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);












share|improve this question































    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);












    share|improve this question



























      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);












      share|improve this 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);








      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 7 at 12:52









      Uwe Keim

      27.7k32135216




      27.7k32135216










      asked Mar 7 at 10:48









      DaliborDalibor

      520318




      520318
























          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.






          share|improve this answer


























          • 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



















          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































            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

































              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


























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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.






                share|improve this answer


























                • 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
















                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.






                share|improve this answer


























                • 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














                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.






                share|improve this answer















                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);






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 7 at 11:03

























                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 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



















                • 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

















                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













                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




























                  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


























                    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);








                    share|improve this answer













                    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);






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Mar 7 at 10:55









                    Grégory NEUTGrégory NEUT

                    9,73721941




                    9,73721941























                        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






























                          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




























                            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()






                            share|improve this answer















                            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














                            share|improve this answer



                            share|improve this answer








                            edited Mar 7 at 11:08

























                            answered Mar 7 at 10:54









                            Charlie HCharlie H

                            9,77042953




                            9,77042953























                                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






























                                  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




























                                    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);








                                    share|improve this answer















                                    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);






                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited Mar 7 at 22:56









                                    dwirony

                                    4,65731434




                                    4,65731434










                                    answered Mar 7 at 10:59









                                    Khyati SharmaKhyati Sharma

                                    837




                                    837






























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