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

Using "includes" to test if the element already exists.

var arr = [1, 1, 4, 5, 5], darr = [], duplicates = [];

for(var i = 0; i < arr.length; i++){
  if(darr.includes(arr[i]) && !duplicates.includes(arr[i]))
    duplicates.push(arr[i])
  else
    darr.push(arr[i]);
}

console.log(duplicates);

<h3>Array with duplicates</h3>
<p>[1, 1, 4, 5, 5]</p>
<h3>Array with distinct elements</h3>
<p>[1, 4, 5]</p>
<h3>duplicate values are</h3>
<p>[1, 5]</p>

Answer 2

I have just figured out a simple way to achieve this using an Array filter

    var list = [9, 9, 111, 2, 3, 4, 4, 5, 7];
    
    // Filter 1: to find all duplicates elements
    var duplicates = list.filter(function(value,index,self) {
       return self.indexOf(value) !== self.lastIndexOf(value) && self.indexOf(value) === index;
    });
    
    console.log(duplicates);

Answer 3

Just to add some theory to the above.

Finding duplicates has a lower bound of O(n*log(n) in the comparison model. SO theoretically, you cannot do any better than first sorting then going through the list sequentially removing any duplicates you find.

If you want to find the duplicates in linear (O(n)) expected time, you could hash each element of the list; if there is a collision, remove/label it as a duplicate, and continue.

Answer 4

To solve the above in O(n) time complexity (without sorting).

var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];

var obj={};

for(var i=0;i<arr.length;i++){
    if(!obj[arr[i]]){
        obj[arr[i]]=1;
    } else {
        obj[arr[i]]=obj[arr[i]]+1;
    }
}
var result=[]
for(var key in obj){
    if(obj[key]>1){
        result.push(Number(key)) // change this to result.push(key) to find duplicate strings in an array
    }
}

console.log(result)

Answer 5

var arr = [2, 1, 2, 2, 4, 4, 2, 5];

function returnDuplicates(arr) {
  return arr.reduce(function(dupes, val, i) {
    if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
      dupes.push(val);
    }
    return dupes;
  }, []);
}

alert(returnDuplicates(arr));

This function avoids the sorting step and uses the reduce() method to push duplicates to a new array if it doesn't already exist in it.

Answer 6

You can add this function, or tweak it and add it to Javascript's Array prototype:

Array.prototype.unique = function () {
    var r = new Array();
    o:for(var i = 0, n = this.length; i < n; i++)
    {
        for(var x = 0, y = r.length; x < y; x++)
        {
            if(r[x]==this[i])
            {
                alert('this is a DUPE!');
                continue o;
            }
        }
        r[r.length] = this[i];
    }
    return r;
}

var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);