根据@Jonas_Wilms的回答，如果你不想输入所有的字段:

    var result = {};

    for ( let { first_field, ...fields } of your_data ) 
    { 
       result[first_field] = result[first_field] || [];
       result[first_field].push({ ...fields }); 
    }

我没有做任何基准测试，但我相信使用for循环会比这个答案中建议的任何方法都更有效。

这是一个通用函数，将返回Array groupBy自己的键。

const getSectionListGroupedByKey = < T > ( property: keyof T, List: Array < T > ): Array < { title: T[keyof T];data: Array < T > } > => { const sectionList: Array < { title: T[keyof T];data: Array < T > } > = []; if (!property || !List ? .[0] ? .[property]) { return []; } const groupedTxnListMap: Map < T[keyof T], Array < T >> = List.reduce((acc, cv) => { const keyValue: T[keyof T] = cv[property]; if (acc.has(keyValue)) { acc.get(keyValue) ? .push(cv); } else { acc.set(keyValue, [cv]); } return acc; }, new Map < T[keyof T], Array < T >> ()); groupedTxnListMap.forEach((value, key) => { sectionList.push({ title: key, data: value }); }); return sectionList; }; // Example const cars = [{ 'make': 'audi', 'model': 'r8', 'year': '2012' }, { 'make': 'audi', 'model': 'rs5', 'year': '2013' }, { 'make': 'ford', 'model': 'mustang', 'year': '2012' }, { 'make': 'ford', 'model': 'fusion', 'year': '2015' }, { 'make': 'kia', 'model': 'optima', 'year': '2012' }, ]; const result = getSectionListGroupedByKey('make', cars); console.log('result: ', result)

我用REAL GROUP BY作为JS数组的例子和这个任务完全一样

const inputArray = [ { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } ]; var outObject = inputArray.reduce(function(a, e) { // GROUP BY estimated key (estKey), well, may be a just plain key // a -- Accumulator result object // e -- sequentally checked Element, the Element that is tested just at this itaration // new grouping name may be calculated, but must be based on real value of real field let estKey = (e['Phase']); (a[estKey] ? a[estKey] : (a[estKey] = null || [])).push(e); return a; }, {}); console.log(outObject);

这是另一个解决方案。按照要求。

我想创建一个新的汽车对象数组，由make分组:

function groupBy() {
  const key = 'make';
  return cars.reduce((acc, x) => ({
    ...acc,
    [x[key]]: (!acc[x[key]]) ? [{
      model: x.model,
      year: x.year
    }] : [...acc[x[key]], {
      model: x.model,
      year: x.year
    }]
  }), {})
}

输出:

console.log('Grouped by make key:',groupBy())

const reGroup = (list, key) => {
    const newGroup = {};
    list.forEach(item => {
        const newItem = Object.assign({}, item);
        delete newItem[key];
        newGroup[item[key]] = newGroup[item[key]] || [];
        newGroup[item[key]].push(newItem);
    });
    return newGroup;
};

const animals = [
  {
    type: 'dog',
    breed: 'puddle'
  },
  {
    type: 'dog',
    breed: 'labradoodle'
  },
  {
    type: 'cat',
    breed: 'siamese'
  },
  {
    type: 'dog',
    breed: 'french bulldog'
  },
  {
    type: 'cat',
    breed: 'mud'
  }
];
console.log(reGroup(animals, 'type'));

const cars = [
  {
      'make': 'audi',
      'model': 'r8',
      'year': '2012'
  }, {
      'make': 'audi',
      'model': 'rs5',
      'year': '2013'
  }, {
      'make': 'ford',
      'model': 'mustang',
      'year': '2012'
  }, {
      'make': 'ford',
      'model': 'fusion',
      'year': '2015'
  }, {
      'make': 'kia',
      'model': 'optima',
      'year': '2012'
  },
];

console.log(reGroup(cars, 'make'));

下面是您自己的groupBy函数，它是来自https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore的代码的泛化

函数groupBy(xs, f) { 返回x。减少((r, v, i, a、k = f (v)) = > ((r [k] | | (r [k] = [])) .push (v), r), {}); ｝ Const cars = [{make: 'audi'，型号:'r8'，年份:'2012'}，{make: 'audi'，型号:'rs5'，年份:'2013'}，{make: 'ford'，型号:'mustang'，年份:'2012'}，{make: 'ford'，型号:'fusion'，年份:'2015'}，{make: 'kia'，型号:'optima'，年份:'2012'}]; const result = groupBy(cars， (c) => c.make); console.log(结果);