Immutable - change elements in array with slice (no splice)

Question

How is possible to change 3/4 elements? Expected output is [1,2,4,3,5]

let list = [1,2,3,4,5];
const removeElement = list.indexOf(3); // remove number 3
list         


        

Answer 1

The easer solution might be using filter instead of splice or slice. According to documentation https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter

The filter() method creates a new array with all elements that pass the test implemented by the provided function.

It means the original array stays immutable. The only difference is that in this case, you have to know the value you want to delete instead of index.

let list = [1,2,3,4,5];
list.filter((item) => item !== 3);

Answer 2

var list = [1,2,3,4,5];
var numToRemove = 3;

var removeElementIndex = list.indexOf(numToRemove);
var afterRemoveElement = list[removeElementIndex+1];

list.slice(0, removeElementIndex).concat(afterRemoveElement).concat(numToRemove).concat(list.slice(removeElementIndex+2)) // [1,2,4,3,5]

Answer 3

slice doesn't mutate the array on which it operates so you need to assign a value to what it returns

let list = [1,2,3,4,5];
const removeElement = list.indexOf(3); // remove number 3
var newList = list.slice(0, removeElement).concat(list.slice(removeElement+1)) // [1,2,4,5]

If you are prepared to use ES2015 syntax, you can use the spread operator as follows:

const removeElement = list.indexOf(3); // remove number 3
var es6List = [
  ...list.slice(0, removeElement),
  ...list.slice(removeElement+1)
];
console.log(es6List);

fiddle

Answer 4

Object.assign actually works here

const newList = Object.assign([], list, {
  2: list[3],
  3: list[2],
});

list // [1,2,3,4,5]
newList // [1,2,4,3,5]
newList === list // false

Answer 5

Arrays are objects, use Object.assign() and access elements with property name expressions.

var numToMove = 2;
console.log(Object.assign(list, {[numToMove]: list[numToMove+1]}, 
{[numToMove+1]: list[numToMove]}));
// [1, 2, 4, 3, 5]

Answer 6

The simplest way to write this is to use the spread operator:

let newList = [...list.slice(0, 2), list[4], list[3], ...list.slice(4)];