Is there a simpler way to swap two elements in an array?
var a = list[x], b = list[y];
list[y] = a;
list[x] = b;
37 Answers
In place swap:
// Array methods
function swapInArray(arr, i1, i2){
let t = arr[i1];
arr[i1] = arr[i2];
arr[i2] = t;
}
function moveBefore(arr, el){
let ind = arr.indexOf(el);
if(ind !== -1 && ind !== 0){
swapInArray(arr, ind, ind - 1);
}
}
function moveAfter(arr, el){
let ind = arr.indexOf(el);
if(ind !== -1 && ind !== arr.length - 1){
swapInArray(arr, ind + 1, ind);
}
}
// DOM methods
function swapInDom(parentNode, i1, i2){
parentNode.insertBefore(parentNode.children[i1], parentNode.children[i2]);
}
function getDomIndex(el){
for (let ii = 0; ii < el.parentNode.children.length; ii++){
if(el.parentNode.children[ii] === el){
return ii;
}
}
}
function moveForward(el){
let ind = getDomIndex(el);
if(ind !== -1 && ind !== 0){
swapInDom(el.parentNode, ind, ind - 1);
}
}
function moveBackward(el){
let ind = getDomIndex(el);
if(ind !== -1 && ind !== el.parentNode.children.length - 1){
swapInDom(el.parentNode, ind + 1, ind);
}
}
Comments
var letters = ["a", "b", "c", "d", "e", "f"];
function swap (arr, from, to) {
let temp = arr[from];
arr.splice(from, 1, arr[to]);
arr.splice(to, 1, temp);
}
swap(letters, 1, 4)
console.log(letters); // output [ 'a', 'e', 'c', 'd', 'b', 'f' ]
1 Comment
Pof
Please don't post only code as answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually more helpful and of better quality, and are more likely to attract upvotes.
Maybe is useless after 30 answer. This is a way to swap last two items of an array. Is not exactly the answer of your question but a solution I've just found for a specific case of needed swap.
const prev = yourArray.pop(); // remove last
yourArray.push(article); // add one
yourArray.push(prev); // add removed one
Comments
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]
Comments
function moveElement(array, sourceIndex, destinationIndex) {
return array.map(a => a.id === sourceIndex ? array.find(a => a.id === destinationIndex): a.id === destinationIndex ? array.find(a => a.id === sourceIndex) : a )
}
let arr = [
{id: "1",title: "abc1"},
{id: "2",title: "abc2"},
{id: "3",title: "abc3"},
{id: "4",title: "abc4"}];
moveElement(arr, "2","4");
1 Comment
Mark Rotteveel
Please don't post only code as answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually more helpful and of better quality, and are more likely to attract upvotes.
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]
Comments
If there is a need to swap the first and last elements only:
array.unshift( array.pop() );
1 Comment
David Archibald
This code is flawed. It takes the last element of the array then puts it at the beginning, that isn't swapping. This code does this:
[1, 2, 3] => [3, 1, 2] instead of [1, 2, 3] => [3, 2, 1].
with()you can do it in one line:list.with(x,list[y]).with(y,list[x])see my answer with an example!