Remove elements from array using spread

Question

I have an array of animals arr = ['cat','dog','elephant','lion','tiger','mouse']

I want to write a function remove(['dog','lion']) which can remove the elements from arr, can we write this function using es6 spread?

example:

arr = ['cat','dog','elephant','lion','tiger','mouse']
remove(['cat', 'lion'])

arr should get changed to

arr = ['dog','elephant','tiger','mouse']

Note: I don't want mutations, so please don't suggest solutions that mutates array.

Answer 1

No, you can't, because they're not next to each other. There's been some discussion of taking spread and rest syntax further, but even in those discussions, I don't think a series of discontiguous selections would be possible.

I think the closest you can get is to call out the first few specifically and then use rest syntax for everything after 'lion' :

 const arr = ['cat','dog','elephant','lion','tiger','mouse']; const arr2 = [arr[1], arr[2], ...arr.slice(4)]; console.log(arr2);

...which I'm sure isn't what you wanted to do. :-)

Answer 2

From what I understand you're looking for a function that has its argument defined using the spread syntax. Here is an example:

 var arr = ['cat','dog','elephant','lion','tiger','mouse']; function remove(...toRemove){ toRemove.forEach(item => { var index = arr.indexOf(item); if(index != -1){ arr.splice(index, 1); } }) } remove('dog', 'lion'); // OR remove(...['dog', 'lion']); console.log(arr);

This is actually changing the original array (it mutates it) you've mentioned that you're not looking for mutation but you've also mentioned this arr should get changed to...

Answer 3

You can use spread if you want to pass lots of strings like remove('lion', 'dog') . Otherwise I don't think spread can help you.

Answer 4

You could take an iterator and exclude the values from iteration with spread operator.

Btw, it is not really advisable.

 function remove(array, items) { array[Symbol.iterator] = function* () { for (let value of Object.values(this)) { if (!items.includes(value)) { yield value; } } }; } var array = ['cat', 'dog', 'elephant', 'lion', 'tiger', 'mouse']; remove(array, ['dog', 'lion']); console.log([...array]); console.log(array);

 .as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 5

I don't understand the selected answer

const items = ['cat', 'dog', 'elephant', 'lion', 'tiger', 'mouse']
const valuesToRemove = ['dog', 'lion']
const filteredItems = items.filter(item => !valuesToRemove.includes(item))

Answer 6

you can use the array.filter method, it will return array with filtered values,

var filterMe = ["cat", "dog", "elephant", "lion", "tiger", "mouse"];
var answer = filterMe.filter((item) => item !== "dog")
                     .filter((item) => item !== "lion");
console.log(answer);

Answer 7

Similar to the previous filter solution, but using a regular expression. Beneficial if the remove array is large...

 let arr = ['cat', 'dog', 'elephant', 'lion', 'tiger', 'mouse']; const removeArr = ['cat', 'lion']; const removeRegExp = new RegExp(removeArr.join('|')); arr = arr.filter(e => !removeRegExp.test(e)); console.log(arr);

Answer 8

If you're using Typescript, you can also use the Omit utility type to remove specific keys from you object/ array.

Assuming an interface like this

interface Todo {
  title: string;
  description: string; // <--- assuming you want to omit `description` key
  completed: boolean;
  createdAt: number;
}
 
type TodoPreview = Omit<Todo, "description">;

the TodoPreview type will now expect an object with the other members included but with decription omitted

This will now be correct

const todo: TodoPreview = {
  title: "Clean room",
  completed: false,
  createdAt: 1615544252770,
};

This will ERROR out

const todo: TodoPreview = {
  title: "Clean room",
  completed: false,
  description: 'this should not be here' // <--- TS compiler will complain
  createdAt: 1615544252770,
};