What is the proper way to retrieve items from an Array that have an index that is evenly divisible by

Question

What is the proper way to retrieve items from an Array that have an index that is evenly divisible by

let numbers = [4, 5, 7, 8, 14, 45, 76];

function getEvenElements(arr) {
  let evenArray = [];

  for (let i = 0; i < arr.length / 2; i++) {
    evenArray.push(arr[2 * i]);
  }

  return evenArray.filter(num => num !== undefined);
}

alert(getEvenElements(numbers));

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

I attempted to extract only the even-indexed elements from an array using this code. It seems to be working, but it includes empty values as well. How can I modify the code to output existing elements only?

javascript arrays

Answer 1

Answer №1

To extract every second element from an array, you have two options:

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

Another approach is to increment i by 2 in each iteration to skip every other element:

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

Note: Be cautious as the provided code retrieves elements with odd indexes instead of even ones. To address this, use i % 2 === 1 or initialize i = 1 at the beginning of the loop.

Answer 2

To extract every second element from an array, you have two options:

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

Another approach is to increment i by 2 in each iteration to skip every other element:

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

Note: Be cautious as the provided code retrieves elements with odd indexes instead of even ones. To address this, use i % 2 === 1 or initialize i = 1 at the beginning of the loop.

Answer 3

Answer №2

To ensure compatibility with Internet Explorer 9 and above, you can utilize the Array.filter method.

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

If you need a fallback for older versions of Internet Explorer, the following code snippet can be added:

if (!Array.prototype.filter)
{
  Array.prototype.filter = function(fun /*, thisp */) // Fallback for IE
  {
    "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[1];
    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;
  };
}

Answer 4

To ensure compatibility with Internet Explorer 9 and above, you can utilize the Array.filter method.

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

If you need a fallback for older versions of Internet Explorer, the following code snippet can be added:

if (!Array.prototype.filter)
{
  Array.prototype.filter = function(fun /*, thisp */) // Fallback for IE
  {
    "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[1];
    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;
  };
}

Answer 5

Answer №3

Make this code work for the year 2018 :)

Extract the elements at odd indexes by using filter method

var arr = [4, 5, 7, 8, 14, 45, 76, 5];
let filtered = arr.filter((a,i) => i%2===1);
console.log(filtered);

Answer 6

Make this code work for the year 2018 :)

Extract the elements at odd indexes by using filter method

var arr = [4, 5, 7, 8, 14, 45, 76, 5];
let filtered = arr.filter((a,i) => i%2===1);
console.log(filtered);

Answer 7

Answer №4

Although this question may be considered old, I want to share a one-liner filter tip:
For odd numbers: arr.filter((e,i)=>i%2)
For even numbers: arr.filter((e,i)=>i%2-1)
A more conventional approach for finding even numbers: arr.filter((e,i)=>!(i%2))

It is unnecessary to use ===1 as suggested by sumit. Utilizing mod 2 already yields either 0 or 1, which can be treated as boolean values.

Instead of using i%2, you can opt for i&1, which may enhance performance on large arrays but is restricted to 31-bit integers.

Answer 8

Although this question may be considered old, I want to share a one-liner filter tip:
For odd numbers: arr.filter((e,i)=>i%2)
For even numbers: arr.filter((e,i)=>i%2-1)
A more conventional approach for finding even numbers: arr.filter((e,i)=>!(i%2))

It is unnecessary to use ===1 as suggested by sumit. Utilizing mod 2 already yields either 0 or 1, which can be treated as boolean values.

Instead of using i%2, you can opt for i&1, which may enhance performance on large arrays but is restricted to 31-bit integers.

Answer 9

Answer №5

Have you considered using the % operator? It calculates the remainder of a division.

Instead of the existing loop, you could implement:

if ((i % 2) === 0) {
    newArr.push(arr[i]);
}

Answer 10

Have you considered using the % operator? It calculates the remainder of a division.

Instead of the existing loop, you could implement:

if ((i % 2) === 0) {
    newArr.push(arr[i]);
}

Answer 11

Answer №6

let numbers = [2, 6, 8, 14, 23, 45, 67];

function getEvenNumbers(array) {
  let evenArray = [];

  for (let num in array) {
    if ((array[num] % 2) == 0)
      evenArray.push(array[num]);
  }

  return evenArray;
}

alert(getEvenNumbers(numbers));

Answer 12

let numbers = [2, 6, 8, 14, 23, 45, 67];

function getEvenNumbers(array) {
  let evenArray = [];

  for (let num in array) {
    if ((array[num] % 2) == 0)
      evenArray.push(array[num]);
  }

  return evenArray;
}

alert(getEvenNumbers(numbers));

Answer 13

Answer №7

It's important to note why the outcome may not match your expectations when iterating over an array of size N, even though others have found excellent solutions. When iterating over an array of size N and attempting to push elements into a new array, the resulting array will still have size N. However, if only half of the elements are present in the original array, then the remaining slots in the resulting array will be filled with blanks. By checking for conditions like a[2*i] existing or a[i] % 2 == 0 before inserting, the resulting array will contain only the values with even indexes.

Answer 14

It's important to note why the outcome may not match your expectations when iterating over an array of size N, even though others have found excellent solutions. When iterating over an array of size N and attempting to push elements into a new array, the resulting array will still have size N. However, if only half of the elements are present in the original array, then the remaining slots in the resulting array will be filled with blanks. By checking for conditions like a[2*i] existing or a[i] % 2 == 0 before inserting, the resulting array will contain only the values with even indexes.

What is the proper way to retrieve items from an Array that have an index that is evenly divisible by

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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