Javascript swap array elements

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:

Answer

You only need one temporary variable.

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

Edit hijacking top answer 10 years later with a lot of ES6 adoption under our belts:

Given the array arr = [1,2,3,4], you can swap values in one line now like so:

[arr[0], arr[1]] = [arr[1], arr[0]];

This would produce the array [2,1,3,4]. This is destructuring assignment.

Answer

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.

Answer

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.

Answer

Well, you don't need to buffer both values - only one:

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

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]
Answer

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];
Answer

This didn't exist when the question was asked, but ES2015 introduced array destructuring, allowing you to write it as follows:

let a = 1, b = 2;
// a: 1, b: 2
[a, b] = [b, a];
// a: 2, b: 1
Answer

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

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

To swap two consecutive elements of array

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

what about Destructuring_assignment

var arr = [1, 2, 3, 4]
[arr[index1], arr[index2]] = [arr[index2], arr[index1]]

which can also be extended to

[src order elements] => [dest order elements]
Answer

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.

Answer

Consider such a solution without a need to define the third variable:

function swap(arr, from, to) {
  arr.splice(from, 1, arr.splice(to, 1, arr[from])[0]);
}

var letters = ["a", "b", "c", "d", "e", "f"];

swap(letters, 1, 4);

console.log(letters); // ["a", "e", "c", "d", "b", "f"]

Note: You may want to add additional checks for example for array length. This solution is mutable so swap function does not need to return a new array, it just does mutation over array passed into.

Answer

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>

Answer

There is one interesting way of swapping:

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

(ES6 way)

Answer
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];
Answer

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!)

Answer

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))
Answer

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

Answer

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]

Answer

For the sake of brevity, here's the ugly one-liner version that's only slightly less ugly than all that concat and slicing above. The accepted answer is truly the way to go and way more readable.

Given:

var foo = [ 0, 1, 2, 3, 4, 5, 6 ];

if you want to swap the values of two indices (a and b); then this would do it:

foo.splice( a, 1, foo.splice(b,1,foo[a])[0] );

For example, if you want to swap the 3 and 5, you could do it this way:

foo.splice( 3, 1, foo.splice(5,1,foo[3])[0] );

or

foo.splice( 5, 1, foo.splice(3,1,foo[5])[0] );

Both yield the same result:

console.log( foo );
// => [ 0, 1, 2, 5, 4, 3, 6 ]

#splicehatersarepunks:)

Answer

try this function...

$(document).ready(function () {
        var pair = [];
        var destinationarray = ['AAA','BBB','CCC'];

        var cityItems = getCityList(destinationarray);
        for (var i = 0; i < cityItems.length; i++) {
            pair = [];
            var ending_point = "";
            for (var j = 0; j < cityItems[i].length; j++) {
                pair.push(cityItems[i][j]);
            }
            alert(pair);
            console.log(pair)
        }

    });
    function getCityList(inputArray) {
        var Util = function () {
        };

        Util.getPermuts = function (array, start, output) {
            if (start >= array.length) {
                var arr = array.slice(0);
                output.push(arr);
            } else {
                var i;

                for (i = start; i < array.length; ++i) {
                    Util.swap(array, start, i);
                    Util.getPermuts(array, start + 1, output);
                    Util.swap(array, start, i);
                }
            }
        }

        Util.getAllPossiblePermuts = function (array, output) {
            Util.getPermuts(array, 0, output);
        }

        Util.swap = function (array, from, to) {
            var tmp = array[from];
            array[from] = array[to];
            array[to] = tmp;
        }
        var output = [];
        Util.getAllPossiblePermuts(inputArray, output);
        return output;
    }
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer

Swap the first and last element in an array without temporary variable or ES6 swap method [a, b] = [b, a]

[a.pop(), ...a.slice(1), a.shift()]

Answer

Using ES6 it's possible to do it like this...

Imagine you have these 2 arrays...

const a = ["a", "b", "c", "d", "e"];
const b = [5, 4, 3, 2, 1];

and you want to swap the first values:

const [a0] = a;
a[0] = b[0];
b[0] = a0;

and value:

a; //[5, "b", "c", "d", "e"]
b; //["a", 4, 3, 2, 1]
Answer

If need swap first and last elements only:

array.unshift( array.pop() );
Answer
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);
Answer

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.

Answer

If you don't want to use temp variable in ES5, this is one way to swap array elements.

var swapArrayElements = function (a, x, y) {
  if (a.length === 1) return a;
  a.splice(y, 1, a.splice(x, 1, a[y])[0]);
  return a;
};

swapArrayElements([1, 2, 3, 4, 5], 1, 3); //=> [ 1, 4, 3, 2, 5 ]
Answer

var arr = [1, 2];
arr.splice(0, 2, arr[1], arr[0]);
console.log(arr); //[2, 1]

Answer

Typescript solution that clones the array instead of mutating existing one

export function swapItemsInArray<T>(items: T[], indexA: number, indexB: number): T[] {
  const itemA = items[indexA];

  const clone = [...items];

  clone[indexA] = clone[indexB];
  clone[indexB] = itemA;

  return clone;
}

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