Get all non-unique values (i.e.: duplicate/more th

Question

I need to check a JavaScript array to see if there are any duplicate values. What's the easiest way to do this? I just need to find what the duplicated values are - I don't actually need their indexes or how many times they are duplicated.

I know I can loop through the array and check all the other values for a match, but it seems like there should be an easier way. Any ideas? Thanks!

Similar question:

• Get all unique values in an array (remove duplicates)

Answer 1

var a = ["a","a","b","c","c"];

a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})

Answer 2

Simple code with ES6 syntax (return sorted array of duplicates):

let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};

How to use:

duplicates([1,2,3,10,10,2,3,3,10]);

Answer 3

Find duplicate values in an array

This should be one of the shortest ways to actually find duplicate values in an array. As specifically asked for by the OP, this does not remove duplicates but finds them.

var input = [1, 2, 3, 1, 3, 1];

var duplicates = input.reduce(function(acc, el, i, arr) {
  if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);

document.write(duplicates); // = 1,3 (actual array == [1, 3])

This doesn't need sorting or any third party framework. It also doesn't need manual loops. It works with every value indexOf() (or to be clearer: the strict comparision operator) supports.

Because of reduce() and indexOf() it needs at least IE 9.

Answer 4

I think the below is the easiest and fastest O(n) way to accomplish exactly what you asked:

function getDuplicates( arr ) {
  var i, value;
  var all = {};
  var duplicates = [];

  for( i=0; i<arr.length; i++ ) {
    value = arr[i];
    if( all[value] ) {
      duplicates.push( value );
      all[value] = false;
    } else if( typeof all[value] == "undefined" ) {
      all[value] = true;
    }
  }

  return duplicates;
}

Or for ES5 or greater:

function getDuplicates( arr ) {
  var all = {};
  return arr.reduce(function( duplicates, value ) {
    if( all[value] ) {
      duplicates.push(value);
      all[value] = false;
    } else if( typeof all[value] == "undefined" ) {
      all[value] = true;
    }
    return duplicates;
  }, []);
}

Answer 5

var a = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1].sort();
a.filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});

or when added to the prototyp.chain of Array

//copy and paste: without error handling
Array.prototype.unique = 
   function(){return this.sort().filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});}

See here: https://gist.github.com/1305056

Answer 6

The following function (a variation of the eliminateDuplicates function already mentioned) seems to do the trick, returning test2,1,7,5 for the input ["test", "test2", "test2", 1, 1, 1, 2, 3, 4, 5, 6, 7, 7, 10, 22, 43, 1, 5, 8]

Note that the problem is stranger in JavaScript than in most other languages, because a JavaScript array can hold just about anything. Note that solutions that use sorting might need to provide an appropriate sorting function--I haven't tried that route yet.

This particular implementation works for (at least) strings and numbers.

function findDuplicates(arr) {
    var i,
        len=arr.length,
        out=[],
        obj={};

    for (i=0;i<len;i++) {
        if (obj[arr[i]] != null) {
            if (!obj[arr[i]]) {
                out.push(arr[i]);
                obj[arr[i]] = 1;
            }
        } else {
            obj[arr[i]] = 0;            
        }
    }
    return out;
}