Remove duplicates in an object array Javascript

Question

I have an array of objects

list = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}, {x:1,y:2}]

And I'm looking for an efficient way (if possible O(log(n))) to remove duplicates and to end up with

list = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}]

I've tried _.uniq or even _.contains but couldn't find a satisfying solution.

Thanks!

Edit : The question has been identified as a duplicate of another one. I saw this question before posting but it didn't answer my question since it's an array of object (and not a 2-dim array, thanks Aaron), or at least the solutions on the other question weren't working in my case.

Answer 1

Plain javascript (ES2015), using Set

const list = [{ x: 1, y: 2 }, { x: 3, y: 4 }, { x: 5, y: 6 }, { x: 1, y: 2 }];

const uniq = new Set(list.map(e => JSON.stringify(e)));

const res = Array.from(uniq).map(e => JSON.parse(e));

document.write(JSON.stringify(res));

Answer 2

Try using the following:

list = list.filter((elem, index, self) => self.findIndex(
    (t) => {return (t.x === elem.x && t.y === elem.y)}) === index)

Answer 3

Vanilla JS version:

const list = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}, {x:1,y:2}];

function dedupe(arr) {
  return arr.reduce(function(p, c) {

    // create an identifying id from the object values
    var id = [c.x, c.y].join('|');

    // if the id is not found in the temp array
    // add the object to the output array
    // and add the key to the temp array
    if (p.temp.indexOf(id) === -1) {
      p.out.push(c);
      p.temp.push(id);
    }
    return p;

    // return the deduped array
  }, {
    temp: [],
    out: []
  }).out;
}

console.log(dedupe(list));

Answer 4

I would use a combination of Arrayr.prototype.reduce and Arrayr.prototype.some methods with spread operator.

1. Explicit solution. Based on complete knowledge of the array object contains.

list = list.reduce((r, i) => 
  !r.some(j => i.x === j.x && i.y === j.y) ? [...r, i] : r
, [])

Here we have strict limitation on compared objects structure: {x: N, y: M}. And [{x:1, y:2}, {x:1, y:2, z:3}] will be filtered to [{x:1, y:2}].

2. Generic solution, JSON.stringify(). The compared objects could have any number of any properties.

list = list.reduce((r, i) => 
  !r.some(j => JSON.stringify(i) === JSON.stringify(j)) ? [...r, i] : r
, [])

This approach has a limitation on properties order, so [{x:1, y:2}, {y:2, x:1}] won't be filtered.

3. Generic solution, Object.keys(). The order doesn't matter.

list = list.reduce((r, i) => 
  !r.some(j => !Object.keys(i).some(k => i[k] !== j[k])) ? [...r, i] : r
, [])

This approach has another limitation: compared objects must have the same list of keys. So [{x:1, y:2}, {x:1}] would be filtered despite the obvious difference.

4. Generic solution, Object.keys() + .length.

list = list.reduce((r, i) => 
  !r.some(j => Object.keys(i).length === Object.keys(j).length 
    && !Object.keys(i).some(k => i[k] !== j[k])) ? [...r, i] : r
, [])

With the last approach objects are being compared by the number of keys, by keys itself and by key values.

I created a Plunker to play with it.

Answer 5

One liners for ES6+

If you want to find uniq by x and y:

arr.filter((v,i,a)=>a.findIndex(t=>(t.x === v.x && t.y===v.y))===i)

If you want to find uniques by all properties:

arr.filter((v,i,a)=>a.findIndex(t=>(JSON.stringify(t) === JSON.stringify(v)))===i)

Answer 6

The following will work:

var a = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}, {x:1,y:2}];

var b = _.uniq(a, function(v) { 
    return v.x && v.y;
})

console.log(b);  // [ { x: 1, y: 2 }, { x: 3, y: 4 }, { x: 5, y: 6 } ]

Answer 7

Filter the array after checking if already in a temorary object in O(n).

var list = [{ x: 1, y: 2 }, { x: 3, y: 4 }, { x: 5, y: 6 }, { x: 1, y: 2 }],
    filtered = function (array) {
        var o = {};
        return array.filter(function (a) {
            var k = a.x + '|' + a.y;
            if (!o[k]) {
                o[k] = true;
                return true;
            }
        });
    }(list);

document.write('<pre>' + JSON.stringify(filtered, 0, 4) + '</pre>');