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'}]

None of the elements in arr2 are in arr1, they are all different? — adeneo
– adeneo, Commented Jul 16, 2016 at 18:43
you can use loops, filter, reduce.. there are several ways, what have you tried so far? — webdeb
– webdeb, Commented Jul 16, 2016 at 18:45
Define "element is in array". All the objects in your question have different references, so JS considers them different. — Oriol
– Oriol, Commented Jul 16, 2016 at 19:02

Community · Accepted Answer · 2017-05-23 10:28:48Z

1

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 :)

edited May 23, 2017 at 10:28

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answered Jul 16, 2016 at 18:35

Washington Guedes

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1 Comment

Developer Strange Over a year ago

If u dont mind can u explain obj => !~stringArr1.indexOf(JSON.stringify(obj)) this line and tell me the use of "~".

Nina Scholz · Accepted Answer · 2016-07-16 18:57:50Z

0

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);

answered Jul 16, 2016 at 18:57

Nina Scholz

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Comments

guest271314 · Accepted Answer · 2016-07-16 18:58:27Z

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, 2016 at 18:58

answered Jul 16, 2016 at 18:36

guest271314

1

8 Comments

Developer Strange Over a year ago

if i use other objects ,Uncaught TypeError: arr1.every is not a function is comming

guest271314 Over a year ago

"if i use other objects" What do you mean by "other objects"? Can you create a jsfiddle jsfiddle.net to demonstrate?

Developer Strange Over a year ago

i use other data instead of "arr1" and "arr2" data

guest271314 Over a year ago

What is "other data"? Can you fork jsfiddle jsfiddle.net/1neqy2e5 to demonstrate? Include "other data" at Question?

Washington Guedes Over a year ago

@GowthamV. every is a function of Array.prototype, you can't use it with other data.

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