How can I identify and remove duplicate elements from an array of objects?

Question

How can I identify and remove duplicate elements from an array of objects?

elements= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];

I want to find elements with identical name and age. How can I achieve this?

I have attempted the following but it is returning all elements in the array:

data.some(d => {
for (const it of form) {
if (it.name == d.name && it.age == d.age) {
arr.push(it)....

arrays angular typescript

Answer 1

Answer №1

Ensuring that you are not comparing the same item with matching properties is crucial:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];

const res = data.filter(item => data.some(i => i !== item && item.name === i.name && item.age === i.age))
console.log(res)

If excluding the first occurrence is desired, checking against previously processed elements is necessary:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];


const [result, _] = data.reduce(([duplicates, checked], el) => {
    checked.some(i => el.name === i.name && el.age === i.age) && duplicates.push(el)
    return [duplicates, [...checked, el]]
  }, [[], []]
)
console.log(result)

Sort-based strategies prove to be highly effective for such scenarios:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];
const sorted = data.sort( (e1, e2) => e1.name.localeCompare(e2.name) || e1.age - e2.age)
const result = sorted.filter( (el, i) => i > 0 && sorted[i-1].name === el.name && sorted[i-1].age === el.age)
console.log(result)

Answer 2

Ensuring that you are not comparing the same item with matching properties is crucial:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];

const res = data.filter(item => data.some(i => i !== item && item.name === i.name && item.age === i.age))
console.log(res)

If excluding the first occurrence is desired, checking against previously processed elements is necessary:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];


const [result, _] = data.reduce(([duplicates, checked], el) => {
    checked.some(i => el.name === i.name && el.age === i.age) && duplicates.push(el)
    return [duplicates, [...checked, el]]
  }, [[], []]
)
console.log(result)

Sort-based strategies prove to be highly effective for such scenarios:

data= [
  { "id": 0, "name": "name1", "age": 12, "city": "cityA" },
  { "id": 1, "name": "name2", "age": 7, "city": "cityC" },
  { "id": 2, "name": "name3", "age": 23, "city": "cityB" },
  { "id": 2, "name": "name1", "age": 12, "city": "cityA" }
];
const sorted = data.sort( (e1, e2) => e1.name.localeCompare(e2.name) || e1.age - e2.age)
const result = sorted.filter( (el, i) => i > 0 && sorted[i-1].name === el.name && sorted[i-1].age === el.age)
console.log(result)

How can I identify and remove duplicate elements from an array of objects?

Answer №1

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