get element from array which is not in first array of object

Question

I have two array of objects in javascript

var arr1 = [{'a':'1'},{'b':'2'},{'c':'3'}];
var arr2 = [{'a':'1'},{'b':'2'},{'d':'4'}];

I want to get the element from arr2 which is not in the arr1. my output will be [{'d':'4'}]

Possible duplicate of [JavaScript array difference](http://stackoverflow.com/questions/1187518/javascript-array-difference) — Alon Eitan, Jul 16 '16 at 18:31
None of the elements in `arr2` are in `arr1`, they are all different? — adeneo, Jul 16 '16 at 18:43
you can use loops, filter, reduce.. there are several ways, what have you tried so far? — webdeb, Jul 16 '16 at 18:45
http://adripofjavascript.com/blog/drips/object-equality-in-javascript.html — charlietfl, Jul 16 '16 at 18:54
Define "element is in array". All the objects in your question have different references, so JS considers them different. — Oriol, Jul 16 '16 at 19:02

score 1 · Accepted Answer · edited May 23 '17 at 10:28

The easiest method that came my mind is using JSON.stringify:

var arr1 = [{'a':'1'},{'b':'2'},{'c':'3'}];
var arr2 = [{'a':'1'},{'b':'2'},{'d':'4'}];

stringArr1 = JSON.stringify(arr1);
var result = arr2.filter(
  obj => !~stringArr1.indexOf(JSON.stringify(obj))
);

console.log(result);

But there should be better ways.

The equivalent of:

var result = arr2.filter(
  obj => !~stringArr1.indexOf(JSON.stringify(obj))
);

is the common:

var result = arr2.filter(function (obj) {
  var stringObj = JSON.stringify(obj);
  if (stringArr1.indexOf(stringObj) != -1)
    return true;
  else
    return false;
});

The tricks are simple, basically you need to know 3 things:

All non-zero numbers are true. SO link.
The bitwise not (~) turns -1 into 0. MDN link.
Inline arrow functions (=>) does not need return keyword. MDN link.

Hope it helps :)

If u dont mind can u explain `obj => !~stringArr1.indexOf(JSON.stringify(obj))` this line and tell me the use of "~". — Gmv, Jul 18 '16 at 05:16

guest271314 · Answer 2 · 2016-07-16T18:58:27.670

0

You can use $.grep(), .every(), Object.keys()

var arr1 = [{'a':'1'},{'b':'2'},{'c':'3'}];
var arr2 = [{'a':'1'},{'b':'2'},{'d':'4'}];
var res = $.grep(arr2, function(v) {
             return arr1.every(function(_v) {
               var keys = [Object.keys(v)[0], Object.keys(_v)[0]];
               return keys[0] !== keys[1] && v[keys[0]] !== _v[keys[1]]
             })
          });
console.log(res)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.0.0/jquery.min.js"></script>

edited Jul 16 '16 at 18:58

answered Jul 16 '16 at 18:36

guest271314

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if i use other objects ,Uncaught TypeError: arr1.every is not a function is comming – Gmv Jul 16 '16 at 19:18
_"if i use other objects"_ What do you mean by "other objects"? Can you create a jsfiddle http://jsfiddle.net to demonstrate? – guest271314 Jul 16 '16 at 19:20
i use other data instead of "arr1" and "arr2" data – Gmv Jul 16 '16 at 19:22
What is "other data"? Can you fork jsfiddle https://jsfiddle.net/1neqy2e5/ to demonstrate? Include "other data" at Question? – guest271314 Jul 16 '16 at 19:24
@GowthamV. `every` is a function of `Array.prototype`, you can't use it with _other data_. – Washington Guedes Jul 16 '16 at 19:25
i ll add the image in question.please find there – Gmv Jul 16 '16 at 19:32
@GowthamV The image does not provide text of what `arr1` and `arr2` are? Can you fork jsfiddle at link to demonstrate? Also, at image `arr1` and `arr2` appear to be same array? – guest271314 Jul 16 '16 at 19:35
@GowthamV Can you reproduce issue at stacksnippets or jsfiddle? – guest271314 Jul 16 '16 at 19:44

score 0 · Answer 3 · answered Jul 16 '16 at 18:57

You could use JSON.stringify for each element of the arrays and use a hash table for checking

var arr1 = [{ 'a': '1' }, { 'b': '2' }, { 'c': '3' }],
    arr2 = [{ 'a': '1' }, { 'b': '2' }, { 'd': '4' }],
    arr3,
    hash = Object.create(null);

arr1.forEach(function (o) {
    hash[JSON.stringify(o)] = true;
});

arr3 = arr2.filter(function (o) {
    return !hash[JSON.stringify(o)];
});

console.log(arr3);

get element from array which is not in first array of object

3 Answers3