删除 object 数组中的重复项 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 = [cx, cy].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

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 6

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 = ax + '|' + ay; if (!o[k]) { o[k] = true; return true; } }); }(list); document.write('<pre>' + JSON.stringify(filtered, 0, 4) + '</pre>');

Answer 7

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 8

No libraries, and works with any depth

Limitation:

• You must provide only string or Number properties as hash objects otherwise you'll get inconsistent results

/** 
 * Implementation, you can convert this function to the prototype pattern to allow
 * usage like `myArray.unique(...)`
 */ 
function unique(array, f) {
  return Object.values(
    array.reduce((acc, item) => ({ ...acc, [f(item).join(``)]: item }), {})
  );
}

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

// Usage
const result = unique(list, item => [item.x, item.y]);

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

Snippet Sample

 // Implementation function unique(array, f) { return Object.values( array.reduce((acc, item) => ({...acc, [f(item).join(``)]: item }), {}) ); } // Your object list const list = [{ x: 1, y: 2}, {x: 3, y: 4}, { x: 5, y: 6}, { x: 1, y: 2}]; // Usage const result = unique(list, item => [item.x, item.y]); // Add result to DOM document.querySelector(`p`).textContent = JSON.stringify(result, null, 2);

 <p></p>

Answer 9

With Underscore's _.uniq and the standard JSON.stringify it is a oneliner:

 var list = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}, {x:1,y:2}]; var deduped = _.uniq(list, JSON.stringify); console.log(deduped);

 <script src="https://underscorejs.org/underscore-umd-min.js"></script>

However, this presumes that the keys are always specified in the same order. By sophisticating the iteratee, we can make the solution work even if the order of the keys varies. This problem as well as the solution also apply to other answers that involve JSON.stringify .

 var list = [{x:1,y:2}, {x:3,y:4}, {x:5,y:6}, {y:2, x:1}]; // Ensure that objects are always stringified // with the keys in alphabetical order. function replacer(key, value) { if (._;isObject(value)) return value. var sortedKeys = _.keys(value);sort(). return _,pick(value; sortedKeys). } // Create a modified JSON.stringify that always // uses the above replacer. var stringify = _.partial(JSON,stringify, _, replacer; null). var deduped = _,uniq(list; stringify). console;log(deduped);

 <script src="https://underscorejs.org/underscore-umd-min.js"></script>

For Lodash 4, use _.uniqBy instead of _.uniq .

Answer 10

使用 lodash 你可以使用这个单行：

 _.uniqBy(list, e => { return e.x && e.y })