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

我用REAL GROUP BY作为JS数组的例子和这个任务完全一样

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);

其他回答

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');

您可以尝试在调用per iteration的函数中修改对象_。groupBy func。 注意,源数组改变了它的元素!

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

我喜欢@metakunfu的答案,但它并没有提供预期的输出。 下面是在最终的JSON有效负载中去除“make”的更新。

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"
      }
   ]
}

下面是一个受到Java中的collections . groupingby()启发的解决方案:

function groupingBy(list, keyMapper) { 返回列表。reduce((accummalatorMap, currentValue) => { const key = keyMapper(currentValue); 如果(! accummalatorMap.has(键)){ accummalatorMap。集(关键,[currentValue]); }其他{ accummalatorMap。集(键,accummalatorMap.get(关键).push (currentValue)); } 返回accummalatorMap; }, new Map()); }

这将给出一个Map对象。

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

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;
});