Remove object from array based on array of some property of that object
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19
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 Mar 7 at 12:52
Uwe Keim
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
add a comment |
19
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 Mar 7 at 12:52
Uwe Keim
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
add a comment |
19
19
19
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 Mar 7 at 12:52
Uwe Keim
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
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 Mar 7 at 12:52
Uwe Keim
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
edited Mar 7 at 12:52
Uwe Keim
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
edited Mar 7 at 12:52
Uwe Keim
27.7k32135216
edited Mar 7 at 12:52
Uwe Keim
27.7k32135216
edited Mar 7 at 12:52
Uwe Keim
27.7k32135216
27.7k32135216
asked Mar 7 at 10:48
DaliborDalibor
520318
asked Mar 7 at 10:48
DaliborDalibor
520318
asked Mar 7 at 10:48
DaliborDalibor
520318
520318
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
35
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 Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
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 theidsToRemovehas 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);answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
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()
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
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);edited Mar 7 at 22:56
dwirony
4,65731434
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
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
35
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 Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
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 theidsToRemovehas 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 |
35
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 Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
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 theidsToRemovehas 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 |
35
35
35
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 Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
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 Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
edited Mar 7 at 11:03
answered Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
answered Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
answered Mar 7 at 10:51
CertainPerformanceCertainPerformance
98.7k166089
98.7k166089
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 theidsToRemovehas 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 |
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 theidsToRemovehas 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
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
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
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
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
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
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
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
@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);answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
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);answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
add a comment |
4
4
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 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
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 Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
answered Mar 7 at 10:55
Grégory NEUTGrégory NEUT
9,73721941
9,73721941
add a comment |
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()
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
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()
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
add a comment |
2
2
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()
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
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 Mar 7 at 10:54
Charlie HCharlie H
9,77042953
edited Mar 7 at 11:08
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
answered Mar 7 at 10:54
Charlie HCharlie H
9,77042953
9,77042953
add a comment |
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);edited Mar 7 at 22:56
dwirony
4,65731434
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);edited Mar 7 at 22:56
dwirony
4,65731434
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
add a comment |
1
1
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 at 22:56
dwirony
4,65731434
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
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 Mar 7 at 22:56
dwirony
4,65731434
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
edited Mar 7 at 22:56
dwirony
4,65731434
edited Mar 7 at 22:56
dwirony
4,65731434
edited Mar 7 at 22:56
dwirony
4,65731434
4,65731434
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
answered Mar 7 at 10:59
Khyati SharmaKhyati Sharma
837
837
add a comment |
add a comment |
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