Removing all but a few specific elements from a list in javascript?

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Let's say I have a list of elements a = document.getElementsByTag('li'). I want to delete every element in that list except for those at positions keep = [3, 5, 8]. Ordinarily, this could easily be done with a for loop, and just making sure to skip the good positions.

for (i = 0; i < a.length; i++) {
  if (!keep.includes(i)) {
    a[i].remove();
  }
}

Unfortunately, when removing elements from a, that dynamically changes a, and the positions/indexes of everything get shifted. If I wanted to, it's not too hard to account for this shifting. But is there a a more straight-forward / elegant method to do this task? I'd like to learn a more proper method before I just hack things to make it work.

For the record, the new solution I was thinking of was something like

shift = 0;
len = a.length;
for (i = 0; i < len; i++) {
  if (!keep.includes(i)) {
    a[i-shift].remove();
    shift++;
  }
}
getelementsbyclassname list arrays javascript
2 Answers

I think you can use filter instead of looping. I find it quite idiomatic in javascript:

a = a.filter(x => keep.includes(x))
userqf30

If you want to maintain the indexing of the keep values in an array then instead of removing the other values, You can replace them with undefined.

Here you go :

const a = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const keep = [3, 5, 8];

a.forEach((num, index) => {
  if (!keep.includes(index)) {
    a[index] = undefined
  }
});

console.log(a);

Rohìt Jíndal0
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