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

letfinaldata=[]

let data =[{id:1,name:"meet"},{id:2,name:"raj"},{id:1,name:"hari"},{id:3,name:"hari"},{id:2,name:"ram"}]

data = data.map((item)=> 
{
    return {...item,
        name: [item.name]
    }
}) // Converting the name key from string to array


let temp = [];

for(let i =0 ;i<data.length;i++)
{
    const index = temp.indexOf(data[i].id) // Checking if the object id is already present
    if(index>=0)
    {
        letfinaldata[index].name = [...letfinaldata[index].name,...data[i].name] // If present then append the name to the name of that object
    }
    else{
        temp.push(data[i].id); // Push the checked object id
        letfinaldata.push({...data[i]}) // Push the object
    }
}

console.log(letfinaldata)

输出

[ { id: 1, name: [ 'meet', 'hari' ] },
  { id: 2, name: [ 'raj', 'ram' ] },
  { id: 3, name: [ 'hari' ] } ]

其他回答

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}))

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))

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。

letfinaldata=[]

let data =[{id:1,name:"meet"},{id:2,name:"raj"},{id:1,name:"hari"},{id:3,name:"hari"},{id:2,name:"ram"}]

data = data.map((item)=> 
{
    return {...item,
        name: [item.name]
    }
}) // Converting the name key from string to array


let temp = [];

for(let i =0 ;i<data.length;i++)
{
    const index = temp.indexOf(data[i].id) // Checking if the object id is already present
    if(index>=0)
    {
        letfinaldata[index].name = [...letfinaldata[index].name,...data[i].name] // If present then append the name to the name of that object
    }
    else{
        temp.push(data[i].id); // Push the checked object id
        letfinaldata.push({...data[i]}) // Push the object
    }
}

console.log(letfinaldata)

输出

[ { id: 1, name: [ 'meet', 'hari' ] },
  { id: 2, name: [ 'raj', 'ram' ] },
  { id: 3, name: [ 'hari' ] } ]
function groupBy(data, property) {
  return data.reduce((acc, obj) => {
    const key = obj[property];
    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(obj);
    return acc;
  }, {});
}
groupBy(people, 'age');