Javascript, filtering array with an array

Question

I need to filter an array1, with each element in my array 2. Both arrays can have a random number of elements.

 array1 = [1,2,3,4,5];
 array2 = [1,3,5];
 filteredArray = [];

 array2.forEach(function(x){
   filteredArray = array1.filter(function(y){
     return y !== x;
   });
 });
 return filteredArray;
 //should return -> [2,4]
 // returns [1,2,3,4]

How can I filter an array with all elements in another array?

http://stackoverflow.com/questions/1181575/determine-whether-an-array-contains-a-value - you can use this — RaV, Jul 27 '16 at 14:11

Jaromanda X · Accepted Answer · 2016-07-27T14:16:02.690

2

use arrays indexOf method

var array1 = [1,2,3,4,5];
var array2 = [1,3,5];
var filteredArray = array1.filter(function(x) {
    return array2.indexOf(x) < 0;
});

or, for sexy people, use !~ with indexOf

var array1 = [1,2,3,4,5];
var array2 = [1,3,5];
var filteredArray = array1.filter(function(x) {
    return !~array2.indexOf(x);
});

edited Jul 27 '16 at 14:16

answered Jul 27 '16 at 14:11

Jaromanda X

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`!array2.includes(x)` *(ECMAScript 2016)* – Jul 27 '16 at 14:22
yes, with ecmascript 2016 ... no browser is 100% ecmascript 2015 yet, so, hell, let's just jump to ecmascript 2020 :p – Jaromanda X Jul 27 '16 at 14:24
It's just a method. Patching is easy, even without a transpiler. – Jul 27 '16 at 14:26
Thank you so much! – Mc-Ac Jul 27 '16 at 14:28
@squint - I was kidding around :p – Jaromanda X Jul 27 '16 at 14:28
No problem... just sayin'... ;) – Jul 27 '16 at 14:29
1

in fact, array#filter wont work in IE8 or less ... so all those ATM machines would crash with this code :p – Jaromanda X Jul 27 '16 at 14:30

Tarun Dugar · Answer 2 · 2016-07-27T14:16:44.613

1

A much simpler way would be:

var filteredArray = array1.filter(function(item) {
    return !(array2.indexOf(item) >= 0);
});

edited Jul 27 '16 at 14:16

answered Jul 27 '16 at 14:11

Tarun Dugar

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that wont work because 0...n should result in false, and -1 in true – Jaromanda X Jul 27 '16 at 14:13
Still doesn't work. If you put in `return !(array2.indexOf(item) >= 0);` that should do the trick. – romellem Jul 27 '16 at 14:14
sexy people use `!~` – Jaromanda X Jul 27 '16 at 14:15
1

isn't `!( ... >= 0)` just `... < 0` – Jaromanda X Jul 27 '16 at 14:18
1

yes, as will ! >= 0 ... 0 >= 0 is true, !true is false ... trust me ... `!(a.indexOf(b) >= 0)` is 100% identical to `a.indexOf(b) < 0` ... or, more simply ... `!(a >= 0)` === `a < 0` – Jaromanda X Jul 27 '16 at 14:27
@TarunDugar: Think that through for just a moment longer. – Jul 27 '16 at 14:28

score 1 · Answer 3 · answered Jul 27 '16 at 14:13

1

array1 = [1, 2, 3, 4, 5];
array2 = [1, 3, 5];
filteredArray = [];

filteredArray = array1.filter(function (y) {
  return array2.indexOf(y) < 0;
 });
console.log(filteredArray);

answered Jul 27 '16 at 14:13

Emil S. Jørgensen

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score 0 · Answer 4 · answered Jul 27 '16 at 14:13

You can use indexOf() to check if the array2 item is in array1 then only add it to filteredArray if it is:

 array1 = [1,2,3,4,5];
 array2 = [1,3,5];
 filteredArray = [];

 array2.forEach(function(x){
     if (array1.indexOf(array2[x] > -1) {
         filteredArray.push(array2[x]);
     }
 });
 return filteredArray;

score 0 · Answer 5 · answered Jul 27 '16 at 14:22

0

In ES6, you could use Set for it.

var array1 = [1, 2, 3, 4, 5],
    array2 = [1, 3, 5],
    filteredArray = array1.filter((set => a => !set.has(a))(new Set(array2)));

console.log(filteredArray);

answered Jul 27 '16 at 14:22

Nina Scholz

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We can make it even more convoluted..! How about `filteredArray = array1.filter(function(set,a) {return !set.has(a)}.bind(null,new Set(array2)));`..? – Redu Jul 27 '16 at 15:16
in this case, you could use `new Set(...)` as this args. – Nina Scholz Jul 27 '16 at 15:18

score 0 · Answer 6 · answered Jul 27 '16 at 15:05

0

All indexOf no play makes me a dull boy...

var array1 = [1,2,3,4,5],
    array2 = [1,3,5],
  filtered = array1.filter(e => !array2.includes(e));
console.log(filtered);

answered Jul 27 '16 at 15:05

Redu

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Javascript, filtering array with an array

6 Answers6