根据值（本身就是数组）将数组收集到子数组桶中

Question

我有一个 Javascript 对象数组，如下所示。[  {    email: '[e-mail]',    fn: 'Alex',    sn: 'McPherson',    phone: '01233xxxxx',    hours: '40',    rate: '20',    amount: '200',    vat: '60',    agency: 'test',    start: '08/06/2017',    end: '10/06/2017',    shipping: {      addresses: [        {          id: '1234',          area: 'xzy'        },        {          id: '2345',          area: 'uhj'        }      ]    }  },  {    email: '[e-mail]',    fn: 'Mike',    sn: 'Mann',    phone: '01233xxxxx',    hours: '50',    rate: '70',    amount: '500',    vat: '90',    agency: 'test',    start: '08/06/2017',    end: '10/06/2017',    shipping: {      addresses: [        {          id: '1234',          area: 'xzy'        },        {          id: '3456',          area: 'uio'        }      ]    }  },  {    email: '[e-mail]',    fn: 'Fred',    sn: 'Frogg',    phone: '01233xxxxx',    hours: '80',    rate: '90',    amount: '800',    vat: '100',    agency: 'test',    start: '08/06/2017',    end: '10/06/2017',    shipping: {      addresses: [        {          id: '4567',          area: 'asdaf'        },        {          id: '3456',          area: 'uio'        }      ]    }  },  {    email: '[e-mail]',    fn: 'Alex',    sn: 'McPherson',    phone: '01233xxxxx',    hours: '90',    rate: '30',    amount: '900',    vat: '120',    agency: 'test',    start: '08/06/2017',    end: '10/06/2017',    shipping: {      addresses: [        {          id: '4567',          area: 'asdaf'        },        {          id: '5678',          area: 'asdf'        }      ]    }  }]我理想的情况是将那些具有相同值（shipping.addresses.id）的对象分组到自己的对象子数组中。预期结果。我可以使用特定键（下面的代码）使用特定属性对输入数组进行分组，但我似乎无法根据数组本身的键来重新排序数组。Array.from(    data.reduce(         (acc, o) => (acc.get(o.email).push(o), acc),        new Map(data.map( o => [o.email, []] ))    ), ([key, value]) => value)

Answer 1

您可以data使用 as 键将数组缩减为一个对象shipping.addresses.id，并使用 返回一个数组Object.values()。您将需要迭代addresses每个对象的数组，并在遇到每个对象时为每个对象创建一个条目id，并推送到这些条目以获取具有相同id.

const byAddressId = Object.values(

  data.reduce((a, o) => {

    o.shipping.addresses.forEach(({id, area}) => {

      a[id]  = {...a[id] ?? {id: id, area: area, data: []}};

      a[id]['data'].push({...o});

    });

    return a;  

  }, {}));

const data = [{"email": "[e-mail]","fn": "Alex","sn": "McPherson","phone": "01233xxxxx","hours": "40","rate": "20","amount": "200","vat": "60","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": {   "addresses": [  { "id": "1234", "area": "xzy"  },  { "id": "2345", "area": "uhj"  }   ]}},{"email": "[e-mail]","fn": "Mike","sn": "Mann","phone": "01233xxxxx","hours": "50","rate": "70","amount": "500","vat": "90","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": {   "addresses": [  { "id": "1234", "area": "xzy"  },  { "id": "3456", "area": "uio"  }   ]}},{"email": "[e-mail]","fn": "Fred","sn": "Frogg","phone": "01233xxxxx","hours": "80","rate": "90","amount": "800","vat": "100","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": {   "addresses": [  { "id": "4567", "area": "asdaf"  },  { "id": "3456", "area": "uio"  }   ]}},{"email": "[e-mail]","fn": "Alex","sn": "McPherson","phone": "01233xxxxx","hours": "90","rate": "30","amount": "900","vat": "120","agency": "test","start": "08/06/2017","end": "10/06/2017","shipping": {   "addresses": [  { "id": "4567", "area": "asdaf"  },  { "id": "5678", "area": "asdf"  } ]}}];

// return array of Object.values from the accumulator

const byAddressId = Object.values(

  // reduce the data array into an object with shipping.addresses.id as keys

  data.reduce((a, o) => {

    // iterate over all addresses for each element

    o.shipping.addresses.forEach(({id, area}) => {

      // check if an id entry exists, otherwise create one

      a[id]  = {...a[id] ?? {id: id, area: area, data: []}};

      // push the object to the data array of the id object

      a[id]['data'].push({...o});

    });

    return a;  

  }, {}));

console.log(byAddressId);

话虽这么说，与问题中包含的示例map()相比，您可以使用相同的方法来节省两次调用。group by email

const byEmail = Object.values(
    data.reduce((a, o) => (a[o.email] = [...a[o.email] ?? [], {...o}], a), {}));