61

How can the following operation be done without mutating the array:

let array = ['item1'];
console.log(array); // ['item1']
array[2] = 'item2'; // array is mutated
console.log(array); // ['item1', undefined, 'item2']

In the above code, array variable is mutated. How can I perform the same operation without mutating the array?

1

9 Answers 9

117

You can use Object.assign(target, source1, source2, /* …, */ sourceN):

Object.assign([], array, {2: newItem});
8
  • 2
    @DanPrince Not sure about how these sharing structures are implemented. But slice copies the full array so that case is also O(n).
    – Oriol
    Commented Jun 27, 2016 at 19:20
  • 8
    That said, copying an array with slice might be faster because length is copied first, so preallocation is possible. Object.assign(array.slice(), {2: newItem}); is another possibility.
    – Oriol
    Commented Jun 27, 2016 at 19:28
  • 1
    Not sure if this is a one-off fix in a program the original poster is writing or the end result will be a application accessed across multiple devices and browsers but... there is no support for internet explorer or android. You may want to add a disclaimer in your answer and reference a polyfill perhaps. Commented Jun 27, 2016 at 19:49
  • 16
    @cemper93 It can work, you just need to use this specific syntax: Object.assign([], array, { [i]: newItem })
    – Paul
    Commented Oct 20, 2017 at 18:20
  • 1
    Not sure but this might be very bad for engines performance optimization when assigning to indexes that are >= than arrays length?
    – Nabuska
    Commented May 31, 2019 at 15:27
19

The fast way

function replaceAt(array, index, value) {
  const ret = array.slice(0);
  ret[index] = value;
  return ret;
}

See the JSPerf (thanks to @Bless)

Related posts:

1
  • 1
    Yes. +1. Here is a perf comparison - jsperf.com/…
    – Bless
    Commented Dec 22, 2017 at 15:46
14

You can simply set up a new array as such:

const newItemArray = array.slice();

And then set value for the index which you wish to have a value for.

newItemArray[position] = newItem

and return that. The values under the indexes in-between will have undefined.

Or the obviously alternative would be:

Object.assign([], array, {<position_here>: newItem});
12

Here is how I'd like to do it:

function update(array, newItem, atIndex) {
    return array.map((item, index) => index === atIndex ? newItem : item);
}

Generally, Array-spread operation produces few temporary arrays for you, but map doesn't, so it can be faster. You can also look at this discussion as a reference

1
7

Well, technically this wouldn't be replacing as there isn't an item at the index you're changing.

Look at how it's handled in Clojure—a language that's built around canonical implementations for immutable data structures.

(assoc [1] 2 3)
;; IndexOutOfBoundsException

Not only does it fail, but it crashes too. These data structures are designed to be as robust as possible and when you come up against these kinds of errors, it's generally not because you've discovered an edge case, but more likely that you're using the wrong data structure.

If you are ending up with sparse arrays, then consider modelling them with objects or maps instead.

let items = { 0: 1 };
{ ...items, 2: 3 };
// => { 0: 1, 2: 3 }

let items = new Map([ [0, 1] ]);
items(2, 3);
// => Map {0 => 1, 2 => 3}

However, Map is a fundamentally mutable data structure, so you'd need to swap this out for an immutable variant with a library like Immutable.js or Mori.

let items = Immutable.Map([ [0, 2] ]);
items.set(2, 3);
// => Immutable.Map {0 => 1, 2 => 3}

let items = mori.hashMap();
mori.assoc(items, 2, 3);
// => mori.hashMap {0 => 1, 2 => 3}

Of course, there might be a perfectly good reason for wanting to use JavaScript's arrays, so here's a solution for good measure.

function set(arr, index, val) {
  if(index < arr.length) {
    return [
      ...arr.slice(0, position),
      val,
      ...arr.slice(position + 1)
    ];
  } else {
    return [
      ...arr,
      ...Array(index - arr.length),
      val
    ];
  }
}
7

Another way could be to use spread operator with slice as

let newVal = 33, position = 3;
let arr = [1,2,3,4,5];
let newArr = [...arr.slice(0,position - 1), newVal, ...arr.slice(position)];
console.log(newArr); //logs [1, 2, 33, 4, 5]
console.log(arr); //logs [1, 2, 3, 4, 5]

1
  • Note that position in this snippet is not zero based. If you have a zero based index, it is [...arr.slice(0, index), newVal, ...arr.slice(index + 1)]
    – Andre
    Commented Mar 17, 2023 at 9:41
3

There's a new tc39 proposal, which adds a with method to Array that returns a copy of the array and doesn't modify the original:

Array.prototype.with(index, value) -> Array

Example from the proposal:

const correctionNeeded = [1, 1, 3];
correctionNeeded.with(1, 2); // => [1, 2, 3]
correctionNeeded; // => [1, 1, 3]

(note that a RangeError will be thrown if the first argument to with is outside the bounds of the array, so the specific example in the question will not work)

As it's currently in stage 3, it will likely be implemented in browser engines soon, but in the meantime a polyfill is available here or in core-js.

2

var list1 = ['a','b','c'];
var list2 = list1.slice();
list2.splice(2, 0, "beta", "gamma");
console.log(list1);
console.log(list2);

Is this what you want?

2

How about this:

const newArray = [...array]; // make a copy of the original array
newArray[2] = 'item2'; // mutate the copy

I find the intent a bit more clear than this one-liner:

const newArray = Object.assign([...array], {2: 'item2'});
1
  • I love the latter form, it really reads well!
    – gnclmorais
    Commented Apr 14, 2023 at 12:26

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