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Convert an array of objects to array of the objects' values
Remove object from array based on array of some property of that object
2019 Community Moderator ElectionIs Set.has() method O(1) and Array.indexOf O(n)?Detecting an undefined object propertyCreate ArrayList from arrayHow can I merge properties of two JavaScript objects dynamically?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?PHP: Delete an element from an arraySort array of objects by string property valueHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?Iterate through object properties
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
edited 19 hours ago
Uwe Keim
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
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
edited 19 hours ago
Uwe Keim
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
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
edited 19 hours ago
Uwe Keim
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
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);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);javascript arrays performance
javascript arrays performance
edited 19 hours ago
Uwe Keim
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
edited 19 hours ago
Uwe Keim
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
edited 19 hours ago
Uwe Keim
27.6k32132213
edited 19 hours ago
Uwe Keim
27.6k32132213
edited 19 hours ago
Uwe Keim
27.6k32132213
27.6k32132213
asked 21 hours ago
DaliborDalibor
490318
asked 21 hours ago
DaliborDalibor
490318
asked 21 hours ago
DaliborDalibor
490318
490318
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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.
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
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
2 hours 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 ?
– Grégory NEUT
51 secs ago
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);answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
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()
answered 21 hours ago
Charlie HCharlie H
9,51642653
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);edited 9 hours ago
dwirony
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
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
2 hours 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 ?
– Grégory NEUT
51 secs ago
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.
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
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
2 hours 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 ?
– Grégory NEUT
51 secs ago
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.
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
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.
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);answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
edited 20 hours ago
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
answered 21 hours ago
CertainPerformanceCertainPerformance
91.6k165380
91.6k165380
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
2 hours 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 ?
– Grégory NEUT
51 secs ago
add a comment |
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
2 hours 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 ?
– Grégory NEUT
51 secs 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/…
– Charlie H
2 hours 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/…
– Charlie H
2 hours 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 ?
– Grégory NEUT
51 secs 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 ?
– Grégory NEUT
51 secs ago
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);answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
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);answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
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);answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
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);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);answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
answered 21 hours ago
Grégory NEUTGrégory NEUT
9,39421940
9,39421940
add a comment |
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()
answered 21 hours ago
Charlie HCharlie H
9,51642653
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()
answered 21 hours ago
Charlie HCharlie H
9,51642653
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()
answered 21 hours ago
Charlie HCharlie H
9,51642653
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()
answered 21 hours ago
Charlie HCharlie H
9,51642653
edited 20 hours ago
answered 21 hours ago
Charlie HCharlie H
9,51642653
answered 21 hours ago
Charlie HCharlie H
9,51642653
answered 21 hours ago
Charlie HCharlie H
9,51642653
9,51642653
add a comment |
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);edited 9 hours ago
dwirony
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
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);edited 9 hours ago
dwirony
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
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);edited 9 hours ago
dwirony
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
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);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);edited 9 hours ago
dwirony
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
edited 9 hours ago
dwirony
4,27231334
edited 9 hours ago
dwirony
4,27231334
edited 9 hours ago
dwirony
4,27231334
4,27231334
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
answered 20 hours ago
Khyati SharmaKhyati Sharma
1387
1387
add a comment |
add a comment |
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