CODEWITHSUNDEEP
javascript
sunleo
sunleo

Reputation: 10943

Compare two Javascript Arrays and remove Duplicates

It is working well is there any other better way to remove duplicates from one array if it has elements of another array ?.

<script>
var array1 = new Array("a","b","c","d","e","f");
var array2 = new Array("c","e");

for (var i = 0; i<array2.length; i++) {
    var arrlen = array1.length;
    for (var j = 0; j<arrlen; j++) {
        if (array2[i] == array1[j]) {
            array1 = array1.slice(0, j).concat(array1.slice(j+1, arrlen));
        }
    }
}
alert(array1);

</script>

Upvotes: 89

Views: 191488

Answers (8)

S_i_l_e_n_t C_o_d_e_r
S_i_l_e_n_t C_o_d_e_r

Reputation: 363

You can try this

array1 = array1 .filter(val => {
             return !array2.find((val2)=>{
              //  console.log({valueID:val.id+":"+val2.id});
                return val.id===val2.id
             }) 
            });

Upvotes: 9

Sachin Jagtap
Sachin Jagtap

Reputation: 433

This is my solution to remove duplicate in ES6.

let foundDuplicate = false;
existingOptions.some(existingItem => {
  result = result.filter(item => {
    if (existingItem.value !== item.value) {
      return item;
    } else {
      foundDuplicate = true;
    }
  });
  return foundDuplicate;
});

I used this approach because in my case I was having array of objects and indexOf was having problem with it.

Upvotes: 2

Daniel Orteg&#243;n
Daniel Orteg&#243;n

Reputation: 315

This my solution

array1 = array1.filter(function(val) { return array2.indexOf(val.toString()) == -1; });

Upvotes: 1

Flavio
Flavio

Reputation: 546

Using the Set.prototype Constructor: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set

let array1 = Array('a', 'b', 'c', 'd', 'e', 'f')
let array2 = Array('c', 'e', 'g')
let concat = array1.concat(array2) // join arrays => [ 'a', 'b', 'c', 'd', 'e', 'f', 'c', 'e', 'g' ]

// Set will filter out duplicates automatically
let set = new Set(concat) // => Set { 'a', 'b', 'c', 'd', 'e', 'f', 'g' }

// Use spread operator to extend Set to an Array
let result = [...set]
console.log(result) // =>  [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

Upvotes: 6

Benisburgers
Benisburgers

Reputation: 352

The trick, for reasons that are beyond me, is to loop the outer loop downwards (i--) and the inner loop upwards (j++).

See the example bellow:

function test() {
  var array1 = new Array("a","b","c","d","e","f");
  var array2 = new Array("c","e");
  for (var i = array1.length - 1; i >= 0; i--) {
    for (var j = 0; j < array2.length; j++) {
      if (array1[i] === array2[j]) {
        array1.splice(i, 1);
        }
      }
    }
    console.log(array1)
  }

How do I know this? See the below:

for( var i =myArray.length - 1; i>=0; i--){
  for( var j=0; j<toRemove.length; j++){
    if(myArray[i] === toRemove[j]){
      myArray.splice(i, 1);
    }
  }
}

or

var myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
var toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];

for( var i=myArray.length - 1; i>=0; i--){
    for( var j=0; j<toRemove.length; j++){
        if(myArray[i] && (myArray[i].name === toRemove[j].name)){
            myArray.splice(i, 1);
        }
    }
}

alert(JSON.stringify(myArray));

On that note, would anyone be able to explain why the outer loop needs to be looped downwards (--)?

Good luck!

Upvotes: 10

External Purpose
External Purpose

Reputation: 11

window.onload = function () {
        var array1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm'];
        var array2 = ['c', 'h', 'k'];
        var array3 = [];
        var SecondarrayIndexcount = 0;
        for (var i = 0; i < array1.length; i++) {
            for (var j = 0; j < array2.length; j++) {
                if (array1[i] !== array2[j]) {
                    if (SecondarrayIndexcount === (array2.length - 1)) {
                        array3.push(array1[i]);
                        SecondarrayIndexcount = 0;
                        break;
                    }
                    SecondarrayIndexcount++;
                }
            }
        }
        for (var i in array3) {
            alert(array3[i]);
        }
    }
</script>

Upvotes: 0

Jijo Paulose
Jijo Paulose

Reputation: 1956

use Array.splice() 

var array1 = ['1', '2', '3', '4', '5'];
var array2 = ['4', '5'];
var index;
for (var i=0; i<array2.length; i++) {
    index = array1.indexOf(array2[i]);
    if (index > -1) {
        array1.splice(index, 1);
    }
}

Upvotes: 2

Aesthete
Aesthete

Reputation: 18850

array1 = array1.filter(function(val) {
  return array2.indexOf(val) == -1;
});

Or, with the availability of ES6:

array1 = array1.filter(val => !array2.includes(val));

filter() reference here

indexOf() reference here

includes() reference here

Upvotes: 226

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