Updating objects in an array (matching by value)

Question

Have first array (a) and second with updated values (b). Need to update matching by IDs objects and get result array (c). How can I make it simple and fast?

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ]
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
  ]
//Result array:
c = [ 
  { id: 1, activated: '0' },
  { id: 2, activated: '1' },
  { id: 3, activated: '1' },
  { id: 4, activated: '0' },
]

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– SherylHohman, Commented Aug 12, 2020 at 7:36

Ele · Accepted Answer · 2020-08-11 21:45:39Z

2

I recommend you to create an object where keys are the ids for fast access.

With the newly created object you can access the specific objects from b using the id as key and update the objects.

Finally, get the array c using the function Object.values.

This is assuming that b has ids that exist in a.

let a = [    { id: 1, activated: '0' },    { id: 2, activated: '0' },    { id: 3, activated: '0' },    { id: 4, activated: '0' }]
let b = [    { id: 2, activated: '1' },    { id: 3, activated: '1' }];
let newA = a.reduce((a, {id, ...rest}) => ({...a, ...{[id]: {id, ...rest}}}), {});

b.forEach(({id, activated}) => newA[id].activated = activated);

let c = Object.values(newA);
console.log(c);

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

answered Aug 11, 2020 at 21:45

Ele

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2 Comments

Максим Юрченко Over a year ago

Great, Thank you!

Ele Over a year ago

@МаксимЮрченко yw!

kimlianlopez · Accepted Answer · 2020-08-11 22:10:55Z

1

You can use Array.map, then foreach object, you can use Array.find to check if it exist in arrayB by ID and return if it does.

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
];

let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
];

const c = a.map(e => {
    let newValue = b.find(n => e.id === n.id);

    if (!!newValue) {
        return newValue;
    }

    return e;
});

answered Aug 11, 2020 at 22:10

kimlianlopez

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Comments

B45i · Accepted Answer · 2020-08-11 22:11:24Z

1

This is how I would approach it:

function mergeArray(...toMerge) {
  let output = {};
  toMerge.forEach(arr => {
    arr.forEach(item => {
      output[item.id] = item;
    });
  });
  return Object.values(output);
}

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ];
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
  ];

console.dir(mergeArray(a, b)) // can merge n-number of arrays.

answered Aug 11, 2020 at 22:11

B45i

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Comments

jamomani · Accepted Answer · 2020-08-11 21:52:27Z

0

let c = a.map( aObj => {
  let found = b.find(bObj => bObj.id === aObj.id);
    if(found) return found;
    else return aObj;
});

answered Aug 11, 2020 at 21:52

jamomani

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Comments

trincot · Accepted Answer · 2020-08-11 21:54:59Z

0

You could create a map of the second array, keyed by id, and then when you make a copy of a objects into c, use that map to merge matching b objects into them.

let a = [{ id: 1, activated: '0' },{ id: 2, activated: '0' },{ id: 3, activated: '0' },{ id: 4, activated: '0' }]
let b = [{ id: 2, activated: '1' },{ id: 3, activated: '1' }];

let map = new Map(b.map(o => [o.id, o]));
let c = a.map(o => ({...o, ...map.get(o.id)}));

console.log(c);

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answered Aug 11, 2020 at 21:54

trincot

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Comments

Sascha A. · Accepted Answer · 2020-08-11 22:05:21Z

Building a hash from the ids of array a. Iterate over array b by looking in the hash for the index b.id, if found actualisize in a with the founded index the property activated. Otherwise (in example id: 5) there don't exists in the old array a for this id an entry so create a new one and actualisize the hash.

let a = [
    { id: 1, activated: '0' },
    { id: 2, activated: '0' },
    { id: 3, activated: '0' },
    { id: 4, activated: '0' },
  ];
let b = [
    { id: 2, activated: '1' },
    { id: 3, activated: '1' },
    { id: 5, activated: '0' },
  ];

let hash = a.flatMap(el => el.id);

b.forEach(elem => {
    index = hash.indexOf(elem.id);
    if (index!=-1) 
        a[index].activated=elem.activated;
     else {
        a.push(elem);
        hash.push(elem.id);
     }
});
  
console.log(a);

Varun Suresh Kumar · Accepted Answer · 2020-08-11 21:42:15Z

-1

Try this

let a = [{
    id: 1,
    activated: '0'
  },
  {
    id: 2,
    activated: '0'
  },
  {
    id: 3,
    activated: '0'
  },
  {
    id: 4,
    activated: '0'
  },
]
let b = [{
    id: 2,
    activated: '1'
  },
  {
    id: 3,
    activated: '1'
  },
]
//Result array:
c = [{
    id: 1,
    activated: '0'
  },
  {
    id: 2,
    activated: '1'
  },
  {
    id: 3,
    activated: '1'
  },
  {
    id: 4,
    activated: '0'
  },
]

let result = a.map(e => {
  let p = b.find(be => be.id === e.id)
  if (p) e.activated = p.activated;
  return e
})

console.log(result)

answered Aug 11, 2020 at 21:42

Varun Suresh Kumar

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2 Comments

Varun Suresh Kumar Over a year ago

can anybody explain why this would be not ideal solution? im curious to know about perf optimisation.

trincot Over a year ago

Not my downvote, but find represents a loop, and so you have a nested loop giving a total time complexity of O(n²), while this can be done in linear time.

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