JS（ES6）：基于id和连接子数组合并数组

Question

我有两个数组，看起来像这样：

const persons = [
  {
    id: 1,
    name: 'Peter',
    job: 'Programmer'
  },
  {
    id: 2,
    name: 'Jeff',
    job: 'Architect'
  },
];

const salaries = [
  {
    id: 1,
    salary: 3000,
    departments: ['A', 'B'] 
  },
  {
    id: 1,
    salary: 4000,
    departments: ['A', 'C']
  },
  {
    id: 2,
    salary: 4000,
    departments: ['C', 'D']
  }
];

现在我需要以某种方式将这个数组合并为一个，这样每个id只存在一次。 应该替换相同的键，除了它是一个数组，然后我希望它们添加/连接。 所以期望的结果看起来像这样：

const result = [
  {
    id: 1,
    name: 'Peter',
    job: 'Programmer',
    salary: 4000,
    departments: ['A', 'B', 'C'] 
  },
  {
    id: 2,
    name: 'Jeff',
    job: 'Architect',
    salary: 4000,
    departments: ['C', 'D']
  }
];

我已经尝试过了：

// double id's, arrays get replaced
Object.assign({}, persons, salaries)

// loadsh: double id's, arrays get concatenated
_.mergeWith(persons, salaries, (objValue, srcValue) => {
    if (_.isArray(objValue)) {
        return objValue.concat(srcValue);
    }
});

// gives me a map but replaces arrays
new Map(salaries.map(x => [x.id, x])

有谁知道如何做到这一点？

Answer 1

您可以使用Map并迭代所有属性并检查类型以向数组添加唯一值。

 var persons = [{ id: 1, name: 'Peter', job: 'Programmer' }, { id: 2, name: 'Jeff', job: 'Architect' }], salaries = [{ id: 1, salary: 3000, departments: ['A', 'B'] }, { id: 1, salary: 4000, departments: ['A', 'C'] }, { id: 2, salary: 4000, departments: ['C', 'D'] }], result = Array.from( salaries .reduce( (m, o) => { var t = m.get(o.id) || {}; Object.keys(o).forEach(k => { if (Array.isArray(o[k])) { t[k] = t[k] || []; o[k].forEach(v => t[k].includes(v) || t[k].push(v)); } else if (t[k] !== o[k]) { t[k] = o[k]; } }); return m; }, persons.reduce((m, o) => m.set(o.id, Object.assign({}, o)), new Map) ) .values() ); console.log(result); 

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Answer 2

您可以连接数组，而不是使用Array.reduce()和Map组合具有相同id所有项。

要组合具有相同id的对象，请从Map获取对象。 使用Array.forEach()迭代新的Object.entries() Array.forEach() 。 检查现有值是否为数组，如果不是则赋值。 如果是数组，请组合数组，并使用带数组传播的Set使项目唯一。

要将Map转换回数组，可以传播Map.values()迭代器。

 const persons = [{"id":1,"name":"Peter","job":"Programmer"},{"id":2,"name":"Jeff","job":"Architect"}]; const salaries = [{"id":1,"salary":3000,"departments":["A","B"]},{"id":1,"salary":4000,"departments":["A","C"]},{"id":2,"salary":4000,"departments":["C","D"]}]; const result = [...persons.concat(salaries) .reduce((r, o) => { r.has(o.id) || r.set(o.id, {}); const item = r.get(o.id); Object.entries(o).forEach(([k, v]) => item[k] = Array.isArray(item[k]) ? [...new Set([...item[k], ...v])] : v ); return r; }, new Map()).values()]; console.log(result); 

Answer 3

您可以使用map() ， filter() ， reduce() ， Object.assign()和Spread syntax来实现所需的结果。

DEMO

 const persons = [{ id: 1, name: 'Peter', job: 'Programmer' }, { id: 2, name: 'Jeff', job: 'Architect' }], salaries = [{ id: 1, salary: 3000, departments: ['A', 'B'] }, { id: 1, salary: 4000, departments: ['A', 'C'] }, { id: 2, salary: 4000, departments: ['C', 'D'] }]; let output = persons.map(obj => { let filter = salaries.filter(v => v.id == obj.id); if (filter) { let departments = filter.reduce((r, v) => [...v.departments, ...r], []); Object.assign(obj, { salary: filter[filter.length - 1].salary, departments: departments.filter((item, pos) => departments.indexOf(item) == pos).sort() }); } return obj; }); console.log(output)