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
– Rayon, Commented Oct 24, 2018 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
– Prasad Telkikar, Commented Oct 24, 2018 at 11:22
@Prasadtelkikar there's not even one answer written in TS :) — ponury-kostek
– ponury-kostek, Commented Oct 24, 2018 at 11:26
@ponury-kostek, good catch.. I updated my comment.. Solutions are working like a charm..Cheers — Prasad Telkikar
– Prasad Telkikar, Commented Oct 24, 2018 at 11:28

Nikhil Aggarwal · Accepted Answer · 2018-10-24 11:22:09Z

1

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);

edited Oct 24, 2018 at 11:22

answered Oct 24, 2018 at 11:14

Nikhil Aggarwal

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3 Comments

Rayon Over a year ago

If only single object is expected then Array#find could be used instead of Array#filter. Good use of Set ;)

Nikhil Aggarwal Over a year ago

@Rayon - Agreed. I will update the answer covering both the scenarios. Thanks!

Prasad Telkikar Over a year ago

I tried this one working as expected.. Thanks Nikhil

Raphael Aleixo · Accepted Answer · 2018-10-24 11:16:49Z

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, 2018 at 11:16

Raphael Aleixo

6167 silver badges19 bronze badges

1 Comment

Prasad Telkikar Over a year ago

More readable and easy to understand. I tried this as well.. working as expected. but sorry I can accept only one answer... Cheers

philip_nunoo · Accepted Answer · 2018-10-24 11:20:29Z

0

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);

answered Oct 24, 2018 at 11:20

philip_nunoo

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Comments

Mr. Polywhirl · Accepted Answer · 2018-10-24 11:31:39Z

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; }

Prasad Telkikar · Accepted Answer · 2021-01-19 05:49:30Z

0

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}]

edited Jan 19, 2021 at 5:49

Prasad Telkikar

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answered Jan 19, 2021 at 0:21

Ahmed Alaa

11 silver badge

Comments

Christopher Moore · Accepted Answer · 2018-10-24 11:52:07Z

-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, 2018 at 11:52

Christopher Moore

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answered Oct 24, 2018 at 11:23

Sinane

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