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