { return true;}
if (arrayA.length != arrayB.length) { return false; }
// sort modifies original array
// (which are passed by reference to our method!)
// so clone the arrays before sorting
a = jQuery.extend(true, [], arrayA);
b = jQuery.extend(true, [], arrayB);
a.sort();
b.sort();
for (i = 0, l = a.length; i < l; i+=1) {
a_type = Object.prototype.toString.apply(a[i]);
b_type = Object.prototype.toString.apply(b[i]);
if (a_type !== b_type) {
return false;
}
if ($.compare(a[i],b[i]) === false) {
return false;
}
}
return true;
}
});
jQuery.extend({
compareObject : function(objA,objB) {
var i,a_type,b_type;
// Compare if they are references to each other
if (objA === objB) { return true;}
if (Object.keys(objA).length !== Object.keys(objB).length) { return false;}
for (i in objA) {
if (objA.hasOwnProperty(i)) {
if (typeof objB[i] === 'undefined') {
return false;
}
else {
a_type = Object.prototype.toString.apply(objA[i]);
b_type = Object.prototype.toString.apply(objB[i]);
if (a_type !== b_type) {
return false;
}
}
}
if ($.compare(objA[i],objB[i]) === false){
return false;
}
}
return true;
}
});
Testing
var a={a : {a : 1, b: 2}},
b={a : {a : 1, b: 2}},
c={a : {a : 1, b: 3}},
d=[1,2,3],
e=[2,1,3];
console.debug('a and b = ' + $.compare(a,b)); // a and b = true
console.debug('b and b = ' + $.compare(b,b)); // b and b = true
console.debug('b and c = ' + $.compare(b,c)); // b and c = false
console.debug('c and d = ' + $.compare(c,d)); // c and d = false
console.debug('d and e = ' + $.compare(d,e)); // d and e = true

In my case compared arrays contain only numbers and strings. This solution worked for me:
function are_arrs_equal(arr1, arr2){
return arr1.sort().toString() === arr2.sort().toString()
}
Let's test it!
arr1 = [1, 2, 3, 'nik']
arr2 = ['nik', 3, 1, 2]
arr3 = [1, 2, 5]
console.log (are_arrs_equal(arr1, arr2)) //true
console.log (are_arrs_equal(arr1, arr3)) //false

I don't think there's a good "jQuery " way to do this, but if you need efficiency, map one of the arrays by a certain key (one of the unique object fields), and then do comparison by looping through the other array and comparing against the map, or associative array, you just built.
If efficiency is not an issue, just compare every object in A to every object in B. As long as |A| and |B| are small, you should be okay.

Well, if you want to compare only the contents of arrays, there's a useful jQuery function $.inArray()
var arr = [11, "String #1", 14, "String #2"];
var arr_true = ["String #1", 14, "String #2", 11]; // contents are the same as arr
var arr_false = ["String #1", 14, "String #2", 16]; // contents differ
function test(arr_1, arr_2) {
var equal = arr_1.length == arr_2.length; // if array sizes mismatches, then we assume, that they are not equal
if (equal) {
$.each(arr_1, function (foo, val) {
if (!equal) return false;
if ($.inArray(val, arr_2) == -1) {
equal = false;
} else {
equal = true;
}
});
}
return equal;
}
alert('Array contents are the same? ' + test(arr, arr_true)); //- returns true
alert('Array contents are the same? ' + test(arr, arr_false)); //- returns false

Change array to string and compare
var arr = [1,2,3],
arr2 = [1,2,3];
console.log(arr.toString() === arr2.toString());

The nice one liner from Sudhakar R as jQuery global method.
/**
* Compare two arrays if they are equal even if they have different order.
*
* #link https://stackoverflow.com/a/7726509
*/
jQuery.extend({
/**
* #param {array} a
* First array to compare.
* #param {array} b
* Second array to compare.
* #return {boolean}
* True if both arrays are equal, otherwise false.
*/
arrayCompare: function (a, b) {
return $(a).not(b).get().length === 0 && $(b).not(a).get().length === 0;
}
});

I also found this when looking to do some array comparisons with jQuery. In my case I had strings which I knew to be arrays:
var needle = 'apple orange';
var haystack = 'kiwi orange banana apple plum';
But I cared if it was a complete match or only a partial match, so I used something like the following, based off of Sudhakar R's answer:
function compareStrings( needle, haystack ){
var needleArr = needle.split(" "),
haystackArr = haystack.split(" "),
compare = $(haystackArr).not(needleArr).get().length;
if( compare == 0 ){
return 'all';
} else if ( compare == haystackArr.length ) {
return 'none';
} else {
return 'partial';
}
}

If duplicates matter such that [1, 1, 2] should not be equal to [2, 1] but should equal [1, 2, 1], here is a reference counting solution:
const arrayContentsEqual = (arrayA, arrayB) => {
if (arrayA.length !== arrayB.length) {
return false}
const refCount = (function() {
const refCountMap = {};
const refCountFn = (elt, count) => {
refCountMap[elt] = (refCountMap[elt] || 0) + count}
refCountFn.isZero = () => {
for (let elt in refCountMap) {
if (refCountMap[elt] !== 0) {
return false}}
return true}
return refCountFn})()
arrayB.map(eltB => refCount(eltB, 1));
arrayA.map(eltA => refCount(eltA, -1));
return refCount.isZero()}
Here is the fiddle to play with.

var arr1 = [
{name: 'a', Val: 1},
{name: 'b', Val: 2},
{name: 'c', Val: 3}
];
var arr2 = [
{name: 'c', Val: 3},
{name: 'x', Val: 4},
{name: 'y', Val: 5},
{name: 'z', Val: 6}
];
var _isEqual = _.intersectionWith(arr1, arr2, _.isEqual);// common in both array
var _difference1 = _.differenceWith(arr1, arr2, _.isEqual);//difference from array1
var _difference2 = _.differenceWith(arr2, arr1, _.isEqual);//difference from array2
console.log(_isEqual);// common in both array
console.log(_difference1);//difference from array1
console.log(_difference2);//difference from array2
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>

Try this
function check(c,d){
var a = c, b = d,flg = 0;
if(a.length == b.length)
{
for(var i=0;i<a.length;i++)
a[i] != b[i] ? flg++ : 0;
}
else
{
flg = 1;
}
return flg = 0;
}