Remove all items that have duplicates in array Javascript

Question

I have searched on here and have not found a solution. Obviously I will be corrected if I am wrong. What I am trying to do is return values that do not have a duplicates in an array.

Examples:

myArr = [2,1,2,3] // answer [1,3]

myArr = [3,1,2,2,3] // answer [1]

I would post some code but I have not been able to figure this out myself and the only code examples I have found are for removing any duplicate values.

The possible solution above is to return no duplicates... I am trying to return values that are don't have duplicates.

possible duplicate: http://stackoverflow.com/questions/1960473/unique-values-in-an-array — 42shadow42, May 13 '16 at 22:35
@42shadow42: This is not about unique, though; see the examples — David Hedlund, May 13 '16 at 22:35
Oh my mistake, you want all copies of the duplicate removed not all but one thanks for correcting me. — 42shadow42, May 13 '16 at 22:36
`myArr.filter((x, i, a) => a.indexOf(x) === a.lastIndexOf(x))` should do it — Bergi, May 13 '16 at 22:40
no it does not. @Bergi has a ES6 solution, I do believe... Maybe an ES5 solution would help me more.. — Chuck Reynolds, May 13 '16 at 22:42
@c.reynolds: The arrow function was only for brevity, you can do the same with a function expression — Bergi, May 13 '16 at 22:46
You can avoid a good bit of the index searching if you only search forward from `i + 1` and track the failures in an object so that you don't search those at all. Like this: https://jsfiddle.net/36ezypzw/ — , May 13 '16 at 22:58

score 4 · Accepted Answer · answered May 13 '16 at 22:42

One option is to use the optional second argument to indexOf to find duplicate indexes. Consider that for a given element e and an index i:

if e is the first of two identical elements in the array, indexOf(e) will return i and indexOf(e, i + 1) will return the index of the second element.
if e is the second of two identical elements in the array, indexOf(e) will return the index of the first element, and indexOf(e, i + 1) will return -1
if e is a unique element, indexOf(e) will return i and indexOf(e, i + 1) will return -1.

Therefore:

myArr.filter(function (e, i, a) {
  return a.indexOf(e) === i && a.indexOf(e, i + 1) === -1
});

code_monk · Answer 2 · 2016-05-13T22:47:36.223

2

var isUnique = function(v,i,arr){
   // return true if the first occurrence is the last occurrence 
   return ( arr.indexOf(v) === arr.lastIndexOf(v) ); 
};

var uniqueVals = myArr.filter(isUnique);

console.log( uniqueVals );

edited May 13 '16 at 22:47

answered May 13 '16 at 22:44

code_monk

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score 0 · Answer 3 · answered May 13 '16 at 23:09

If is not an associative array (your case):

var myArr = [1,2,2,3,4,4,1,5];
var myNewArr = [];

if (myArr.length > 0 )
{
    myNewArr[0] = myArr[myArr.length-1];    
}

var count = 1;

myArr.sort();

for (var i = myArr.length - 2; i >= 0; i--) {

    if(myArr[i] != myArr[i-1])
    {
        myNewArr[count] = myArr[i];
        count++;
    }
}

Harsha pps · Answer 4 · 2016-11-24T07:06:32.313

0

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

        var uniqueValues = [];
        $.each(yourArray, function (i, value) { //taking each 'value' from yourArray[]
            if ($.inArray(value, uniqueValues) === -1) { 
                 uniqueValues.push(value); // Pushing the non - duplicate value into the uniqueValues[] 
            }
        });            
        console.log(uniqueValues);

Result: [1,2,3];

edited Nov 24 '16 at 07:06

answered Nov 22 '16 at 13:05

Harsha pps

2,012
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Remove all items that have duplicates in array Javascript

4 Answers4