Move object value from one array of objects to another

Question

I have two arrays of objects and need to move the value from the 2nd array in to the first array object with the same id.

array1 = [{id:1, location: 'A'},{id:2, location: 'B'},{id:3, location: 'C'},{id:4, location: 'D'}]

array2 = [{id:1, value: 123},{id:2, value: 5466},{id:3, value: 89484},{id:4, value: -4.215}]

Ive tried merging them but i just end up with duplicate objects.

how i want it to end up looking like

array1 = [{id:1, location: 'A', value: 123},{id:2, location: 'B', value: 5466},{id:3, location: 'C', value: 89484},{id:4, location: 'D', value: -4.215}]

Yevhen Horbunkov · Accepted Answer · 2020-07-08 12:34:53Z

1

You may traverse your base array with Array.prototype.map(), using Array.prototype.find() along the way to lookup another array for matching id:

const array1 = [{id:1, location: 'A'},{id:2, location: 'B'},{id:3, location: 'C'},{id:4, location: 'D'}],
      array2 = [{id:1, value: 123},{id:2, value: 5466},{id:3, value: 89484},{id:4, value: -4.215}],
      
      result = array1.map(o => ({...o, ...array2.find(_o => _o.id == o.id)}))
      
console.log(result)

.as-console-wrapper{min-height:100%;}

edited Jul 8, 2020 at 12:34

answered Jul 8, 2020 at 11:08

Yevhen Horbunkov

15.6k3 gold badges27 silver badges46 bronze badges

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

Comments

Stark Buttowski · Accepted Answer · 2020-07-08 11:10:34Z

0

Please find below snippet

array1 = [{id:1, location: 'A'},{id:2, location: 'B'},{id:3, location: 'C'},{id:4, location: 'D'}]

array2 = [{id:1, value: 123},{id:2, value: 5466},{id:3, value: 89484},{id:4, value: -4.215}]

let result = array1.map(obj => {
  return {
        ...array2.filter(obj2 => obj2.id === obj.id)[0],
        ...obj
   }
})

console.log(result)

answered Jul 8, 2020 at 11:10

Stark Buttowski

1,8492 gold badges11 silver badges21 bronze badges

1 Comment

Yevhen Horbunkov Over a year ago

This will loop unnecessarily throughout the entire array2 even when the match (unique) was already found. So, .find()-approach is more optimal from performance standpoint.

Alberto · Accepted Answer · 2020-07-08 11:11:48Z

0

This should work:

array1.map( // build a second array with the new objects
       el => Object.assign( // create the merged object
               el, //          from the current object
               array2.find( // with the object in array2 with the same id
                    obj => obj.id == el.id
               )
       )
)

answered Jul 8, 2020 at 11:11

Alberto

13.1k3 gold badges31 silver badges72 bronze badges

1 Comment

Yevhen Horbunkov Over a year ago

This would also mutate array1 behind the scenes, so using .map() rather than .forEach() doesn't really make much sense. If you need to stay immutable, you would need to do Object.assign({}, el, array2.find..

sumit · Accepted Answer · 2020-07-08 11:38:11Z

0

If index are same you can map and spread the object like below

const array1 = [{id:1, location: 'A'},{id:2, location: 'B'},{id:3, location: 'C'},{id:4, location: 'D'}];

const array2 = [{id:1, value: 123},{id:2, value: 5466},{id:3, value: 89484},{id:4, value: -4.215}];

const merged = array1.map((a,i)=>({...a, ...array2[i]}));
 
console.log(merged);

If the index are not same then you can do like below

const array1 = [{id:1, location: 'A'},{id:2, location: 'B'},{id:3, location: 'C'},{id:4, location: 'D'}];

const array2 = [{id:1, value: 123},{id:2, value: 5466},{id:3, value: 89484},{id:4, value: -4.215}];
  
const merged = array1.map(a=>({...a, ...array2.find(a2=>a2.id==a.id)}));

console.log(merged);

edited Jul 8, 2020 at 11:38

answered Jul 8, 2020 at 11:24

sumit

15.5k14 gold badges75 silver badges113 bronze badges

Collectives™ on Stack Overflow

Move object value from one array of objects to another

4 Answers 4

Comments

1 Comment

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related