Question

我正在尝试合并共享相同模块密钥值的对象，即“信用卡”。我需要为每个唯一的模块返回一个对象，其中包含一系列不同的公司。 id和prod键并不重要。

这是我需要转换的对象：

const data = [
   {module: "credit_cards", id: "1", company: "ABC", prod: "1"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "credit_cards", id: "3", company: "EFG", prod: "2"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "mortgages", id: "4", company: "EFG", prod: "3"}
]

我需要返回以下内容：

const result = [
{module: "credit_cards", company: ["ABC", "XYZ", "EFG"]}, 
{module: "mortgages", company: ["EFG"]}
]

到目前为止，我已经尝试了此功能：

const result = data.reduce(function(r, e) {
  return Object.keys(e).forEach(function(k) {
    if(!r[k]) r[k] = [].concat(e[k])
    else r[k] = r[k].concat(e[k])
  }), r
}, {})

但是它只是将所有键/值连接在一起...

任何帮助都会很棒，谢谢：）

Answer 1

您可以使用reduce将数组汇总为Map。将Set用作唯一值。

const data = [{"module":"credit_cards","id":"1","comapany":"ABC","prod":"1"},{"module":"credit_cards","id":"2","comapany":"XYZ","prod":"2"},{"module":"credit_cards","id":"3","comapany":"EFG","prod":"2"},{"module":"credit_cards","id":"2","comapany":"XYZ","prod":"2"},{"module":"mortgages","id":"4","comapany":"EFG","prod":"3"}]

const result = Array.from(data.reduce((c, {module,comapany}) => {
  if (!c.has(module)) c.set(module, new Set);
  c.get(module).add(comapany);
  return c;
}, new Map)).map(([module, comapany]) => ({module,comapany: [...comapany]}));

console.log(result);

Answer 2

简化为以module索引的对象，然后获取该对象的值。因为只需要唯一的company值，所以请先使用Set，然后再将所有项的company集转换回数组：

const data = [
   {module: "credit_cards", id: "1", company: "ABC", prod: "1"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "credit_cards", id: "3", company: "EFG", prod: "2"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "mortgages", id: "4", company: "EFG", prod: "3"}
]

const result = Object.values(data.reduce(
  (a, { module, company }) => {
    if (!a[module]) a[module] = { module, company: new Set() };
    a[module].company.add(company);
    return a;
  },
  {}
));

result.forEach((item) => {
  item.company = [...item.company];
});
console.log(result);

Answer 3

您可以先使用Array.reduce()通过Set上的companies键将module分组。第二步，您可以对生成的对象array.Map()使用.entries()以获得最终的期望结果：

const data = [
   {module: "credit_cards", id: "1", company: "ABC", prod: "1"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "credit_cards", id: "3", company: "EFG", prod: "2"}, 
   {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, 
   {module: "mortgages", id: "4", company: "EFG", prod: "3"}
];

let res = data.reduce((acc, {module, company}) =>
{
    acc[module] = acc[module] || new Set();
    acc[module].add(company);
    return acc;
}, {})

res = Object.entries(res).map(
    ([module, companies]) => ({module, company: [...companies]})
);

console.log(res);

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Answer 4

这里是使用Array.reduce并进行内部检查以确保公司列表唯一且没有重复的另一种方法。

const data = [ {module: "credit_cards", id: "1", company: "ABC", prod: "1"}, {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, {module: "credit_cards", id: "3", company: "EFG", prod: "2"}, {module: "credit_cards", id: "2", company: "XYZ", prod: "2"}, {module: "mortgages", id: "4", company: "EFG", prod: "3"} ]

let result = data.reduce((r,{module, company}) => {
  r[module] = r[module] || { module, company: [] }
  if(!r[module].company.includes(company))
    r[module].company.push(company)
  return r
}, {})

console.log(Object.values(result))

Answer 5

   const data = [
  {module: 'credit_cards', id: '1', company: 'ABC', prod: '1'},
  {module: 'credit_cards', id: '2', company: 'XYZ', prod: '2'},
  {module: 'credit_cards', id: '3', company: 'EFG', prod: '2'},
  {module: 'credit_cards', id: '2', company: 'XYZ', prod: '2'},
  {module: 'mortgages', id: '4', company: 'EFG', prod: '3'}
  ];



 const result = data.reduce((array, item) => {
     const id = array.findIndex(el => el.module === item.module);
     if (id > -1) {
       array[id] = {...array[id], company : [...array[id].company, item.company]};
       return array;
     }
     return [...array, {...item, company : [ item.company]}];
    }, []);

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