Faster way to remove array elements with list of IDs?

Question

I'm trying to remove elements in an array using a list of IDs to remove.

const dataArray = [];
const idsToRemove = new Set();

for (let i=0; i<10; i++) {
  dataArray.push({id: "myID" + i, value: i});
}

for (let i=0; i<10; i+=2) {
  idsToRemove.add("myID" + i);
}

const newArray = dataArray.filter(obj => !idsToRemove.has(obj.id));
console.log(newArray);

Result:

{id:"myID1", value:1}, {id:"myID3",value:3}, {id:"myID5", value:5}, {id:"myID7", value:7}, {id:"myID9",value:9}

Is this an ideal solution? Or would there be a more performant solution than this?

ADDED: Would there be a way to avoid making a newArray but just directly remove elements from dataArray?

The solution seems ideal to me.... also we can use Map instead of Set to store idsToRemove so that it's time complexity would improve... — Abhisar Tripathi
– Abhisar Tripathi, Commented May 3, 2020 at 5:29
I found two solutions that are faster than mine: jsben.ch/ld9op — Zack Lee
– Zack Lee, Commented May 3, 2020 at 7:05

Carsten Massmann · Accepted Answer · 2020-05-03 05:41:36Z

1

You can save some computational effort by preparing the remove list as an object with the target IDs as key:

let dataArray=[];
for (let i=0; i<10; i++) {
  dataArray.push({id: "myID" + i, value: i});
}
let idsToRemove={};
for (i=0;i<10;i+=2) idsToRemove["myID"+i]=1;
var filtered=dataArray.filter(o=>!idsToRemove[o.id]);   
console.log(filtered);

edited May 3, 2020 at 5:41

answered May 3, 2020 at 5:35

Carsten Massmann

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Comments

Amir Popovich · Accepted Answer · 2020-05-03 05:54:56Z

1

Here's a modern way of coding this (not performance oriented):

const dataArray = new Array(10).fill(0).map((item, idx) => idx);

const idsToRemove = new Set(dataArray.filter(i => i % 2 === 0));

const newArray = dataArray.filter(i => !idsToRemove.has(i)).map(i => ({id: `myID${i}`, value: i}));

console.log(newArray);

edited May 3, 2020 at 5:54

answered May 3, 2020 at 5:33

Amir Popovich

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2 Comments

Nick Over a year ago

How does this not have loops? You have fill, map and filter all of which are shorthand for loops.

Amir Popovich Over a year ago

@Nick - Behind the scenes it does loop obviously. But this is a more modern approach of coding these stuff today if you aren't performance oriented.

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