Move an array element from one array position to another

Move an array element from one array position to another

I’m having a hard time figuring out how to move an array element. For example, given the following:
var arr = [ ‘a’, ‘b’, ‘c’, ‘d’, ‘e’];

How can I write a function to move ‘d’ before ‘b’?
Or ‘a’ after ‘c’?
After the move, the indices of the rest of the elements should be updated. This means in the first example after the move arr[0] would = ‘a’, arr[1] = ‘d’ arr[2] = ‘b’, arr[3] = ‘c’, arr[4] = ‘e’
This seems like it should be pretty simple, but I can’t wrap my head around it.


Solution 1:

If you’d like a version on npm, array-move is the closest to this answer, although it’s not the same implementation. See its usage section for more details. The previous version of this answer (that modified Array.prototype.move) can be found on npm at array.prototype.move.

I had fairly good success with this function:

function array_move(arr, old_index, new_index) {
    if (new_index >= arr.length) {
        var k = new_index - arr.length + 1;
        while (k--) {
    arr.splice(new_index, 0, arr.splice(old_index, 1)[0]);
    return arr; // for testing

// returns [2, 1, 3]
console.log(array_move([1, 2, 3], 0, 1)); 

Note that the last return is simply for testing purposes: splice performs operations on the array in-place, so a return is not necessary. By extension, this move is an in-place operation. If you want to avoid that and return a copy, use slice.

Stepping through the code:

  1. If new_index is greater than the length of the array, we want (I presume) to pad the array properly with new undefineds. This little snippet handles this by pushing undefined on the array until we have the proper length.
  2. Then, in arr.splice(old_index, 1)[0], we splice out the old element. splice returns the element that was spliced out, but it’s in an array. In our above example, this was [1]. So we take the first index of that array to get the raw 1 there.
  3. Then we use splice to insert this element in the new_index’s place. Since we padded the array above if new_index > arr.length, it will probably appear in the right place, unless they’ve done something strange like pass in a negative number.

A fancier version to account for negative indices:

function array_move(arr, old_index, new_index) {
    while (old_index < 0) {
        old_index += arr.length;
    while (new_index < 0) {
        new_index += arr.length;
    if (new_index >= arr.length) {
        var k = new_index - arr.length + 1;
        while (k--) {
    arr.splice(new_index, 0, arr.splice(old_index, 1)[0]);
    return arr; // for testing purposes
// returns [1, 3, 2]
console.log(array_move([1, 2, 3], -1, -2));

Which should account for things like array_move([1, 2, 3], -1, -2) properly (move the last element to the second to last place). Result for that should be [1, 3, 2].

Either way, in your original question, you would do array_move(arr, 0, 2) for a after c. For d before b, you would do array_move(arr, 3, 1).

Solution 2:

Here’s a one liner I found on JSPerf….

Array.prototype.move = function(from, to) {
    this.splice(to, 0, this.splice(from, 1)[0]);

which is awesome to read, but if you want performance (in small data sets) try…

 Array.prototype.move2 = function(pos1, pos2) {
    // local variables
    var i, tmp;
    // cast input parameters to integers
    pos1 = parseInt(pos1, 10);
    pos2 = parseInt(pos2, 10);
    // if positions are different and inside array
    if (pos1 !== pos2 && 0 <= pos1 && pos1 <= this.length && 0 <= pos2 && pos2 <= this.length) {
      // save element from position 1
      tmp = this[pos1];
      // move element down and shift other elements up
      if (pos1 < pos2) {
        for (i = pos1; i < pos2; i++) {
          this[i] = this[i + 1];
      // move element up and shift other elements down
      else {
        for (i = pos1; i > pos2; i--) {
          this[i] = this[i - 1];
      // put element from position 1 to destination
      this[pos2] = tmp;

I can’t take any credit, it should all go to Richard Scarrott. It beats the splice based method for smaller data sets in this performance test. It is however significantly slower on larger data sets as Darwayne points out.

Solution 3:

I like this way. It works, it’s concise and elegant.

function arraymove(arr, fromIndex, toIndex) {
    var element = arr[fromIndex];
    arr.splice(fromIndex, 1);
    arr.splice(toIndex, 0, element);

Note: always remember to check your array bounds.

Here’s a jsFiddle to test:

Solution 4:

The splice() method adds/removes items to/from an array, and returns the removed item(s).

Note: This method changes the original array. /w3schools/

Array.prototype.move = function(from,to){
  return this;

var arr = [ 'a', 'b', 'c', 'd', 'e'];
arr.move(3,1);//["a", "d", "b", "c", "e"]

var arr = [ 'a', 'b', 'c', 'd', 'e'];
arr.move(0,2);//["b", "c", "a", "d", "e"]

as the function is chainable this works too:


demo here

Solution 5:

My 2c. Easy to read, it works, it’s fast, it doesn’t create new arrays.

function move(array, from, to) {
  if( to === from ) return array;

  var target = array[from];                         
  var increment = to < from ? -1 : 1;

  for(var k = from; k != to; k += increment){
    array[k] = array[k + increment];
  array[to] = target;
  return array;

Solution 6:

Got this idea from @Reid of pushing something in the place of the item that is supposed to be moved to keep the array size constant. That does simplify calculations. Also, pushing an empty object has the added benefits of being able to search for it uniquely later on. This works because two objects are not equal until they are referring to the same object.

({}) == ({}); // false

So here’s the function which takes in the source array, and the source, destination indexes. You could add it to the Array.prototype if needed.

function moveObjectAtIndex(array, sourceIndex, destIndex) {
    var placeholder = {};
    // remove the object from its initial position and
    // plant the placeholder object in its place to
    // keep the array length constant
    var objectToMove = array.splice(sourceIndex, 1, placeholder)[0];
    // place the object in the desired position
    array.splice(destIndex, 0, objectToMove);
    // take out the temporary object
    array.splice(array.indexOf(placeholder), 1);