Prototype version using ES6 as well. Basically this uses the reduce function to pass in an accumulator and current item, which then uses this to build your "grouped" arrays based on the passed in key. the inner part of the reduce may look complicated but essentially it is testing to see if the key of the passed in object exists and if it doesn't then create an empty array and append the current item to that newly created array otherwise using the spread operator pass in all the objects of the current key array and append current item. Hope this helps someone!.

Array.prototype.groupBy = function(k) {
  return this.reduce((acc, item) => ((acc[item[k]] = [...(acc[item[k]] || []), item]), acc),{});
};

const projs = [
  {
    project: "A",
    timeTake: 2,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 4,
    desc: "this is a description"
  },
  {
    project: "A",
    timeTake: 12,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 45,
    desc: "this is a description"
  }
];

console.log(projs.groupBy("project"));

提莫的答案是我会怎么做。简单的_。groupBy，并允许在分组结构中的对象中有一些重复。

然而，OP还要求删除重复的make键。如果你想从头到尾:

var grouped = _.mapValues(_.groupBy(cars, 'make'),
                          clist => clist.map(car => _.omit(car, 'make')));

console.log(grouped);

收益率:

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

如果你想使用Underscore.js来实现这个功能，请注意它的_. js版本。mapValues被称为_.mapObject。

你也可以像这样使用数组#forEach()方法:

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' }]; let newcars = {} cars.forEach(car => { newcars[car.make] ? // check if that array exists or not in newcars object newcars[car.make].push({model: car.model, year: car.year}) // just push : (newcars[car.make] = [], newcars[car.make].push({model: car.model, year: car.year})) // create a new array and push }) console.log(newcars);

对象的分组数组在typescript中:

groupBy (list: any[], key: string): Map<string, Array<any>> {
    let map = new Map();
    list.map(val=> {
        if(!map.has(val[key])){
            map.set(val[key],list.filter(data => data[key] == val[key]));
        }
    });
    return map;
});

我制定了一个基准测试不使用外部库的每个解决方案的性能。

JSBen.ch

由@Nina Scholz发布的reduce()选项似乎是最佳选项。

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)