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;
});

另一个解决方案:

Var汽车= [ {“使”:“奥迪”,“模型”:“r8”,“年”:“2012”},{“使”:“奥迪”,“模型”:“生活费”,“年”:“2013”}, {“使”:“福特”,“模型”:“野马”,“年”:“2012”},{“使”:“福特”,“模型”:“融合”,“年”:“2015”}, {'make': 'kia'，'model': 'optima'，'year': '2012'}， ]; const reducedCars =汽车。Reduce ((acc， {make, model, year}) => ( ｛ acc, [make]: acc[make] ?[…Acc [make]， {model, year}]: [{model, year}]， ｝ ), {}); console.log (reducedCars);

同意除非经常使用这些库，否则不需要外部库。虽然有类似的解决方案，但我发现其中一些很难遵循。如果您试图理解正在发生的事情，这里有一个带有注释的解决方案的要点。

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' }, ]; /** * Groups an array of objects by a key an returns an object or array grouped by provided key. * @param array - array to group objects by key. * @param key - key to group array objects by. * @param removeKey - remove the key and it's value from the resulting object. * @param outputType - type of structure the output should be contained in. */ const groupBy = ( inputArray, key, removeKey = false, outputType = {}, ) => { return inputArray.reduce( (previous, current) => { // Get the current value that matches the input key and remove the key value for it. const { [key]: keyValue } = current; // remove the key if option is set removeKey && keyValue && delete current[key]; // If there is already an array for the user provided key use it else default to an empty array. const { [keyValue]: reducedValue = [] } = previous; // Create a new object and return that merges the previous with the current object return Object.assign(previous, { [keyValue]: reducedValue.concat(current) }); }, // Replace the object here to an array to change output object to an array outputType, ); }; console.log(groupBy(cars, 'make', true))

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);

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。