Delete array element based on elements in an array as index

Question

Suppose I have an array as

let arr1 = [0,2] This is always sorted

Note: These elements in array represent indexes to be deleted from another array.

I have another array as:

let arrOvj = [1,4,6,7,21,17,12]

I want to delete element of arrObj, based on indexes present in arr1.

So after deletion expected OP should be,

[4,7,21,17,12].

So how can I achieve this.

I tried as:

for(let i=0;i<arr1.length;i++){
   arrObj.splice(arr1[i],1)
}

But this is giving incorrect result.

Example: If arr1=[0]

It deletes first two elements instead of deleting element at index 0 of arrObj.

What other alternative can I follow so it deletes from provided index value only

Please let me know if anyone need any further information.

Answer 1

You can loop backwards over the indices to delete so that elements will never be shifted into a location that has not been already processed.

 let arr1 = [0,2] let arrOvj = [1,4,6,7,21,17,12] for(let i = arr1.length - 1; i >= 0; i--) arrOvj.splice(arr1[i], 1); console.log(arrOvj);

Answer 2

Other way of doing it is using the filter method on the array and removing out the corresponding indexes as follows:

 let arrObj = [1,4,6,7,21,17,12] let indexes = [0,2] let newArr = arrObj.filter((ele, index) => !indexes.includes(index)); console.log(newArr);

Answer 3

You can use splice and reduceRight here.

 let arr1 = [0, 2]; let arrOvj = [1, 4, 6, 7, 21, 17, 12]; arrOvj = arr1.reduceRight((a, c) => (a.splice(c, 1), a), arrOvj); console.log(arrOvj);