Как дифференцировать с JSON на основе ключа и без ключа по loda sh groupBy ()?

Question

Результат loda sh groupBy() должен дать { "withType": [], "withOutType": [] }.

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

Можно ли разделить две группы на основе наличия клавиши type в JSON?

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,type:"teacher"},{_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', 'noneType');
console.log(splittedData);

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Ожидаемый результат

{
  "withType": [],
  "withOutType": []
}

Answer 1

Используйте _.partition(), чтобы создать две группы - одну для истинных значений, а другую для ложных. Разрушаем и создаем объект:

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,type:"teacher"},{_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 [withType, withoutType] = _.partition(datas, 'type');
console.log({ withType, withoutType });

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Answer 2

Вы можете предоставить функцию в качестве второго аргумента для _.groupBy(), который вернет 'withType', если текущий объект имеет свойство type (проверено с помощью _.has()), или он вернет 'withoutType' если нет:

const data = [{_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,type:"teacher"},{_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(data, o => _.has(o, 'type') ? 'withType' : 'withoutType'); 
console.log(splittedData);

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Answer 3

Вы можете проверить, существует ли type в объекте, и получить объект, сгруппированный по true или false.

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,type:"teacher"},{_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 => 'type' in o ? "withType" : "withOutType");
console.log(splittedData);

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Answer 4

Я вижу решение, подобное этому:

const splits = [
    [...datas.filter(element => element.type)],
    [...datas.filter(element => !element.type)]
]