return duplicates in a new array

Question

I need to return duplicates in a new array and get this output: [1, 2, 5] but get two times 1. why?

const array = [1, 2, 3, 1, 5, 6, 1, 2, 5];
let newArrFiltered = array.filter(
  (item, index) => array.indexOf(item) !== index
);
console.log(newArrFiltered); // [1, 1, 2, 5]

Note that`filter` returns a new array. In this instance making a copy of the array is unnecessary. — Andy, Jul 22 '23 at 13:28
Have you debugged this? There are exactly two `1` in your input array for which `array.indexOf(item) !== index` is true. Why does this surprise you? Just check what the value is of `array.indexOf(1)` and then check what `index` is for each occurrence of 1... — trincot, Jul 22 '23 at 13:28
For how to use `indexOf` for this purpose, see [here](https://stackoverflow.com/a/8315486/5459839). — trincot, Jul 22 '23 at 13:36

score 0 · Answer 1 · answered Jul 22 '23 at 14:04

Maybe like this:

var array = [1,2,3,1,5,6,1,2,5];
var test_arr = [];
var exist_arr = [];
for(var key in array){ 
   if(test_arr.indexOf(array[key]) === -1){
        test_arr.push(array[key]);
   }else{
       if(exist_arr.indexOf(array[key]) === -1){
           exist_arr.push(array[key]);
       }
   }

}

console.log(exist_arr);

return duplicates in a new array

1 Answers1