Javascript 对嵌套数组进行分组

我在

d3.js

中有一个由

nest

方法生成的数组，有人能推荐一种javascript/jQuery方法，按

类别对
值
属性中的对象进行分组吗

var data = [
  {
    "key": "0",
    "values": [
      {
        "category": "phone",
        "brand": "Sony",
        "brand_id": "0",
        "name": "item A",
        "id": "551",
        "point": "600"
      },
      {
        "category": "software",
        "brand": "Sony",
        "brand_id": "0",
        "name": "item D",
        "id": "51",
        "point": "800"
      },
      {
        "category": "phone",
        "brand": "Sony",
        "brand_id": "0",
        "name": "item E",
        "id": "52",
        "point": "1800"
      }
    ]
  },
  {
    "key": "0",
    "values": [
      {
        "category": "software",
        "brand": "Microsoft",
        "brand_id": "1",
        "name": "item B",
        "id": "54",
        "point": "800"
      },
      {
        "category": "phone",
        "brand": "Microsoft",
        "brand_id": "1",
        "name": "item B",
        "id": "60",
        "point": "200"
      },
      {
        "category": "phone",
        "brand": "Microsoft",
        "brand_id": "1",
        "name": "item T",
        "id": "30",
        "point": "600"
      },
      {
        "category": "software",
        "brand": "Microsoft",
        "brand_id": "1",
        "name": "item N",
        "id": "90",
        "point": "500"
      }
    ]
  }
];

下面是一个期望的结果：

[
  {
    "key": "0",
    "values": [
      {
        "phone": [
          {
            "category": "phone",
            "brand": "Sony",
            "brand_id": "0",
            "name": "item A",
            "id": "551",
            "point": "600"
          },
          {
            "category": "phone",
            "brand": "Sony",
            "brand_id": "0",
            "name": "item E",
            "id": "52",
            "point": "1800"
          }
        ],
        "software": [
          {
            "category": "software",
            "brand": "Sony",
            "brand_id": "0",
            "name": "item D",
            "id": "51",
            "point": "800"
          }
        ]
      }
    ]
  },
  {
    "key": "0",
    "values": [
      {
        "phone": [
          {
            "category": "phone",
            "brand": "Microsoft",
            "brand_id": "0",
            "name": "item B",
            "id": "80",
            "point": "54"
          },
          {
            "category": "phone",
            "brand": "Microsoft",
            "brand_id": "0",
            "name": "item D",
            "id": "52",
            "point": "1800"
          }
        ],
        "software": [
          {
            "category": "software",
            "brand": "Microsoft",
            "brand_id": "0",
            "name": "item G",
            "id": "41",
            "point": "800"
          },
          {
            "category": "software",
            "brand": "Microsoft",
            "brand_id": "0",
            "name": "item L",
            "id": "42",
            "point": "900"
          }
        ]
      }
    ]
  }
]

以下代码不起作用。我想这是因为
grouped[t.category]=[]
是在每个循环中定义的。上一项将被最后一项覆盖。如果我没有定义它，我会得到一个
grouped[t.category]
undefined错误：

grouped = {};
data.forEach(function(d){
  d.values.map(function(t){ 
     grouped[t.category] = [];       
     grouped[t.category].push({name:t.name,tid:t.id,point:parseInt(t.point)});
   })
})

试试这个

var grouped;

data.forEach(function(d) {
    grouped = {};

    d.values.forEach(function(t) {
         if (!grouped[t.category]) {
             grouped[t.category] = [];
         }

         grouped[t.category].push({
             name:  t.name,
             tid:   t.id, 
             point: parseInt(t.point)
         });
     })

    d.values = grouped;
});

您可以尝试以下方法：

您可以使用或类似的库来解决此问题

_.map(data, function(x){
  return _.assign({}, x, {values: _.groupBy(x.values, 'category')})
})

我熟悉：

希望它能帮助你

_.map(data, function(x){
  return _.assign({}, x, {values: _.groupBy(x.values, 'category')})
})

var desireResult = _.map(data, function(obj){
    obj.values = _.groupBy(obj.values, 'category');
    return obj;
});