# How do I extract even elements of an Array?

``````var arr = [4, 5, 7, 8, 14, 45, 76];

function even(a) {
var ar = [];

for (var i = 0; i < a.length; i++) {
ar.push(a[2 * i + 1]);
}

return ar;
}

``````

http://jsbin.com/unocar/2/edit

I have tried this code in order to output even (index) elements of an array. It works, but it also outputs some empty elements. How do I fix this code to output only existing elements? Either use modulus:

``````for (var i = 0; i < a.length; i++) {
if(i % 2 === 0) { // index is even
ar.push(a[i]);
}
}
``````

or skip every second element by incrementing `i` accordingly:

``````for(var i = 0; i < a.length; i += 2) {  // take every second element
ar.push(a[i]);
}
``````

Notice: Your code actually takes the elements with odd indexes from the array. If this is what you want you have to use `i % 2 === 1` or start the loop with `var i = 1` respectively. For IE9+ use `Array.filter`

``````var arr = [4,5,7,8,14,45,76];
var filtered = arr.filter(function(element, index, array) {
return (index % 2 === 0);
});
``````

With a fallback for older IEs, all the other browsers are OK without this fallback

``````if (!Array.prototype.filter)
{
Array.prototype.filter = function(fun /*, thisp */)
{
"use strict";

if (this === void 0 || this === null)
throw new TypeError();

var t = Object(this);
var len = t.length >>> 0;
if (typeof fun !== "function")
throw new TypeError();

var res = [];
var thisp = arguments;
for (var i = 0; i < len; i++)
{
if (i in t)
{
var val = t[i]; // in case fun mutates this
if (fun.call(thisp, val, i, t))
res.push(val);
}
}

return res;
};
}
`````` This will work on 2018 :)

take the odd indexes and apply to filter

``````var arr = [4, 5, 7, 8, 14, 45, 76,5];
let filtered=arr.filter((a,i)=>i%2===1);
console.log(filtered);`````` why don't you try with the % operator. It gives you the remaining of a division.

replace the loop block with

``````if ((i % 2) === 0) {
ar.push(a[i])
}
`````` You need to test the elements for evenness like this:

``````var arr = [4,5,7,8,14,45,76];

function even(a){
var ar = [];

for (var i=0; i<a.length;i++){
if (a[i] % 2 === 0)
{
ar.push(a[i]);
}

}

return ar;
}

``````

%2 is the modulo operator, it returns the remainder of integer division. ``````var arr = [4,5,7,8,14,45,76];

function even(a)
{
var ar = [];

for (x in a)
{

if((a[x]%2)==0)
ar.push(a[x]);

}
return ar;
}

`````` I just wanted to explain why your result is not what you expected since everyone else shows excellent solutions. You are iterating over an array size N so your resulting array will attempt to push elements in an array that will result in size N. Since only N/2 will be found in the original array your resulting array will fill the rest with blanks to fill in the rest of N. So if you checked to see if a[2*i] exists OR checked to see if a[i] % 2 == 0 before inserting, your resulting array will contain only the even indexed values Even if this question is quite old, I would like to add a one-liner filter:
Odd numbers: `arr.filter((e,i)=>i%2)`
Even numbers: `arr.filter((e,i)=>i%2-1)`
A more 'legal' way for even numbers: `arr.filter((e,i)=>!(i%2))`

There's no need to check with `i%2===1` like sumit said; as `mod 2` already returns a 0 or a 1 as numbers, they can be interpreted as boolean values in js.