获取下面的对象数组＆＃34; toursByHotels＆＃34; - 我想获取这些数据并得到每个具有相同名称的对象（酒店的名称，即万豪等）的旅行总和。
在d3中是否有一种比使用相同名称创建5个新数组更简单的方法，然后对这些数组进行总结？
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
}
];
答案 0 :(得分：1)
您可以使用reduce并返回对象，其中每个键都是酒店名称，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)
答案 1 :(得分：1)
使用reduce是最好的方法：
// lets keep track of the total here:
var total = 0;
var totals = a.reduce(function(curr, next) {
curr[next.name] = (curr[next.name] || 0) + next.tours;
// increment total here:
total += next.tours;
return curr;
}, {});
for (var hotel in totals) {
var obj = {};
obj.name = totals[hotel];
obj.percentage = (totals[hotel] / total) * 100 + '%';
totals[hotel] = obj;
}