Как дать имя для неопределенного ключа объекта, который генерируется loda sh groupBy ()?

Question

Результат loda sh groupBy() дает { "student": [], "undefined": [] }.

const splittedData = _.groupBy(datas, 'type');

Можно ли дать имя для ключа undefined?

const datas = [
  {
    _id: "5ea2ecbde53c090b4ba4fa12",
    firstName: "Dillon",
    surname: "Mclaughlin",
    guid: "a727438f-4ddc-4422-af4a-c7396fda73c3",
    isActive: true,
    balance: "$3,331.33",
    picture: "http://placehold.it/32x32",
    age: 37,
    type: "student"
  },
  {
    _id: "5ea2ecbd067a8ad1cb1f9e62",
    firstName: "Fleming",
    surname: "Austin",
    guid: "e228bbea-ef6b-4709-995c-bf485fa3665c",
    isActive: true,
    balance: "$1,869.92",
    picture: "http://placehold.it/32x32",
    age: 35
  },
  {
    _id: "5ea2ecbd0d1a074b255a1349",
    firstName: "Kaufman",
    surname: "Alford",
    guid: "176fb7fe-b7bc-4574-8887-4252a042b24a",
    isActive: true,
    balance: "$2,163.34",
    picture: "http://placehold.it/32x32",
    age: 29,
    type: "student"
  },
  {
    _id: "5ea2ecbd1162ee44119aa0d3",
    firstName: "Sykes",
    surname: "Petty",
    guid: "92b13963-7d32-469c-be80-c88728a7842a",
    isActive: false,
    balance: "$2,728.46",
    picture: "http://placehold.it/32x32",
    age: 37
  },
  {
    _id: "5ea2ecbd297854a2f85909a4",
    firstName: "Stuart",
    surname: "Pickett",
    guid: "322031f9-1a71-44ae-b1cf-cbe7430b78b6",
    isActive: false,
    balance: "$1,725.25",
    picture: "http://placehold.it/32x32",
    age: 25,
    type: "student"
  },
  {
    _id: "5ea2ecbd6ef4a9bee37a223a",
    firstName: "Berg",
    surname: "Hewitt",
    guid: "ab05bed9-29e4-4570-aa12-fead954417b0",
    isActive: false,
    balance: "$2,935.72",
    picture: "http://placehold.it/32x32",
    age: 31
  }
];

const splittedData = _.groupBy(datas, 'type');
console.log(splittedData);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

Демонстрация: https://codepen.io/athimannil/pen/pojePYK

Answer 1

Вы можете использовать функцию _.get(), чтобы получить type или указать запасное значение, если type равно undefined:

const datas = [{"_id":"5ea2ecbde53c090b4ba4fa12","firstName":"Dillon","surname":"Mclaughlin","guid":"a727438f-4ddc-4422-af4a-c7396fda73c3","isActive":true,"balance":"$3,331.33","picture":"http://placehold.it/32x32","age":37,"type":"student"},{"_id":"5ea2ecbd067a8ad1cb1f9e62","firstName":"Fleming","surname":"Austin","guid":"e228bbea-ef6b-4709-995c-bf485fa3665c","isActive":true,"balance":"$1,869.92","picture":"http://placehold.it/32x32","age":35},{"_id":"5ea2ecbd0d1a074b255a1349","firstName":"Kaufman","surname":"Alford","guid":"176fb7fe-b7bc-4574-8887-4252a042b24a","isActive":true,"balance":"$2,163.34","picture":"http://placehold.it/32x32","age":29,"type":"student"},{"_id":"5ea2ecbd1162ee44119aa0d3","firstName":"Sykes","surname":"Petty","guid":"92b13963-7d32-469c-be80-c88728a7842a","isActive":false,"balance":"$2,728.46","picture":"http://placehold.it/32x32","age":37},{"_id":"5ea2ecbd297854a2f85909a4","firstName":"Stuart","surname":"Pickett","guid":"322031f9-1a71-44ae-b1cf-cbe7430b78b6","isActive":false,"balance":"$1,725.25","picture":"http://placehold.it/32x32","age":25,"type":"student"},{"_id":"5ea2ecbd6ef4a9bee37a223a","firstName":"Berg","surname":"Hewitt","guid":"ab05bed9-29e4-4570-aa12-fead954417b0","isActive":false,"balance":"$2,935.72","picture":"http://placehold.it/32x32","age":31}];

const splittedData = _.groupBy(datas, o => _.get(o, 'type', 'not a student') );

console.log(splittedData);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

Answer 2

Вот рабочая ссылка https://codepen.io/sudheer-ranga/pen/LYpWLLZ

Если вы хотите фильтровать только по студентам

const grouped = _(datas)
  .filter(object => _.has(object, 'type'))
  .groupBy('type')
  .value();

console.log(grouped);

Если вы хотите группировать студентов и не студентов

const nonStudents = _.groupBy(datas, object => _.get(object, 'type', 'no_type'))

console.log(nonStudents)

Answer 3

Вы можете указать функцию iteratee, которая использует требуемое значение по умолчанию, если в объекте вообще не было type:

const splittedData = _.groupBy(datas, o => 'type' in o ? o.type : 'unknown');

Разница между этим и последующим "переименованием" "undefined" в результате, опция «переименование» не сможет различить type: "undefined" и вообще никакого типа, в то время как эта делает.

Answer 4

Конечно, просто скопируйте неопределенное значение во что-то другое и удалите оригинал.

const splittedData = _.groupBy(datas, 'type');
splittedData["default-type"] = splittedData["undefined"];
delete splittedData["undefined"];