对于key可以为null的情况，我们希望将它们分组为其他

var 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'},
            {'make':'kia','model':'optima','year':'2033'},
            {'make':null,'model':'zen','year':'2012'},
            {'make':null,'model':'blue','year':'2017'},

           ];


 result = cars.reduce(function (r, a) {
        key = a.make || 'others';
        r[key] = r[key] || [];
        r[key].push(a);
        return r;
    }, Object.create(null));

const groupBy = (array, callback) => {
  const groups = {};
  
  array.forEach((element) => {
    const groupName = callback(element);
    if (groupName in groups) {
      groups[groupName].push(element);
    } else {
      groups[groupName] = [element];
    }
  });
  
  return groups;
};

或者花哨的裤子:

(() => {
  Array.prototype.groupBy = function (callback) {
    const groups = {};
    this.forEach((element, ...args) => {
      const groupName = callback(element, ...args);
      if (groupName in groups) {
        groups[groupName].push(element);
      } else {
        groups[groupName] = [element];
      }
    });

    return groups;
  };
})();

const res = [{ name: 1 }, { name: 1 }, { name: 0 }].groupBy(({ name }) => name);

// const res = {
//   0: [{name: 0}],
//   1: [{name: 1}, {name: 1}]
// }

这是一个MDN阵列的填充。groupBy函数。

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))

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;
});

我喜欢写它没有依赖/复杂性，只是纯粹的简单js。

const mp = {} const cars = [ { model: 'Imaginary space craft SpaceX model', year: '2025' }, { 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' } ] cars.forEach(c => { if (!c.make) return // exit (maybe add them to a "no_make" category) if (!mp[c.make]) mp[c.make] = [{ model: c.model, year: c.year }] else mp[c.make].push({ model: c.model, year: c.year }) }) console.log(mp)

完全没有理由下载第三方库来解决这个简单的问题，就像上面的解决方案所建议的那样。

在es6中按特定键对对象列表进行分组的单行版本:

const groupByKey = (list, key) => list.reduce((hash, obj) => ({...hash, [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}), {})

较长的版本过滤掉没有键的对象:

function groupByKey(array, key) { return array .reduce((hash, obj) => { if(obj[key] === undefined) return hash; return Object.assign(hash, { [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}) }, {}) } var 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(groupByKey(cars, 'make'))

注意:原来的问题似乎是问如何按制造商对汽车进行分组，但省略了每组中的制造商。因此，如果没有第三方库，简单的回答是这样的:

const groupByKey = (list, key, {omitKey=false}) => list.reduce((hash, {[key]:value, ...rest}) => ({...hash, [value]:( hash[value] || [] ).concat(omitKey ? {...rest} : {[key]:value, ...rest})} ), {}) var 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(groupByKey(cars, 'make', {omitKey:true}))