如何从对象列表中分组和收集明确引用的特定值，每个值都通过它的键？

Question

我有一个像

const dd = [{
  keyone: "test",
  two: "you",
  three: 'op',
}, {
  keyone: "youuuu",
  two: "ttt",
  three: 'op',
}, {
  keyone: "operation",
  two: "test",
  three: 'op',
}];

我希望能够将 keyone 和两个拉出到 object 中，如下所示

const obj = { keyone: ['test', 'youuuu', 'operation'], two: ['you', 'ttt', 'test']}

我已经使用两张地图并将它们组合起来实现了这一点，但如果可能的话，我只想使用一个循环。

编辑：

我目前正在使用解构来提取值：

const mapOne = dd.map(({ keyone }) => keyone);
const mapTwo = dd.map(({ two }) => two);
const test = {
  keyone: mapOne,
  two: mapTwo,
};

Answer 1

你可以这样做：

 const dd = [ { keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' }]; const result = dd.reduce((prev, acc, arr ) => { if (acc['keyone']) { prev['keyone'].push(acc['keyone']) } if (acc['two']) { prev['two'].push(acc['two']) } return prev; }, {keyone: [], two: []}); console.log(result);

Answer 2

您可以使用reduce对数组进行一次迭代，并在每次迭代中通过Object.keys和forEach对 object 进行迭代：

 const data = [ { keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' }]; const targetKeys = ['keyone', 'two'] const result = data.reduce((acc, obj)=> { Object.keys(obj).forEach(k => { if(targetKeys.includes(k)) (acc[k] = acc[k] || []).push(obj[k]) }); return acc; }, {}); console.log(result);

Answer 3

您可以获取一组键并将值组合在一起。

 const data = [{ keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' }], keys = ['keyone', 'two'], grouped = data.reduce((r, o) => { keys.forEach(k => (r[k]??= []).push(o[k])); return r; }, {}); console.log(grouped);

Answer 4

一个通用实现，因此可重用和可配置的 function，它完全符合 OP 的要求，将接近下一个提供的示例代码......

 function groupAndCollectSpecifcEntriesOnly(collector, item) { const { keyList, result } = collector; keyList.forEach(key => { if (item.hasOwnProperty(key)) { (result[key]??= []).push(item[key]) } }); return collector; } const sampleData = [{ keyone: "test", two: "you", three: 'op', }, { keyone: "youuuu", two: "ttt", three: 'op', }, { keyone: "operation", two: "test", three: 'op', }]; console.log( sampleData.reduce(groupAndCollectSpecifcEntriesOnly, { keyList: ['keyone', 'two'], result: {}, }).result ); console.log( sampleData.reduce(groupAndCollectSpecifcEntriesOnly, { keyList: ['three', 'keyone'], result: {}, }).result );

 .as-console-wrapper { min-height: 100%;important: top; 0; }

Answer 5

 const dd = [{ keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' } ]; const pullOut = (...keys) => { return keys.map(key => { const val = dd.reduce((pV, cV, cI) => { pV.push(cV[key]); return pV; }, []); return { [key]: val }; }); }; console.log(pullOut('keyone', 'two'));

Answer 6

你要求一个循环：

const test = { keyone:[], two:[] }
dd.map( obj => {
  if ( obj.hasOwnProperty( "keyone" )){
    test[ "keyone" ].push( obj[ "keyone" ])
  }
  if ( obj.hasOwnProperty( "two" )){
    test[ "two" ].push( obj[ "two" ])
  }
})

Answer 7

我对 JavaScript 相当陌生，也许我的解决方案不是最好的。 我创建了一个 function，您可以在其中添加要从中获取信息的键的数量作为可选参数。 我假设您不知道键的名称。

const dd = [
  {
    keyone: "test",
    two: "you",
    three: "op",
  },
  {
    keyone: "youuuu",
    two: "ttt",
    three: "op",
  },
  {
    keyone: "operation",
    two: "test",
    three: "op",
  },
];

getterInformation(dd);

function getterInformation(object, keyCount = Object.keys(object).length) {
  let objectWithDesiredKey = {};

  for (let index = 0; index < keyCount; index++) {
    objectWithDesiredKey[`${Object.keys(object[0])[index]}`] = [];
  }

  object.forEach((items) => {
    for (const key in items) {
      if (key in objectWithDesiredKey) {
        objectWithDesiredKey[key].push(items[key]);
      }
    }
  });

  console.log(objectWithDesiredKey);
}

Answer 8

你要求一个循环，但我免费给你三个；-)

 function objReducer( arrayOfObjects, ...keys ) { const result = {} keys.forEach( key => { const arrayOfKeyValues = [] result[key] = arrayOfKeyValues }) arrayOfObjects.forEach( obj => { keys.forEach( key => { if ( obj.hasOwnProperty( key )) { result[key].push( obj[key] ) } }) }) return result } const dd = [{ keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' } ] console.log( objReducer( dd, 'keyone', 'two' ))

Answer 9

我认为这是您正在寻找的答案：
（使用object 解构从每个 object 中提取“keyone”和“two”并将它们push送到同一个map循环中）

 const data = [ { keyone: "test", two: "you", three: 'op' }, { keyone: "youuuu", two: "ttt", three: 'op' }, { keyone: "operation", two: "test", three: 'op' }]; const test = { keyone:[], two:[] } data.map(({ keyone, two }) => { test.keyone.push( keyone ); test.two.push( two ) }) console.log( test )