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

17

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);

javascript arrays performance

share|improve this question

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

add a comment | 

17

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);

javascript arrays performance

share|improve this question

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

add a comment | 

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);

javascript arrays performance

share|improve this question

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

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);

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);

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);

share|improve this question

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

share|improve this question

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

edited Mar 7 at 12:52

Uwe Keim

27.6k32132214

asked Mar 7 at 10:48

DaliborDalibor

513318

asked Mar 7 at 10:48

DaliborDalibor

513318

4 Answers
4

active

oldest

votes

34

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.

share|improve this answer

edited Mar 7 at 11:03

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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/…
  
  – Charlie H
  Mar 8 at 5:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 ?
  
  – Grégory NEUT
  Mar 8 at 7:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You're right, if the idsToRemove has a very small number of elements, using a Set instead doesn't provide much of a benefit.
  
  – CertainPerformance
  Mar 8 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you explain me the difference between your syntax and this syntax: const filtered = objList.filter(o => !set.has(o.id));
  
  – Dalibor
  Mar 9 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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.
  
  – CertainPerformance
  Mar 9 at 11:48
  

  
  
  
  

add a comment | 

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);

share|improve this answer

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

add a comment | 

2

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()

share|improve this answer

edited Mar 7 at 11:08

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

add a comment | 

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);

share|improve this answer

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

add a comment | 

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34

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.

share|improve this answer

edited Mar 7 at 11:03

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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/…
  
  – Charlie H
  Mar 8 at 5:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 ?
  
  – Grégory NEUT
  Mar 8 at 7:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You're right, if the idsToRemove has a very small number of elements, using a Set instead doesn't provide much of a benefit.
  
  – CertainPerformance
  Mar 8 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you explain me the difference between your syntax and this syntax: const filtered = objList.filter(o => !set.has(o.id));
  
  – Dalibor
  Mar 9 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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.
  
  – CertainPerformance
  Mar 9 at 11:48
  

  
  
  
  

add a comment | 

34

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.

share|improve this answer

edited Mar 7 at 11:03

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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/…
  
  – Charlie H
  Mar 8 at 5:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 ?
  
  – Grégory NEUT
  Mar 8 at 7:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You're right, if the idsToRemove has a very small number of elements, using a Set instead doesn't provide much of a benefit.
  
  – CertainPerformance
  Mar 8 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Can you explain me the difference between your syntax and this syntax: const filtered = objList.filter(o => !set.has(o.id));
  
  – Dalibor
  Mar 9 at 11:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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.
  
  – CertainPerformance
  Mar 9 at 11:48
  

  
  
  
  

add a comment | 

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.

share|improve this answer

edited Mar 7 at 11:03

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

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);

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);

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);

share|improve this answer

edited Mar 7 at 11:03

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

answered Mar 7 at 10:51

CertainPerformanceCertainPerformance

92.9k165384

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);

share|improve this answer

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

add a comment | 

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);

share|improve this answer

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

add a comment | 

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);

share|improve this answer

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

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);

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);

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);

share|improve this answer

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

answered Mar 7 at 10:55

Grégory NEUTGrégory NEUT

9,58721941

2

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()

share|improve this answer

edited Mar 7 at 11:08

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

add a comment | 

2

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()

share|improve this answer

edited Mar 7 at 11:08

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

add a comment | 

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()

share|improve this answer

edited Mar 7 at 11:08

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

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

share|improve this answer

edited Mar 7 at 11:08

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

answered Mar 7 at 10:54

Charlie HCharlie H

9,56342653

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);

share|improve this answer

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

add a comment | 

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);

share|improve this answer

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

add a comment | 

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);

share|improve this answer

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

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);

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);

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);

share|improve this answer

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

edited Mar 7 at 22:56

dwirony

4,42731334

edited Mar 7 at 22:56

dwirony

4,42731334

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837

answered Mar 7 at 10:59

Khyati SharmaKhyati Sharma

837