Question

我有3组对象，其中1组可以通过唯一键映射到相应的组。可以通过不同的键将同一组映射到第三组。我需要能够将所有这些映射到一个新的组合集中。这些集合都具有不同的属性。

唯一计数[用户] [部门]
许多发票[p]

每个唯一的[用户]都属于（comitID）找到的特定[部门]，但是同一[用户]可以具有许多[发票]。如果您熟悉关系数据库，则为一对多字段

const users= [ // Unique Entries
    {name:'user1', comitId: 'aa1'},
    {name:'user2', comitId: 'aa2'}
]

const sector= [ // Unique Entries
    {comitID: 'aa1', department: 'finance'},
    {comitID: 'aa2', department: 'marketing'},
    {comitID: 'aa3', department: 'engineering'}
]

const invoices= [ // Multiple Entries
    {name: 'user1' : statementDate: '2/1/2019'},
    {name: 'user1' : statementDate: '2/14/2019'},
    {name: 'user2' : statementDate: '2/1/2019'}
]

新设置应如下所示。无法包含对帐单日期的列表，它们每个都需要是一个新对象。

const results = [
    {name: 'user1', comitId: 'aa1', department: 'finance', statementDate: '2/1/2019'},
    {name: 'user1', comitId: 'aa1', department: 'finance', statementDate: '2/14/2019'},
    {name: 'user2', comitId: 'aa2', department: 'marketing', statementDate: '2/1/2019'}
]

我一直在Excel中使用vlookups和公式进行尝试。这些文件的不重复计数通常为10k，发票为40k。

Answer 1

您可以在invoices和Array.map()上使用Array.find()，以在users然后sectors中获得相应的条目：

const users = [ // Unique Entries
  {name:'user1', comitId: 'aa1'},
  {name:'user2', comitId: 'aa2'}
];

const sectors = [ // Unique Entries
  {comitID: 'aa1', department: 'finance'},
  {comitID: 'aa2', department: 'marketing'},
  {comitID: 'aa3', department: 'engineering'}
];

const invoices = [ // Multiple Entries
  {name: 'user1', statementDate: '2/1/2019'},
  {name: 'user1', statementDate: '2/14/2019'},
  {name: 'user2', statementDate: '2/1/2019'}
];

const result = invoices.map(invoice => {
  const user = users.find(u => u.name === invoice.name) || {};
  const sector = sectors.find(s => s.comitID === user.comitId) || {};
  return { ...invoice, ...sector };
});

console.log(result);

Answer 2

我建议您只是遍历发票，并使用具有唯一条目的集合中的条目充实条目。

这样的事情（没有测试代码，但我希望你能理解这个想法）

const data = invoices
    .map(entry => {...entry, ...{
        users.find(user => user.name === entry.name)
            .map(user => {...user,
                sector.find(sec=> sec.comitID === user.commitID)
             })
        }}
     )

当您首先从集合中创建地图，然后仅查找联接属性而不是搜索它们时，您可以提高速度

const userMap = users.reduce((map, user) => {...map, ...{user.name: user}}, {})
const sectorMap = sector.reduce((map, sec) => {...map, ...{sector.comitID: sec}}), {})

const data = invoices.map(invoice => {...invoice, ...userMap[invoice.name], ...sector[userMap[invoice.name].comitID]})

Answer 3

您可以将用户和部门项目移到地图中，并在必要时使用此对象。

const
    users = [{ name: 'user1', comitId: 'aa1' }, { name: 'user2', comitId: 'aa2' }],
    sector = [{ comitID: 'aa1', department: 'finance' }, { comitID: 'aa2', department: 'marketing' }, { comitID: 'aa3', department: 'engineering' }],
    invoices = [{ name: 'user1', statementDate: '2/1/2019'}, { name: 'user1', statementDate: '2/14/2019' }, { name: 'user2', statementDate: '2/1/2019' }],
    setMap = k => (m, o) => m.set(o[k], o),
    userMap = users.reduce(setMap('name'), new Map),
    sectorMap = sector.reduce(setMap('comitID'), new Map),
    result = invoices.map(({ name, statementDate }) => {
        var { comitId } = userMap.get(name),
            { department } = sectorMap.get(comitId);
        return { name, comitId, department, statementDate };
    });

console.log(result);

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Answer 4

这是一个可行的基本脚本。

const users = [ // Unique Entries
    {name:'user1', comitId: 'aa1'},
    {name:'user2', comitId: 'aa2'}
]

const sectors = [ // Unique Entries
    {comitID: 'aa1', department: 'finance'},
    {comitID: 'aa2', department: 'marketing'},
    {comitID: 'aa3', department: 'engineering'}
]

const invoices = [ // Multiple Entries
    {name: 'user1', statementDate: '2/1/2019'},
    {name: 'user1', statementDate: '2/14/2019'},
    {name: 'user2', statementDate: '2/1/2019'}
]

sectors.forEach(sector => {
    const user = users.find(user => sector.comitID === user.comitId);
    if (user) {
        user.department = sector.department;
    }
});

const results = invoices.map(invoice => {
    const user = users.find(user => invoice.name === user.name);
    return Object.assign({}, user, { statementDate: invoice.statementDate });
});

console.log(results);

Answer 5

您可以使用map和filter

const users = [ // Unique Entries
  {
    name: 'user1',
    comitId: 'aa1'
  },
  {
    name: 'user2',
    comitId: 'aa2'
  }
]

const sector = [ // Unique Entries
  {
    comitID: 'aa1',
    department: 'finance'
  },
  {
    comitID: 'aa2',
    department: 'marketing'
  },
  {
    comitID: 'aa3',
    department: 'engineering'
  }
]

const invoices = [ // Multiple Entries
  {
    name: 'user1',
    statementDate: '2/1/2019'
  },
  {
    name: 'user1',
    statementDate: '2/14/2019'
  },
  {
    name: 'user2',
    statementDate: '2/1/2019'
  }
]
let newArray = invoices.map(function(item) {
  // this value will be use to find match between users & sectors
  let cId = users.filter(user => user.name === item.name)[0].comitId;
  return {
    name: item.name,
    statementDate: item.statementDate,
    comitId: cId,
    department: sector.filter(sector => sector.comitID === cId)[0].department
  }
});

console.log(newArray)

通过JavaScript中的键映射3个对象数组

5 个答案: