Remove all unique values and keep all duplicates in an array

April 3, 2023

I have an object array and I want to filter it so that it only contains duplicates (by duplicates I mean objects that have the same value for the key number) and remove objects that have a unique number value. I want to keep all duplicate objects (not just identify what the duplicate values are!). The duplicates will always be in consecutive order.

My array:

const array = [
  {
   number: 1,
   color: "red"
  },
  {
   number: 2,
   color: "blue"
  },
  {
   number: 2,
   color: "yellow"
  },
  {
   number: 3,
   color: "black"
  },
  {
   number: 3,
   color: "orange"
  },
  {
   number: 4,
   color : "white"
  }
]

Expected output :

[
  {
   number: 2,
   color: "blue"
  },
  {
   number: 2,
   color: "yellow"
  },
  {
   number: 3,
   color: "black"
  },
  {
   number: 3,
   color: "orange"
  }
]

I have searched high and low but all answers seem to either remove or identify duplicates and since I’m new to javascript I don’t have the skills to modify those codes to work for me. Any help is greatly appreciated!

>Solution :

A straightforward approach is to filter the array and retain all elements for which you can find some element with the same number at a different position in the array:

const result = array.filter(
  (v1, i1, a) => a.some((v2, i2) => v1.number === v2.number && i1 !== i2)
);

Complete snippet:

const array = [
  {
   number: 1,
   color: "red"
  },
  {
   number: 2,
   color: "blue"
  },
  {
   number: 2,
   color: "yellow"
  },
  {
   number: 3,
   color: "black"
  },
  {
   number: 3,
   color: "orange"
  },
  {
   number: 4,
   color : "white"
  }
];

const result = array.filter(
  (v1, i1, a) => a.some((v2, i2) => v1.number === v2.number && i1 !== i2)
);

console.log(result);