转换数据

Question

我有以下数据结构：

const data = [
  {
    name: 'ABC',
    salesData: [
      {
        timestamp: '2017-09-01',
        value: 10
      },
      {
        timestamp: '2017-09-02',
        value: 2
      }
    ]
  },
  {
    name: 'DEF',
    salesData: [
      {
        timestamp: '2017-09-01',
        value: 8
      },
      {
        timestamp: '2017-09-02',
        value: 3
      }
    ]
  }
];

我想将其转换为：

[
  {
    name: 'ABC',
    '2017-09-01': 10,
    '2017-09-02': 2
  },
  {
    name: 'CDE',
    '2017-09-01': 8,
    '2017-09-02': 3
  }
]

我正在尝试使用Underscore的链和地图，这让我很困惑。到目前为止，我有以下内容，不知道如何根据需要编写convertedSalesData进行转换：

_.map(data, function(item) {
    let name = item.name;
    let salesData = item.salesData;
    let convertedSalesData = ?
})

Answer 1

使用ES6，您可以使用扩展语法...来获得此结果。

＆＃13;

＆＃13;

const data = [{"name":"ABC","salesData":[{"timestamp":"2017-09-01","value":10},{"timestamp":"2017-09-02","value":2}]},{"name":"DEF","salesData":[{"timestamp":"2017-09-01","value":8},{"timestamp":"2017-09-02","value":3}]}]


var result = data.map(function({name, salesData}) {
  return {name, ...Object.assign({}, ...salesData.map(({timestamp, value}) => ({[timestamp]: value})))}
})
console.log(result)

＆＃13;

＆＃13;

＆＃13;

Answer 2

＆＃13;

＆＃13;

const data = [{
    name: 'ABC',
    salesData: [{
        timestamp: '2017-09-01',
        value: 10
      },
      {
        timestamp: '2017-09-02',
        value: 2
      }
    ]
  },
  {
    name: 'DEF',
    salesData: [{
        timestamp: '2017-09-01',
        value: 8
      },
      {
        timestamp: '2017-09-02',
        value: 3
      }
    ]
  }
];

var res = data.map(function(a) {
  var obj = {
    name: a.name
  };
  a.salesData.forEach(function(x) {
    obj[x.timestamp] = x.value;
  })
  return obj;
})

console.log(res);

＆＃13;

＆＃13;

＆＃13;

Answer 3

与@Nenad Vracar相似。我更愿意使用＆＃39; reduce＆＃39;：

data.map(({ name, salesData }) => ({
  name,
  ...salesData.reduce(
    (record, { timestamp, value }) => {
      record[timestamp] = value
      return record
    },
    Object.create(null)
  )
}))