javascript array filter duplicates by object key

Question

I have an array of objects in Javascript

I am trying to filter out duplicates by object value

get duplicate
get unique

example arrays:

let array = [
{ 
    id: '001',
    name: 'aaa',
    type_id: 'a111'
},{ 
    id: '002',
    name: 'bbb',
    type_id: 'a111'
},{
    id: '003',
    name: 'ccc',
    type_id: 'a222'
},{
    id: '004',
    name: 'ddd',
    type_id: 'a333'
}];

This is what i want my array to turn out like:

function 1(array, 'type_id')
[{ 
    id: '001',
    name: 'aaa',
    type_id: 'a111'
},{ 
    id: '002',
    name: 'bbb',
    type_id: 'a111'
}];

function 2(array, 'type_id')
[{ 
    id: '003',
    name: 'ccc',
    type_id: 'a222'
},{ 
    id: '004',
    name: 'ddd',
    type_id: 'a333'
}];

Please share your attempt as a working snippet using <> demonstrating where you are stuck. — gurvinder372
– gurvinder372, Commented Feb 7, 2018 at 13:49
Says: "This is what i want my array to turn out like" But gives an example which is not an Array. — Niladri Basu
– Niladri Basu, Commented Feb 7, 2018 at 13:55

Belian · Accepted Answer · 2018-02-07 14:02:49Z

I've written some code to do those things you wanted to achieve:

let array = [{
  id: '001',
  name: 'aaa',
  type_id: 'a111'
}, {
  id: '002',
  name: 'bbb',
  type_id: 'a111'
}, {
  id: '003',
  name: 'ccc',
  type_id: 'a222'
}, {
  id: '004',
  name: 'ddd',
  type_id: 'a333'
}];

var duplicates = [];
var uniques = [];
var valueCounts = {};


function getDuplicatesAndUniques(array, value) {
  for (var i = 0; i < array.length; i++) {
    if (!valueCounts[array[i][value]])
      valueCounts[array[i][value]] = 1;
    else
      valueCounts[array[i][value]]++;
  }

  for (var i = 0; i < array.length; i++) {
    if (valueCounts[array[i][value]] == 1) {
      uniques.push(array[i]);
    } else {
      duplicates.push(array[i]);
    }
  }

}

getDuplicatesAndUniques(array, "type_id");

console.log(duplicates);
console.log(uniques);

https://jsfiddle.net/hbpx08xh/1/

Hope this helps! :)

Best Regards

Nina Scholz · Accepted Answer · 2018-02-15 07:44:53Z

You could take a Map for same type_id and collect the objects and then push the mapped items to either unique or dupes, based of the length of the array.

The line

map.forEach(a => [unique, dupes][a.length - 1 && 1].push(...a));

takes the map and iterates all elements of it and uses a target array of unique and dupes,

                 [unique, dupes]                                 // targets

takes the length of the maped array, for separating the target array

                                [a.length - 1 && 1]              // select target

Cosider this list for getting an index, 0 for unique items and 1 for duplicates. A lenght of zero does not exist, because if a map entry is made, the length is one.

length  length - 1  length - 1 && 1  comment
------  ----------  ---------------  --------
   1         0             0         take `length - 1`, because of falsy 1st operand
   2         1             1         take 2nd operand
   3         2             1         take 2nd operand
   4         3             1         take 2nd operand for all other greater length

Then with spread syntax ..., all items of the array a are used as parameters for Array#push.

                                                   .push(...a)   // push all itmes of array

var array = [{ id: '001', name: 'aaa', type_id: 'a111' }, { id: '002', name: 'bbb', type_id: 'a111' }, { id: '003', name: 'ccc', type_id: 'a222' }, { id: '004', name: 'ddd', type_id: 'a333' }],
    key = 'type_id',
    map = new Map,
    unique = [],
    dupes = [];

array.forEach(o => map.set(o[key], (map.get(o[key]) || []).concat(o)));
map.forEach(a => [unique, dupes][a.length - 1 && 1].push(...a));

console.log(unique);
console.log(dupes);

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

Hi! Can you please add some explanations to your answer? I don't understand what you did here: map.forEach(a => [unique, dupes][a.length - 1 && 1].push(...a));. PS: I'm relatively new to this. Thanks in advance :)

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