如何通过键对对象数组进行分组

2020/10/28 08:42 · javascript ·  · 0评论

有谁知道(如果可能的话,也用破折号)通过对象键对对象数组进行分组然后根据分组创建新的对象数组的方法吗?例如,我有一系列汽车对象:

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'
    },
];

我想创建一个新的汽车对象阵列,按以下类别分组make

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

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

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

蒂莫的答案是我会怎么做。简单_.groupBy,并允许在分组结构中的对象中进行一些重复。

但是,OP还要求删除重复的make密钥。如果您想一路走下去:

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

console.log(grouped);

产量:

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

如果要使用Underscore.js进行此操作,请注意,其版本_.mapValues称为_.mapObject

在普通的Javascript中,您可以Array#reduce与对象一起使用

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

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

您正在寻找_.groupBy()

如果需要,从对象中删除要分组的属性应该很简单:

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'},];

var grouped = _.groupBy(cars, function(car) {
  return car.make;
});

console.log(grouped);
<script src='https://cdn.jsdelivr.net/lodash/4.17.2/lodash.min.js'></script>

另外,使用ES6箭头函数可以获得更好的语法:

const grouped = _.groupBy(cars, car => car.make);

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

在es6中list按某个对象对对象a进行分组的单行版本key

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

较长的版本,可过滤掉不带的对象key

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:似乎原始问题询问如何按品牌对汽车进行分组,但在每个组中都省略了品牌。因此,没有第3方库的简短答案如下所示:

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}))

这是您自己的groupBy函数,它是对代码的概括:https : //github.com/you-dont-need/You-Dont-Need-Lodash-Underscore

function groupBy(xs, f) {
  return xs.reduce((r, v, i, a, k = f(v)) => ((r[k] || (r[k] = [])).push(v), r), {});
}

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 = groupBy(cars, (c) => c.make);
console.log(result);
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'
}].reduce((r, car) => {

  const {
    model,
    year,
    make
  } = car;

  r[make] = [...r[make] || [], {
    model,
    year
  }];

  return r;
}, {});

console.log(cars);

一个简单的for循环也是可能的

 const result = {};

 for(const {make, model, year} of cars) {
   if(!result[make]) result[make] = [];
   result[make].push({ model, year });
 }

我将在这里REAL GROUP BY使用JS Arrays示例完全相同

const inputArray = [ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
];

var outObject = inputArray.reduce(function(a, e) {
  // GROUP BY estimated key (estKey), well, may be a just plain key
  // a -- Accumulator result object
  // e -- sequentally checked Element, the Element that is tested just at this itaration

  // new grouping name may be calculated, but must be based on real value of real field
  let estKey = (e['Phase']); 

  (a[estKey] ? a[estKey] : (a[estKey] = null || [])).push(e);
  return a;
}, {});

console.log(outObject);

您可以尝试通过_.groupBy func在每次迭代调用的函数内修改对象。请注意,源数组更改了他的元素!

var res = _.groupBy(cars,(car)=>{
    const makeValue=car.make;
    delete car.make;
    return makeValue;
})
console.log(res);
console.log(cars);

创建可以重复使用的方法

Array.prototype.groupBy = function(prop) {
      return this.reduce(function(groups, item) {
        const val = item[prop]
        groups[val] = groups[val] || []
        groups[val].push(item)
        return groups
      }, {})
    };

然后,您可以在下面按任何条件分组

const groupByMake = cars.groupBy('make');
        console.log(groupByMake);

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'
    },
];
  //re-usable method
Array.prototype.groupBy = function(prop) {
	  return this.reduce(function(groups, item) {
		const val = item[prop]
		groups[val] = groups[val] || []
		groups[val].push(item)
		return groups
	  }, {})
	};
  
 // initiate your groupBy. Notice the recordset Cars and the field Make....
  const groupByMake = cars.groupBy('make');
		console.log(groupByMake);
    
    //At this point we have objects. You can use Object.keys to return an array

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));

原型版本也使用ES6。基本上,这使用reduce函数来传递累加器和当前项,然后使用该函数根据传入的键来构建“分组”数组。reduce的内部看起来可能很复杂,但实际上它正在测试以查看传入对象的键是否存在,如果不存在,则创建一个空数组并将当前项附加到该新创建的数组,否则使用散点图运算符传入当前键数组的所有对象并追加当前项。希望这对某人有帮助!

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"));

您还可以使用如下array#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);
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');

我喜欢编写无依赖/复杂性的纯简单js脚本。

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)

我做了一个基准测试,以测试每个不使用外部库的解决方案的性能。

JSBench

reduce()@Nina Scholz发布选项似乎是最佳选择。

我喜欢@metakunfu答案,但它没有提供确切的预期输出。这是更新的版本,在最终的JSON有效负载中消除了“ make”。

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'
    },
];

result = cars.reduce((h, car) => Object.assign(h, { [car.make]:( h[car.make] || [] ).concat({model: car.model, year: car.year}) }), {})

console.log(JSON.stringify(result));

输出:

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

只需尝试一下,对我来说效果很好。

let grouped = _.groupBy(cars, 'make');

使用lodash / fp,您可以_.flow()按键创建第一个组的功能,然后映射每个组,并从每个项目中省略一个键:

const { flow, groupBy, mapValues, map, omit } = _;

const groupAndOmitBy = key => flow(
  groupBy(key),
  mapValues(map(omit(key)))
);

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 groupAndOmitMake = groupAndOmitBy('make');

const result = groupAndOmitMake(cars);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
<script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>

如果您不想在所有字段中输入内容,请以@Jonas_Wilms答案为基础:

    var result = {};

    for ( let { first_field, ...fields } of your_data ) 
    { 
       result[first_field] = result[first_field] || [];
       result[first_field].push({ ...fields }); 
    }

我没有做任何基准测试,但我相信使用for循环也比此答案中建议的效率更高

const reGroup = (list, key) => {
    const newGroup = {};
    list.forEach(item => {
        const newItem = Object.assign({}, item);
        delete newItem[key];
        newGroup[item[key]] = newGroup[item[key]] || [];
        newGroup[item[key]].push(newItem);
    });
    return newGroup;
};
const animals = [
  {
    type: 'dog',
    breed: 'puddle'
  },
  {
    type: 'dog',
    breed: 'labradoodle'
  },
  {
    type: 'cat',
    breed: 'siamese'
  },
  {
    type: 'dog',
    breed: 'french bulldog'
  },
  {
    type: 'cat',
    breed: 'mud'
  }
];
console.log(reGroup(animals, 'type'));
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'
  },
];

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

在打字稿中将对象分组数组如下:

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;
});

这是另一种解决方案。按照要求。

我想制作一个新的汽车对象阵列,按make分组:

function groupBy() {
  const key = 'make';
  return cars.reduce((acc, x) => ({
    ...acc,
    [x[key]]: (!acc[x[key]]) ? [{
      model: x.model,
      year: x.year
    }] : [...acc[x[key]], {
      model: x.model,
      year: x.year
    }]
  }), {})
}

输出:

console.log('Grouped by make key:',groupBy())

这是一个受Java中Collectors.groupingBy()启发的解决方案:

function groupingBy(list, keyMapper) {
  return list.reduce((accummalatorMap, currentValue) => {
    const key = keyMapper(currentValue);
    if(!accummalatorMap.has(key)) {
      accummalatorMap.set(key, [currentValue]);
    } else {
      accummalatorMap.set(key, accummalatorMap.get(key).push(currentValue));
    }
    return accummalatorMap;
  }, new Map());
}

这将提供一个Map对象。

// Usage

const carMakers = groupingBy(cars, car => car.make);
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