如何在JavaScript中同时从不同的索引中删除数组中的多个元素？

Question

I have an array of integers, which I'm using the .push() method to add to.

I know that splice() method adds/removes items to/from an array, and returns the removed item(s).

But is there is a way to remove multiple elements from the array at different indexes at the same time?

I know we can also use filter() method but The filter() method creates a new array with all elements that pass the test implemented by the provided function which I don't want.

Ex:-

var array = [1,2,3,4,5];

I know to remove 3 and 5 I can follow following steps:-

1. array.splice(2, 1);
2. array.splice(3, 1);

But Can I achieve this in a single step without using splice() method twice?

Answer 1

If you know the exact values you want removed you can use array.filter() .

This removes more than one element from an array at different indexes at same time.

Example in your case:

 var filtered = [1,2,3,4,5].filter(value => { return value !== 3 && value !== 5 }); console.log(filtered) 

Answer 2

You can use trailing comma at destructuring assignment to select specific elements from array, assign resulting values within array to original array reference.

 var array = [1,2,3,4,5]; {let [a,b,,c,,] = array; array = [a,b,c]}; console.log(array); 

You can use object destructuring on an array to get only specific indexes from array

 var array = [1,2,3,4,5]; {let {0:a,1:b,3:c} = array; array = [a,b,c]}; console.log(array); 

You can also use .forEach() to iterate an array of elements to match, .indexOf() , .splice() to remove elements which match value of element within array

 var array = [1,2,3,4,5]; [3, 5].forEach(p => array.splice(array.indexOf(p), 1)); console.log(array); 

Using for..of loop, Array.prototype.findIndex() , Array.prototype.splice()

 var array = [1,2,3,4,5]; var not = [3, 5]; for (let n of not) array.splice(array.findIndex(v => v === n), 1); console.log(array); 

Using for..of loop with Array.prototype.entries() , Array.prototype.some() , Array.prototype.splice()

 var array = [1,2,3,4,5]; var not = [3,5]; for (let [k, p] of array.entries()) not.some(n => !(np)) && array.splice(k, 1); console.log(array); 

You can also use for loop, Object.assign() to set indexes to keep at beginning of array, call .splice() with parameter set to - .length of the number of elements to remove from array

 var array = [1,2,3,4,5]; var not = [3,5]; for (var o = {}, k = -1, i = 0; i < array.length; i++) { if (!not.some(n => n === array[i])) o[++k] = array[i]; } Object.assign(array, o).splice(-not.length); console.log(array); 

Using Array.prototype.reduce() with Object.assign()

 var array = [1,2,3,4,5]; var not = [3,5]; Object.assign(array, array.reduce(([o, not, k], p) => [Object.assign(o, !not.some(n => n === p) ? {[++k]:p} : void 0), not, k] , [{}, not, -1]).shift() ).splice(-not.length); console.log(array); 

Another option, if the numbers are unique, is to use Set , .delete() , convert Set object to Array using rest element at destructuring assignment

 var array = new Set; var not = [3, 5]; for (let n = 1; n <= 5; n++) array.add(n); for (let n of not) array.delete(n); [...array] = array; console.log(array); // `array = new Set(array)` 

Answer 3

Hard code when you know the specific values you need to remove and when the array is small:

 var array = [1,2,3,4,5,6,7]; array.splice(2,3,4); console.log(array); 

Answer 4

May be you can do like this in place;

 var arr = [1,2,3,4,5,17,23,37,61,15], itemsToGetRidOff = [3,17,15,1,23]; arr.reduceRight((_,e,i,a) => itemsToGetRidOff.includes(e) && a.splice(i,1),[]); console.log(arr);