Move an array element from one array position to another

04/15/2020 01:00:01

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.

Verified Answer (657 Votes)

03/15/2011 07:23:38

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).


Answer #2 (263 Votes)

08/24/2011 22:59:50

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.


Answer #3 (219 Votes)

06/24/2011 21:25:14

I like this way. It's concise and it works.

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.

Run Snippet on jsFiddle

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