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I have an array like

const arr = [
{id: '1', name: 'name 1', age: 20},
{id: '2', name: 'name 2', age: 21},
{id: '2', name: 'name 3', age: 22},
{id: '2', name: 'name 4', age: 23},
{id: '3', name: 'name 5', age: 24},
{id: '3', name: 'name 6', age: 25},
{id: '4', name: 'name 7', age: 26},
{id: '5', name: 'name 8', age: 27},
{id: '5', name: 'name 9', age: 28},
{id: '5', name: 'name 10', age: 29},
];

I know I have duplicates(looking only id property) in the array and I don't want to remove duplicates instead, I need to append something to differentiate like.

arr = [
{id: '1_0', name: 'name 1', age: 20},
{id: '2_0', name: 'name 2', age: 21},
{id: '2_1', name: 'name 3', age: 22},
{id: '2_2', name: 'name 4', age: 23},
{id: '3_0', name: 'name 5', age: 24},
{id: '3_1', name: 'name 6', age: 25},
{id: '4_0', name: 'name 7', age: 26},
{id: '5_0', name: 'name 8', age: 27},
{id: '5_1', name: 'name 9', age: 28},
{id: '5_2', name: 'name 10', age: 29},
];

Thanks in advance.

I have an array like

const arr = [
{id: '1', name: 'name 1', age: 20},
{id: '2', name: 'name 2', age: 21},
{id: '2', name: 'name 3', age: 22},
{id: '2', name: 'name 4', age: 23},
{id: '3', name: 'name 5', age: 24},
{id: '3', name: 'name 6', age: 25},
{id: '4', name: 'name 7', age: 26},
{id: '5', name: 'name 8', age: 27},
{id: '5', name: 'name 9', age: 28},
{id: '5', name: 'name 10', age: 29},
];

I know I have duplicates(looking only id property) in the array and I don't want to remove duplicates instead, I need to append something to differentiate like.

arr = [
{id: '1_0', name: 'name 1', age: 20},
{id: '2_0', name: 'name 2', age: 21},
{id: '2_1', name: 'name 3', age: 22},
{id: '2_2', name: 'name 4', age: 23},
{id: '3_0', name: 'name 5', age: 24},
{id: '3_1', name: 'name 6', age: 25},
{id: '4_0', name: 'name 7', age: 26},
{id: '5_0', name: 'name 8', age: 27},
{id: '5_1', name: 'name 9', age: 28},
{id: '5_2', name: 'name 10', age: 29},
];

Thanks in advance.

• javascript
• arrays
• typescript
• ecmascript-6

Share Improve this question Follow edited Nov 8, 2021 at 14:55 Arjun Kumar asked Nov 8, 2021 at 14:49 Arjun KumarArjun Kumar 4177 bronze badges 7

• 2_1 is just 21, no need for underscores – MrMythical Commented Nov 8, 2021 at 14:51
• 3 @MrMythical what is 210 then? Is it ID 21, instance zero, or is it ID 2 ,instance ten? – VLAZ Commented Nov 8, 2021 at 14:51
• @VLAZ well JavaScript pretty much ignores the underscore. It would be better to use strings for this – MrMythical Commented Nov 8, 2021 at 14:53
• 1 @MrMythical OK, I missed that the ID wasn't a string. I assume OP did mean to use strings, though, given the phrasing for "append". (also, IDs shouldn't be given to consumers as numbers but that's a separate gripe of mine). – VLAZ Commented Nov 8, 2021 at 14:54
• sorry, the property 'id' value should be a string updated the question – Arjun Kumar Commented Nov 8, 2021 at 14:57

 |  Show 2 more ments

5 Answers 5

Sorted by: Reset to default 4

1) You can easily achieve the result using Map

const arr = [
  { id: "1", name: "name 1", age: 20 },
  { id: "2", name: "name 2", age: 21 },
  { id: "2", name: "name 3", age: 22 },
  { id: "2", name: "name 4", age: 23 },
  { id: "3", name: "name 5", age: 24 },
  { id: "3", name: "name 6", age: 25 },
  { id: "4", name: "name 7", age: 26 },
  { id: "5", name: "name 8", age: 27 },
  { id: "5", name: "name 9", age: 28 },
  { id: "5", name: "name 10", age: 29 },
];

const map = new Map();
arr.forEach((o) => map.has(o.id) ? map.get(o.id).push(o) : map.set(o.id, [o]));

const result = [];
for (let [, arr] of map) {
  arr.forEach((o, i) => result.push({ ...o, id: `${o.id}_${i}` }));
}

console.log(result);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
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2) You can also use here map with Map (Thank to VLAZ)

const arr = [
  { id: "1", name: "name 1", age: 20 },
  { id: "2", name: "name 2", age: 21 },
  { id: "2", name: "name 3", age: 22 },
  { id: "2", name: "name 4", age: 23 },
  { id: "3", name: "name 5", age: 24 },
  { id: "3", name: "name 6", age: 25 },
  { id: "4", name: "name 7", age: 26 },
  { id: "5", name: "name 8", age: 27 },
  { id: "5", name: "name 9", age: 28 },
  { id: "5", name: "name 10", age: 29 },
];

const map = new Map();
const result = arr.map((curr) => {
  map.set(curr.id, (map.get(curr.id) ?? -1) + 1);
  return { ...curr, id: `${curr.id}_${map.get(curr.id)}` };
});

console.log(result);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

You could take an object for counting and map new objects with subversions.

This answer features a closure

(ids => o => ({ ...o, id: `${o.id}_${ids[o.id] ??= 0, ids[o.id]++}` }))
({})

where the function is called with an object and returns another function

        o => ({ ...o, id: `${o.id}_${ids[o.id] ??= 0, ids[o.id]++}` })

for mapping.

The mapping function returns a new object with a replacement of id property with part of the old id and a new value which starts with zero for any unseen id.

const
    array = [{ id: '1', name: 'name 1', age: 20 }, { id: '2', name: 'name 2', age: 21 }, { id: '2', name: 'name 3', age: 22 }, { id: '2', name: 'name 4', age: 23 }, { id: '3', name: 'name 5', age: 24 }, { id: '3', name: 'name 6', age: 25 }, { id: '4', name: 'name 7', age: 26 }, { id: '5', name: 'name 8', age: 27 }, { id: '5', name: 'name 9', age: 28 }, { id: '5', name: 'name 10', age: 29 }],
    result = array.map(
        (ids => o => ({ ...o, id: `${o.id}_${ids[o.id] ??= 0, ids[o.id]++}` }))
        ({})
    );

console.log(result);

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

const arr = [
  { id: "1", name: "name 1", age: 20 },
  { id: "2", name: "name 2", age: 21 },
  { id: "2", name: "name 3", age: 22 },
  { id: "2", name: "name 4", age: 23 },
  { id: "3", name: "name 5", age: 24 },
  { id: "3", name: "name 6", age: 25 },
  { id: "4", name: "name 7", age: 26 },
  { id: "5", name: "name 8", age: 27 },
  { id: "5", name: "name 9", age: 28 },
  { id: "5", name: "name 10", age: 29 },
];

let obj = {};

for (let i = 0; i < arr.length; i++) {
  if (obj[arr[i].id] == undefined) {
    obj[arr[i].id] = 0;
  } else {
    obj[arr[i].id] = obj[arr[i].id] + 1;
  }
  arr[i].id = arr[i].id + "_" + obj[arr[i].id];
}
console.log(arr);

This will do

Single iteration needed:

const arr = [
  {id: '1', name: 'name 1', age: 20},
  {id: '2', name: 'name 2', age: 21},
  {id: '2', name: 'name 3', age: 22},
  {id: '2', name: 'name 4', age: 23},
  {id: '3', name: 'name 5', age: 24},
  {id: '3', name: 'name 6', age: 25},
  {id: '4', name: 'name 7', age: 26},
  {id: '5', name: 'name 8', age: 27},
  {id: '5', name: 'name 9', age: 28},
  {id: '5', name: 'name 10', age: 29},
];

//keep track of how many times each id has been encountered
const counterPerRecord = new Map();

for (const record of arr) {
  //get count or default to zero
  const next = counterPerRecord.get(record.id) ?? 0;
  //update the count for next time we encounter the id
  counterPerRecord.set(record.id, next+1);
  
  //update the id
  record.id += `_${next}`;
}

console.log(arr);

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

You only need a map to keep track of what the next instance for the next number to append to each id property. The first time it is zero, and it is increased every time you encounter the same id.

You can use somthing like this:

const arr = [
  { id: "1", name: "name 1", age: 20 },
  { id: "2", name: "name 2", age: 21 },
  { id: "2", name: "name 3", age: 22 },
  { id: "2", name: "name 4", age: 23 },
  { id: "3", name: "name 5", age: 24 },
  { id: "3", name: "name 6", age: 25 },
  { id: "4", name: "name 7", age: 26 },
  { id: "5", name: "name 8", age: 27 },
  { id: "5", name: "name 9", age: 28 },
  { id: "5", name: "name 10", age: 29 },
];

for (let i = arr.length - 1; i >= 0; i--) {
   arr[i].id = `${arr[i].id}_${arr.filter(el => el.id === arr[i].id).length - 1}`;
}

console.log(arr);

Notice: as @MrMythical mentioned, 1_0 is just 10, 1_1 is 11 and so on. So I transform ids into strings in example to avoid id duplications.

Warning: as @VLAZ and @Jeremy Thille noticed, this is expensive solution (O(n^2)) which is not acceptable for large data. For well-optimized solution (O(n)), use Map.

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