javascript - 按名称组织对象

Question

获取“toursByHotels”下方的对象数组 — 我想获取此数据并获取具有相同名称(酒店名称，即 Marriott 等)的每个对象的旅游总和。

在 d3 中有没有比必须创建 5 个具有相同名称的新数组，然后对这些数组求​​和更简单的方法？

var toursByHotel = [
        {
          "name": "Marriott",
          "month": 1,
          "tours": 10
        },
        {
          "name": "Marriott",
          "month": 2,
          "tours": 15
        },
        {
          "name": "Marriott",
          "month": 3,
          "tours": 8
        },
        {
          "name": "Marriott",
          "month": 4,
          "tours": 12
        },
        {
          "name": "Marriott",
          "month": 5,
          "tours": 18
        },
        {
          "name": "Marriott",
          "month": 6,
          "tours": 25
        },
        {
          "name": "Marriott",
          "month": 7,
          "tours": 40
        },
        {
          "name": "Marriott",
          "month": 8,
          "tours": 33
        },
        {
          "name": "Marriott",
          "month": 9,
          "tours": 25
        },
        {
         "name": "Marriott",
          "month": 10,
          "tours": 21
        },
        {
         "name": "Marriott",
          "month": 11,
          "tours": 18
        },
        {
         "name": "Marriott",
          "month": 12,
          "tours": 14
        },
        {
          "name": "Springhill",
          "month": 1,
          "tours": 10
        },
        {
          "name": "Springhill",
          "month": 2,
          "tours": 15
        },
        {
          "name": "Springhill",
          "month": 3,
          "tours": 8
        },
        {
          "name": "Springhill",
          "month": 4,
          "tours": 12
        },
        {
          "name": "Springhill",
          "month": 5,
          "tours": 18
        },
        {
          "name": "Springhill",
          "month": 6,
          "tours": 25
        },
        {
          "name": "Springhill",
          "month": 7,
          "tours": 40
        },
        {
          "name": "Springhill",
          "month": 8,
          "tours": 33
        },
        {
          "name": "Springhill",
          "month": 9,
          "tours": 25
        },
        {
         "name": "Springhill",
          "month": 10,
          "tours": 21
        },
        {
         "name": "Springhill",
          "month": 11,
          "tours": 18
        },
        {
         "name": "Springhill",
          "month": 12,
          "tours": 14
        },
        {
          "name": "Residence",
          "month": 1,
          "tours": 10
        },
        {
          "name": "Residence",
          "month": 2,
          "tours": 15
        },
        {
          "name": "Residence",
          "month": 3,
          "tours": 8
        },
        {
          "name": "Residence",
          "month": 4,
          "tours": 12
        },
        {
          "name": "Residence",
          "month": 5,
          "tours": 18
        },
        {
          "name": "Residence",
          "month": 6,
          "tours": 25
        },
        {
          "name": "Residence",
          "month": 7,
          "tours": 40
        },
        {
          "name": "Residence",
          "month": 8,
          "tours": 33
        },
        {
          "name": "Residence",
          "month": 9,
          "tours": 25
        },
        {
         "name": "Residence",
          "month": 10,
          "tours": 21
        },
        {
         "name": "Residence",
          "month": 11,
          "tours": 18
        },
        {
         "name": "Residence",
          "month": 12,
          "tours": 14
        },
        {
          "name": "Courtyard",
          "month": 1,
          "tours": 10
        },
        {
          "name": "Courtyard",
          "month": 2,
          "tours": 15
        },
        {
          "name": "Courtyard",
          "month": 3,
          "tours": 8
        },
        {
          "name": "Courtyard",
          "month": 4,
          "tours": 12
        },
        {
          "name": "Courtyard",
          "month": 5,
          "tours": 18
        },
        {
          "name": "Courtyard",
          "month": 6,
          "tours": 25
        },
        {
          "name": "Courtyard",
          "month": 7,
          "tours": 40
        },
        {
          "name": "Courtyard",
          "month": 8,
          "tours": 33
        },
        {
          "name": "Courtyard",
          "month": 9,
          "tours": 25
        },
        {
         "name": "Courtyard",
          "month": 10,
          "tours": 21
        },
        {
         "name": "Courtyard",
          "month": 11,
          "tours": 18
        },
        {
         "name": "Courtyard",
          "month": 12,
          "tours": 14
        },
        {
          "name": "Renaissance",
          "month": 1,
          "tours": 10
        },
        {
          "name": "Renaissance",
          "month": 2,
          "tours": 15
        },
        {
          "name": "Renaissance",
          "month": 3,
          "tours": 8
        },
        {
          "name": "Renaissance",
          "month": 4,
          "tours": 12
        },
        {
          "name": "Renaissance",
          "month": 5,
          "tours": 18
        },
        {
          "name": "Renaissance",
          "month": 6,
          "tours": 25
        },
        {
          "name": "Renaissance",
          "month": 7,
          "tours": 40
        },
        {
          "name": "Renaissance",
          "month": 8,
          "tours": 33
        },
        {
          "name": "Renaissance",
          "month": 9,
          "tours": 25
        },
        {
         "name": "Renaissance",
          "month": 10,
          "tours": 21
        },
        {
         "name": "Renaissance",
          "month": 11,
          "tours": 18
        },
        {
         "name": "Renaissance",
          "month": 12,
          "tours": 14
        }
      ];

Answer 1

您可以使用 reduce并返回对象，其中每个键是酒店名称，值是旅游总和。 DEMO

data = data.reduce(function(obj, e) {
  obj[e.name] = (obj[e.name] || 0) + e.tours;
  return obj;
}, {});

console.log(data)

如果您想获得每家酒店的总和和百分比，您可以这样做 DEMO

var result = {}

var total = data.reduce((a, b) => {return a + b.tours }, 0);
data.forEach(function(e) {
  if(!this[e.name]) {
    this[e.name] = {sum: e.tours, percentage: 0}
    result[e.name] = this[e.name];
  }
  this[e.name].sum += e.tours;
  this[e.name].percentage = (this[e.name].sum / total)*100;
}, {});

console.log(result)