let x  = [
  {
    "id": "6",
    "name": "SMD L13",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "7",
    "name": "SMD L15",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "2",
    "name": "SMD L1",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  }
];

function groupBy(array, property) {
  return array.reduce((accumulator, current) => {
    const object_property = current[property];
    delete current[property]

    let classified_element = accumulator.find(x => x.id === object_property.id);
    let other_elements = accumulator.filter(x => x.id !== object_property.id);

   if (classified_element) {
     classified_element.children.push(current)
   } else {
     classified_element = {
       ...object_property, 
       'children': [current]
     }
   }
   return [classified_element, ...other_elements];
 }, [])
}

console.log( groupBy(x, 'equipmentType') )

/* output 

[
  {
    "id": "1",
    "name": "SMD",
    "children": [
      {
        "id": "6",
        "name": "SMD L13"
      },
      {
        "id": "7",
        "name": "SMD L15"
      },
      {
        "id": "2",
        "name": "SMD L1"
      }
    ]
  }
]

*/

具有排序功能

export const groupBy = function groupByArray(xs, key, sortKey) {
      return xs.reduce(function(rv, x) {
        let v = key instanceof Function ? key(x) : x[key];
        let el = rv.find(r => r && r.key === v);

        if (el) {
          el.values.push(x);
          el.values.sort(function(a, b) {
            return a[sortKey].toLowerCase().localeCompare(b[sortKey].toLowerCase());
          });
        } else {
          rv.push({ key: v, values: [x] });
        }

        return rv;
      }, []);
    };

示例：

var state = [
    {
      name: "Arkansas",
      population: "2.978M",
      flag:
  "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },{
      name: "Crkansas",
      population: "2.978M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },
    {
      name: "Balifornia",
      population: "39.14M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/0/01/Flag_of_California.svg",
      category: "city"
    },
    {
      name: "Florida",
      population: "20.27M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Florida.svg",
      category: "airport"
    },
    {
      name: "Texas",
      population: "27.47M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Texas.svg",
      category: "landmark"
    }
  ];
console.log(JSON.stringify(groupBy(state,'category','name')));

通常，我使用Lodash JavaScript实用程序库和预先构建的groupBy（）方法。它非常容易使用，请在此处查看更多详细信息。

您可以使用Alasql JavaScript库来实现：

var data = [ { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
             { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }];

var res = alasql('SELECT Phase, Step, SUM(CAST([Value] AS INT)) AS [Value] \
                  FROM ? GROUP BY Phase, Step',[data]);

在jsFiddle尝试这个示例。

BTW：在大型阵列（100000条记录及以上）上，Alasql比Linq更快。参见jsPref中的测试。

评论：

这里我将Value放在方括号中，因为Value是SQL中的关键字我必须使用CAST（）函数将字符串值转换为数字类型。

let groupbyKeys = function(arr, ...keys) {
  let keysFieldName = keys.join();
  return arr.map(ele => {
    let keysField = {};
    keysField[keysFieldName] = keys.reduce((keyValue, key) => {
      return keyValue + ele[key]
    }, "");
    return Object.assign({}, ele, keysField);
  }).reduce((groups, ele) => {
    (groups[ele[keysFieldName]] = groups[ele[keysFieldName]] || [])
      .push([ele].map(e => {
        if (keys.length > 1) {
          delete e[keysFieldName];
        }
        return e;
    })[0]);
    return groups;
  }, {});
};

console.log(groupbyKeys(array, 'Phase'));
console.log(groupbyKeys(array, 'Phase', 'Step'));
console.log(groupbyKeys(array, 'Phase', 'Step', 'Task'));

data = [{id:1, name:'BMW'}, {id:2, name:'AN'}, {id:3, name:'BMW'}, {id:1, name:'NNN'}]
key = 'id'//try by id or name
data.reduce((previous, current)=>{
    previous[current[key]] && previous[current[key]].length != 0 ? previous[current[key]].push(current) : previous[current[key]] = new Array(current)
    return previous;
}, {})