Get non existing element from array in javascript

Question

I have two arrays

var a = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}, {'Id': 4, 'name':'prasad', 'age':25}];
var b = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}];

I want to compare each element of array a with array b based on ID (Ids are unique always) and want only record which present in array a but not in array b

Here expected output would be

{'Id': 4, 'name':'prasad', 'age':25}

What I have tried so far:

for (var elementA of a) {
    var uniqueElement = true;
    for (var elementB of b) {
         //if condition chekcs elementA.Id and elementB.Id
         //setting values
    }
}

I believe there will be elegant approve to solve this?

Could be done using one liner: `a.find(elA => !b.find(elB => elB.Id === elA.Id))` — Rayon, Oct 24 '18 at 11:21
Thank you all for your overwhelming response, I will try each and every suggestions. It would be great if you help me with explanation as **I am new to typescript, javascript too** — Prasad Telkikar, Oct 24 '18 at 11:22
@Prasadtelkikar there's not even one answer written in TS :) — ponury-kostek, Oct 24 '18 at 11:26
@ponury-kostek, good catch.. I updated my comment.. Solutions are working like a charm..Cheers — Prasad Telkikar, Oct 24 '18 at 11:28

Nikhil Aggarwal · Accepted Answer · 2018-10-24T11:22:09.980

Try following. Create a Set of Ids of b array and then filter array a based on that.

CASE 1: Use Array.filter in case of more than 1 entry

var a = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}, {'Id': 4, 'name':'prasad', 'age':25}];
var b = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}];

var objB = b.reduce((a,c) => a.add(c.Id), new Set());

var result = a.filter(v => !objB.has(v.Id));
console.log(result);

CASE 2: Use Array.find in case of 1 entry

var a = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}, {'Id': 4, 'name':'prasad', 'age':25}];
var b = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}];

var objB = b.reduce((a,c) => a.add(c.Id), new Set());

var result = a.find(v => !objB.has(v.Id));
console.log(result);

If only single object is expected then `Array#find` could be used instead of `Array#filter`. Good use of `Set` ;) — Rayon, Oct 24 '18 at 11:18
@Rayon - Agreed. I will update the answer covering both the scenarios. Thanks! — Nikhil Aggarwal, Oct 24 '18 at 11:20

score 1 · Answer 2 · answered Oct 24 '18 at 11:16

1

With ES6:

const idsFromB = b.map(item => item.Id)
const c = a.filter(item => idsFromB.indexOf(item.Id) < 0)
//outputs [{'Id': 4, 'name':'prasad', 'age':25}]

answered Oct 24 '18 at 11:16

Raphael Aleixo

597
6
18

More readable and easy to understand. I tried this as well.. working as expected. but sorry I can accept only one answer... Cheers – Prasad Telkikar Oct 24 '18 at 11:30

score 0 · Answer 3 · answered Oct 24 '18 at 11:20

var a = [{ 'Id': 1, 'name': 'bob', 'age': 22 }, { 'Id': 2, 'name': 'alice', 'age': 12 }, { 'Id': 3, 'name': 'mike', 'age': 13 }, { 'Id': 4, 'name': 'prasad', 'age': 25 }];
var b = [{ 'Id': 1, 'name': 'bob', 'age': 22 }, { 'Id': 2, 'name': 'alice', 'age': 12 }, { 'Id': 3, 'name': 'mike', 'age': 13 }];

const na = [];

a.filter((_a) => {
    let isNotUnique = true;
    for (var elementB of b) {
        if (_a.Id === elementB.Id) {
            isNotUnique = false;
            break;
        }
    }
    
    if (isNotUnique) {
        na.push(_a);
    }
    
});

console.log(na);

score 0 · Answer 4 · answered Oct 24 '18 at 11:31

This is essentially a set difference problem.

var a = [
  {'Id': 1, 'name':'bob', 'age':22},
  {'Id': 2, 'name':'alice', 'age':12},
  {'Id': 3, 'name':'mike', 'age':13},
  {'Id': 4, 'name':'prasad', 'age':25}
];
var b = [
  {'Id': 1, 'name':'bob', 'age':22},
  {'Id': 2, 'name':'alice', 'age':12},
  {'Id': 3, 'name':'mike', 'age':13}
];

console.log(JSON.stringify(diff(a, b), null, 2));

function diff(a, b) {
  return a.filter(x => !b.some(y => deepCompare(x, y)));
}

// https://stackoverflow.com/a/1144249/1762224
function deepCompare(){var t,e,r,n;function o(t,e){var f;if(isNaN(t)&&isNaN(e)&&"number"==typeof t&&"number"==typeof e)return!0;if(t===e)return!0;if("function"==typeof t&&"function"==typeof e||t instanceof Date&&e instanceof Date||t instanceof RegExp&&e instanceof RegExp||t instanceof String&&e instanceof String||t instanceof Number&&e instanceof Number)return t.toString()===e.toString();if(!(t instanceof Object&&e instanceof Object))return!1;if(t.isPrototypeOf(e)||e.isPrototypeOf(t))return!1;if(t.constructor!==e.constructor)return!1;if(t.prototype!==e.prototype)return!1;if(r.indexOf(t)>-1||n.indexOf(e)>-1)return!1;for(f in e){if(e.hasOwnProperty(f)!==t.hasOwnProperty(f))return!1;if(typeof e[f]!=typeof t[f])return!1}for(f in t){if(e.hasOwnProperty(f)!==t.hasOwnProperty(f))return!1;if(typeof e[f]!=typeof t[f])return!1;switch(typeof t[f]){case"object":case"function":if(r.push(t),n.push(e),!o(t[f],e[f]))return!1;r.pop(),n.pop();break;default:if(t[f]!==e[f])return!1}}return!0}if(arguments.length<1)return!0;for(t=1,e=arguments.length;t<e;t++)if(r=[],n=[],!o(arguments[0],arguments[t]))return!1;return!0}

.as-console-wrapper { top: 0; max-height: 100% !important; }

score 0 · Answer 5 · edited Jan 19 '21 at 05:49

Use Array.filter() to filter the first array (if the second Array.some() has an element with the first array Id).

The result of arr2.some() is inverted to include only the ones that are not in arr1.

    let arr1 = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}, {'Id': 4, 'name':'prasad', 'age':25}];
let arr2 = [{'Id': 1, 'name':'bob', 'age':22}, {'Id': 2, 'name':'alice', 'age':12}, {'Id': 3, 'name':'mike', 'age':13}];
    
let result = arr1.filter(x=>!arr2.some(y=>x.Id===y.Id));
console.log(result)    // output: [{'Id': 4, 'name':'prasad', 'age':25}]

score -1 · Answer 6 · edited Oct 24 '18 at 11:52

-1

var c;

a.forEach((aobj) => {
c = 0;
b.forEach((bobj) => {
if (aobj.Id === bobj.Id) {
  c++;
 }
})
 if (c === 0) console.log(aobj);
})`

edited Oct 24 '18 at 11:52

Christopher Moore

3,071
4
30
46

answered Oct 24 '18 at 11:23

Sinane

1,614
2
12
17

Get non existing element from array in javascript

6 Answers6