remove objects with a specific `key: value` from an array










0















I have an array of object and need to use lodash to remove some of those objects with a specific key: value, for instance:



[
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
]


I need to remove all object with id of 1 and 3.



i know the way below but was wondering if there is a shorter way:



 var array = [
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
];

 _.remove(array, function(n) 
  return _.includes([ 1,3 ], n.id);
);









share|improve this question






















  • Do you really need to mutate the object? it's usually a better idea to work with immutable data.

    – tokland
    Nov 14 '18 at 19:17















0















I have an array of object and need to use lodash to remove some of those objects with a specific key: value, for instance:



[
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
]


I need to remove all object with id of 1 and 3.



i know the way below but was wondering if there is a shorter way:



 var array = [
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
];

 _.remove(array, function(n) 
  return _.includes([ 1,3 ], n.id);
);









share|improve this question






















  • Do you really need to mutate the object? it's usually a better idea to work with immutable data.

    – tokland
    Nov 14 '18 at 19:17













0












0








0








I have an array of object and need to use lodash to remove some of those objects with a specific key: value, for instance:



[
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
]


I need to remove all object with id of 1 and 3.



i know the way below but was wondering if there is a shorter way:



 var array = [
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
];

 _.remove(array, function(n) 
  return _.includes([ 1,3 ], n.id);
);









share|improve this question














I have an array of object and need to use lodash to remove some of those objects with a specific key: value, for instance:



[
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
]


I need to remove all object with id of 1 and 3.



i know the way below but was wondering if there is a shorter way:



 var array = [
id:1,b:22,
id:2,b:44,
id:3,b:56,
id:4,b:-29
];

 _.remove(array, function(n) 
  return _.includes([ 1,3 ], n.id);
);






lodash






share|improve this question













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asked Nov 14 '18 at 0:14









farm commandfarm command

6719




6719












  • Do you really need to mutate the object? it's usually a better idea to work with immutable data.

    – tokland
    Nov 14 '18 at 19:17

















  • Do you really need to mutate the object? it's usually a better idea to work with immutable data.

    – tokland
    Nov 14 '18 at 19:17
















Do you really need to mutate the object? it's usually a better idea to work with immutable data.

– tokland
Nov 14 '18 at 19:17





Do you really need to mutate the object? it's usually a better idea to work with immutable data.

– tokland
Nov 14 '18 at 19:17












1 Answer
1






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oldest

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0














You could use pullAllBy and provide the elements to be removed as the 2nd parameter.






let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

const result = _.pullAllBy(data, [id:1,id:3], 'id')
console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





Note that this method mutates the array. If you do not want that use _.differenceBy



Using ES6 and arrow functions + filter to get the values and not mutate the array will also get you somewhat of a short version:






let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

const result = data.filter(x => ![1,3].includes(x.id))
console.log(result)








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    1 Answer
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    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

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    0














    You could use pullAllBy and provide the elements to be removed as the 2nd parameter.






    let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

    const result = _.pullAllBy(data, [id:1,id:3], 'id')
    console.log(result)

    <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





    Note that this method mutates the array. If you do not want that use _.differenceBy



    Using ES6 and arrow functions + filter to get the values and not mutate the array will also get you somewhat of a short version:






    let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

    const result = data.filter(x => ![1,3].includes(x.id))
    console.log(result)








    share|improve this answer





























      0














      You could use pullAllBy and provide the elements to be removed as the 2nd parameter.






      let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

      const result = _.pullAllBy(data, [id:1,id:3], 'id')
      console.log(result)

      <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





      Note that this method mutates the array. If you do not want that use _.differenceBy



      Using ES6 and arrow functions + filter to get the values and not mutate the array will also get you somewhat of a short version:






      let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

      const result = data.filter(x => ![1,3].includes(x.id))
      console.log(result)








      share|improve this answer



























        0












        0








        0







        You could use pullAllBy and provide the elements to be removed as the 2nd parameter.






        let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

        const result = _.pullAllBy(data, [id:1,id:3], 'id')
        console.log(result)

        <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





        Note that this method mutates the array. If you do not want that use _.differenceBy



        Using ES6 and arrow functions + filter to get the values and not mutate the array will also get you somewhat of a short version:






        let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

        const result = data.filter(x => ![1,3].includes(x.id))
        console.log(result)








        share|improve this answer















        You could use pullAllBy and provide the elements to be removed as the 2nd parameter.






        let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

        const result = _.pullAllBy(data, [id:1,id:3], 'id')
        console.log(result)

        <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





        Note that this method mutates the array. If you do not want that use _.differenceBy



        Using ES6 and arrow functions + filter to get the values and not mutate the array will also get you somewhat of a short version:






        let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

        const result = data.filter(x => ![1,3].includes(x.id))
        console.log(result)








        let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

        const result = _.pullAllBy(data, [id:1,id:3], 'id')
        console.log(result)

        <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





        let data = [id:1,b:22,id:2,b:44,id:3,b:56,id:4,b:-29]

        const result = _.pullAllBy(data, [id:1,id:3], 'id')
        console.log(result)

        <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>





        let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

        const result = data.filter(x => ![1,3].includes(x.id))
        console.log(result)





        let data = [ id: 1, b: 22 , id: 2, b: 44 , id: 3, b: 56 , id: 4, b: -29 ]

        const result = data.filter(x => ![1,3].includes(x.id))
        console.log(result)






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 14 '18 at 17:29

























        answered Nov 14 '18 at 0:49









        AkrionAkrion

        9,42711224




        9,42711224



























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