有人知道一种方法(lodash如果可能的话)通过对象键分组对象数组,然后根据分组创建一个新的对象数组吗?例如,我有一个汽车对象数组:

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'
    },
];

我想创建一个新的汽车对象数组,由make分组:

const cars = {
    'audi': [
        {
            'model': 'r8',
            'year': '2012'
        }, {
            'model': 'rs5',
            'year': '2013'
        },
    ],

    'ford': [
        {
            'model': 'mustang',
            'year': '2012'
        }, {
            'model': 'fusion',
            'year': '2015'
        }
    ],

    'kia': [
        {
            'model': 'optima',
            'year': '2012'
        }
    ]
}

当前回答

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))

其他回答

@metakungfu answer略有不同,主要区别在于它从结果对象中省略了原始键,因为在某些情况下对象本身不再需要它,因为它现在在父对象中可用。

const groupBy = (_k, a) => a.reduce((r, {[_k]:k, ...p}) => ({
    ...r, ...{[k]: (
        r[k] ? [...r[k], {...p}] : [{...p}]
    )}
}), {});

考虑到您的原始输入对象:

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

会导致:

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

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)

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。