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.

CodePudding user response：

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);

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CodePudding user response：

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);

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This will do

CodePudding user response：

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);

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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.

CodePudding user response：

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);

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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);

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CodePudding user response：

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);

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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.