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Javascript swap array elements

Posted by: admin November 30, 2017 Leave a comment

Questions:

Is there any simpler way to swap two elements in an array?

var a = list[x], b = list[y];
list[y] = a;
list[x] = b;
Answers:

You only need one temporary variable.

var b = list[y];
list[y] = list[x];
list[x] = b;

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If you want a single expression, using native javascript,
remember that the return value from a splice operation
contains the element(s) that was removed.

var A = [1, 2, 3, 4, 5, 6, 7, 8, 9], x= 0, y= 1;
A[x] = A.splice(y, 1, A[x])[0];
alert(A); // alerts "2,1,3,4,5,6,7,8,9"

Edit:

The [0] is necessary at the end of the expression as Array.splice() returns an array, and in this situation we require the single element in the returned array.

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This seems ok….

var b = list[y];
list[y] = list[x];
list[x] = b;

Howerver using

var b = list[y];

means a b variable is going to be to be present for the rest of the scope. This can potentially lead to a memory leak. Unlikely, but still better to avoid.

Maybe a good idea to put this into Array.prototype.swap

Array.prototype.swap = function (x,y) {
  var b = this[x];
  this[x] = this[y];
  this[y] = b;
  return this;
}

which can be called like:

list.swap( x, y )

This is a clean approach to both avoiding memory leaks and DRY.

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Well, you don’t need to buffer both values – only one:

var tmp = list[x];
list[x] = list[y];
list[y] = tmp;

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With numeric values you can avoid a temporary variable by using bitwise xor

list[x] = list[x] ^ list[y];
list[y] = list[y] ^ list[x];
list[x] = list[x] ^ list[y];

or an arithmetic sum (noting that this only works if x + y is less than the maximum value for the data type)

list[x] = list[x] + list[y];
list[y] = list[x] - list[y];
list[x] = list[x] - list[y];

Questions:
Answers:

You can swap elements in an array the following way:

list[x] = [list[y],list[y]=list[x]][0]

See the following example:

list = [1,2,3,4,5]
list[1] = [list[3],list[3]=list[1]][0]
//list is now [1,4,3,2,5]

Note: it works the same way for regular variables

var a=1,b=5;
a = [b,b=a][0]

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According to some random person on Metafilter,
“Recent versions of Javascript allow you to do swaps (among other things) much more neatly:”

[ list[x], list[y] ] = [ list[y], list[x] ];

My quick tests showed that this Pythonic code works great in the version of JavaScript
currently used in “Google Apps Script” (“.gs”).
Alas, further tests show this code gives a “Uncaught ReferenceError: Invalid left-hand side in assignment.” in
whatever version of JavaScript (“.js”) is used by
Google Chrome Version 24.0.1312.57 m.

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Digest from http://www.greywyvern.com/?post=265

var a = 5, b = 9;    
b = (a += b -= a) - b;    
alert([a, b]); // alerts "9, 5"

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To swap two consecutive elements of array

array.splice(IndexToSwap,2,array[IndexToSwap+1],array[IndexToSwap]);

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You can swap any number of objects or literals, even of different types, using a simple identity function like this:

var swap = function (x){return x};
b = swap(a, a=b);
c = swap(a, a=b, b=c);

For your problem:

var swap = function (x){return x};
list[y]  = swap(list[x], list[x]=list[y]);

This works in JavaScript because it accepts additional arguments even if they are not declared or used. The assignments a=b etc, happen after a is passed into the function.

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Answers:
var a = [1,2,3,4,5], b=a.length;

for (var i=0; i<b; i++) {
    a.unshift(a.splice(1+i,1).shift());
}
a.shift();
//a = [5,4,3,2,1];

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This didn’t exist 7 years ago but no with the new hot sauce of es2015 and up you can use array destructuring to

var a = 1, b = 2;
[a, b] = [b, a]; --> a = 2, b = 1

Questions:
Answers:

For two or more elements (fixed number)

[list[y], list[x]] = [list[x], list[y]];

No temporary variable required!

I was thinking about simply calling list.reverse().
But then I realised it would work as swap only when list.length = x + y + 1.

For variable number of elements

I have looked into various modern Javascript constructions to this effect, including Map and map, but sadly none has resulted in a code that was more compact or faster than this old-fashioned, loop-based construction:

function multiswap(arr,i0,i1) {/* argument immutable if string */
    if (arr.split) return multiswap(arr.split(""), i0, i1).join("");
    var diff = [];
    for (let i in i0) diff[i0[i]] = arr[i1[i]];
    return Object.assign(arr,diff);
}

Example:
    var alphabet = "abcdefghijklmnopqrstuvwxyz";
    var [x,y,z] = [14,6,15];
    var output = document.getElementsByTagName("code");
    output[0].innerHTML = alphabet;
    output[1].innerHTML = multiswap(alphabet, [0,25], [25,0]);
    output[2].innerHTML = multiswap(alphabet, [0,25,z,1,y,x], [25,0,x,y,z,3]);
<table>
    <tr><td>Input:</td>                        <td><code></code></td></tr>
    <tr><td>Swap two elements:</td>            <td><code></code></td></tr>
    <tr><td>Swap multiple elements:&nbsp;</td> <td><code></code></td></tr>
</table>

Questions:
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There is one interesting way of swapping:

var a = 1;
var b = 2;
[a,b] = [b,a];

(ES6 way)

Questions:
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Here’s a compact version
swaps value at i1 with i2 in arr

arr.slice(0,i1).concat(arr[i2],arr.slice(i1+1,i2),arr[i1],arr.slice(i2+1))

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Here is a variation that first checks if the index exists in the array:

Array.prototype.swapItems = function(a, b){
    if(  !(a in this) || !(b in this) )
        return this;
    this[a] = this.splice(b, 1, this[a])[0];
    return this;
}

It currently will just return this if the index does not exist, but you could easily modify behavior on fail

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Just for the fun of it, another way without using any extra variable would be:

var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];

// swap index 0 and 2
arr[arr.length] = arr[0];   // copy idx1 to the end of the array
arr[0] = arr[2];            // copy idx2 to idx1
arr[2] = arr[arr.length-1]; // copy idx1 to idx2
arr.length--;               // remove idx1 (was added to the end of the array)


console.log( arr ); // -> [3, 2, 1, 4, 5, 6, 7, 8, 9]

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Here’s a one-liner that doesn’t mutate list:

let newList = Object.assign([], list, {[x]: list[y], [y]: list[x]})

(Uses language features not available in 2009 when the question was posted!)

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If need swap first and last elements only:

array.unshift( array.pop() );

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Here is the correct way to do it:

Array.prototype.swap = function(a, b) {
  var temp = this[a];
  this[a] = this[b];
  this[b] = temp;
};

Usage:

var myArray = [0,1,2,3,4...];
myArray.swap(4,1);