Remove object from array based on array of some property of that object

Question

I have an array of objects (objList) that each has "id" property.

I have an array of strings (idsToRemove), representing IDs of the objects to remove from objList.

I find some solution but I fear it's slow, especially with the large list of objects with lots of properties. Is there more efficient way to do this?

var idsToRemove = ["3", "1"];
var objList = [{
    id: "1",
    name: "aaa"
  },
  {
    id: "2",
    name: "bbb"
  },
  {
    id: "3",
    name: "ccc"
  }
];

for (var i = 0, len = idsToRemove.length; i < len; i++) {
  objList = objList.filter(o => o.id != idsToRemove[i]);
}

console.log(objList);

CertainPerformance · Accepted Answer · 2019-03-07 11:03:01Z

37

Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):

var idsToRemove = ["3", "1"];
var objList = [{
    id: "1",
    name: "aaa"
  },
  {
    id: "2",
    name: "bbb"
  },
  {
    id: "3",
    name: "ccc"
  }
];

const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);

Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.

edited Mar 7, 2019 at 11:03

answered Mar 7, 2019 at 10:51

CertainPerformance

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5 Comments

Charlie Over a year ago

Not sure if the claim Set.has is O(1) is correct. Please provide an answer to this question if it is correct. stackoverflow.com/questions/55057200/…

Orelsanpls Over a year ago

Does it not depend on the size of the array? If the array is small, is it not better to through it than creating a new copy in memory then access it ?

CertainPerformance Over a year ago

You're right, if the idsToRemove has a very small number of elements, using a Set instead doesn't provide much of a benefit.

Dalibor Over a year ago

Can you explain me the difference between your syntax and this syntax: const filtered = objList.filter(o => !set.has(o.id));

CertainPerformance Over a year ago

@Dalibor My version uses destructuring in the argument list, whereas that version doesn't. If a function doesn't need the whole object, but only one (or some) of its properties, I like to extract that property immediately rather than having to hold on to the whole object - but both options work just fine.

Orelsanpls · Accepted Answer · 2019-03-07 10:55:08Z

4

You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.

const idsToRemove = ['3', '1'];

const objList = [{
    id: '1',
    name: 'aaa',
  },
  {
    id: '2',
    name: 'bbb',
  },
  {
    id: '3',
    name: 'ccc',
  },
];

const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));

console.log(filteredObjList);

answered Mar 7, 2019 at 10:55

Orelsanpls

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Comments

Charlie · Accepted Answer · 2019-03-07 11:08:53Z

3

You don't need two nested iterators if you use a built-in lookup function

   objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);

Documentation:

Array.prototype.indexOf()

Array.prototype.includes()

edited Mar 7, 2019 at 11:08

answered Mar 7, 2019 at 10:54

Charlie

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Comments

dwirony · Accepted Answer · 2019-03-07 22:56:29Z

1

Simply use Array.filter()

const idsToRemove = ['3', '1'];

const objList = [{
    id: '1',
    name: 'aaa',
  },
  {
    id: '2',
    name: 'bbb',
  },
  {
    id: '3',
    name: 'ccc',
  }
];

const res = objList.filter(value => !idsToRemove.includes(value.id));

console.log("result",res);

edited Mar 7, 2019 at 22:56

dwirony

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answered Mar 7, 2019 at 10:59

Khyati Sharma

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