compare two arrays and remove object by object id javascript

Question

my tried code is :

    const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
    const sasas = [];
    const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
    data2.map((io) => {
     sasas.push(io.id);
     });
    
    sasas.forEach((ik) => {
     for (let i = 0; i < data1.length - 1; i++) {
       if (data1[i].id === ik) {
          data1.splice(i, 1);
            }
          }
      });
    console.log(data1);

i have two arrays and i map the data2 data and getting id values into sasas.

then forEach the sasas data ,

and use forloop to splice the index data of the data1.

then getting the answer is id :1 , id:3, id:5 objects in data1.

but my expected out put is :

const data1 = [{ id: 1, name: 'dd' },{ id: 3, name: 'dd' }];

please help me !

@KamilKiełczewski yes

Ganesh Pandiyan
– Ganesh Pandiyan

2018-12-20 10:29:22 +00:00
Commented Dec 20, 2018 at 10:29 — Ganesh Pandiyan
– Ganesh Pandiyan, Commented Dec 20, 2018 at 10:29

Kamil Kiełczewski · Accepted Answer · 2018-12-20 11:29:45Z

4

Try this

data1.filter(x=> !data2.some(y=> y.id==x.id ))

const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];

let result = data1.filter(x=>!data2.some(y=>y.id==x.id))

console.log(result);

Above short solution time complexity is O(n*m), below O(n+m) version (n,m are arrays lengths):

let h = {}; data2.map(x=>h[x.id]=1); 
let result = data1.filter(x=>!h[x.id]);

const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];

let h = {}; data2.map(x=>h[x.id]=1);     // we use hash map with data2 ids
let result = data1.filter(x=>!h[x.id]);

console.log(result);

edited Dec 20, 2018 at 11:29

answered Dec 20, 2018 at 10:31

Kamil Kiełczewski

93.7k34 gold badges401 silver badges375 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Nikhil Aggarwal · Accepted Answer · 2018-12-20 10:42:27Z

2

You can try following 2 approaches. Also, in place of creating an Array you can use Set (better performance)

Update existing array using while loop

const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];

const sasas = data2.reduce((a,{id}) => a.add(id), new Set())
let i = 0;
while(i < data1.length) {
  if(sasas.has(data1[i].id)) {
    data1.splice(i,1);
  } else i++;
}
console.log(data1);

Create new array using Array.filter

const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];

const sasas = data2.reduce((a,{id}) => a.add(id), new Set())

const result = data1.filter(({id}) => !sasas.has(id));
console.log(result);

edited Dec 20, 2018 at 10:42

answered Dec 20, 2018 at 10:30

Nikhil Aggarwal

28.5k4 gold badges46 silver badges61 bronze badges

2 Comments

HMR Over a year ago

temp = new Set(data2.map(({id}) => id) would be simpler

Nikhil Aggarwal Over a year ago

@HMR - That is simpler to see but I think with this approach total iterations will be 2n whereas the approach I have followed will have n iterations.

dquijada · Accepted Answer · 2018-12-20 10:40:06Z

First of all, you could use that map better, instead of

data2.map((io) => {
    sasas.push(io.id);
});

you could simply:

sasas = data2.map(io => io.id);

Your error:

for (let i = 0; i < data1.length - 1; i++) {

That means you never check the last index of the array, you can either change the < to a <= or simply remove the -1.

This should work:

for (let i = 0; i < data1.length; i++) {

Array.filter

The functionality you are trying to achieve is easily obtainable via .filter() function, like this:

    const data1 = [{ id: 1, name: 'dd' }, { id: 2, name: 'dd' }, { id: 3, name: 'dd' }, { id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
    const data2 = [{ id: 2, name: 'dd' },{ id: 4, name: 'dd' }, { id: 5, name: 'dd' }];
    const sasas = data2.map((io) => io.id);
    
    var result = data1.filter(e => sasas.indexOf(e.id) === -1)
    console.log(result);

Hung Thai · Accepted Answer · 2018-12-20 10:34:11Z

1

Because of this loop, i is stopped at data1.length - 2

for (let i = 0; i < data1.length - 1; i++) {

Variable i should be from 0 to data1.length - 1, which makes the condition become

i < data1.length

answered Dec 20, 2018 at 10:34

Hung Thai

2412 silver badges8 bronze badges

Collectives™ on Stack Overflow

compare two arrays and remove object by object id javascript

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related