GroupBy对象阵列JavaScript

Question

我有一个建议：

var modbusData = [
  { Type: "rtu", IdDevice: "1", Time: "11:01:00", Data: "1,12,23" },
  { Type: "tcp", IdDevice: "2", Time: "11:01:11", Data: "30,40,50" },
  { Type: "rtu", IdDevice: "1", Time: "11:02:00", Data: "5,10,21" },
  { Type: "tcp", IdDevice: "2", Time: "11:02:11", Data: "32,44,53" },
];

我打电话后：

function groupBy(objectArray, property) {
  return objectArray.reduce(function (acc, obj) {
    let key = obj[property]
    if (!acc[key]) {
      acc[key] = []
    }
    acc[key].push(obj)
    return acc
  }, {})
}
let groupedData = groupBy(modbusData, 'Type')
console.log(groupedData)

其结果将是：

{
  rtu: [
    { Type: 'rtu', IdDevice: '1', Time: '11:01:00', Data: '1,12,23' },
    { Type: 'rtu', IdDevice: '1', Time: '11:02:00', Data: '5,10,21' }
  ],
  tcp: [
    { Type: 'tcp', IdDevice: '2', Time: '11:01:11', Data: '30,40,50' },
    { Type: 'tcp', IdDevice: '2', Time: '11:02:11', Data: '32,44,53' }
  ]
}

现在我希望在modbusData中没有“类型”："rtu“或"tcp”时，结果将返回一个空数组，如下所示：

{
  rtu: [
    { Type: 'rtu', IdDevice: '1', Time: '11:01:00', Data: '1,12,23' },
    { Type: 'rtu', IdDevice: '1', Time: '11:02:00', Data: '5,10,21' }
  ],
  tcp: []
}

Answer 1

您需要将预期的组键作为还原器的初级对象传递。

>>> [ 'rtu', 'tcp' ]

<<< Object.fromEntries(expectedGroups.map(k => [k, []])) // { rtu: [], tcp: [] }

​

var modbusData = [
  { Type: "rtu", IdDevice: "1", Time: "11:01:00", Data: "1,12,23" },
  { Type: "rtu", IdDevice: "1", Time: "11:02:00", Data: "5,10,21" }
];

function groupBy(objectArray, property, expectedGroups = []) {
  return objectArray.reduce(function (acc, obj) {
    let key = obj[property]
    if (!acc[key]) {
      acc[key] = []
    }
    acc[key].push(obj)
    return acc
  }, Object.fromEntries(expectedGroups.map(k => [k, []])))
};

let groupedData = groupBy(modbusData, 'Type', [ 'rtu', 'tcp' ]);
console.log(groupedData);

.as-console-wrapper { top: 0; max-height: 100% !important; }

​

这里有一个更精简的版本，它使用ES5/6构造。

​

const modbusData = [
  { Type: "rtu", IdDevice: "1", Time: "11:01:00", Data: "1,12,23" },
  { Type: "rtu", IdDevice: "1", Time: "11:02:00", Data: "5,10,21" }
];

const groupBy = (arr, prop, requiredKeys = []) =>
  arr.reduce((acc, obj) =>
    ({ ...acc,
      [obj[prop]]: [...(acc[obj[prop]] ?? []), obj]
    }), Object.fromEntries(requiredKeys.map(k => [k, []])));

const groupedData = groupBy(modbusData, 'Type', [ 'rtu', 'tcp' ]);

console.log(groupedData);

.as-console-wrapper { top: 0; max-height: 100% !important; }

​

Answer 2

尝试使用扩展运算符合并alltypes

let alltypes = { rtu: [], tcp: [] }
let result = {...alltypes, ...groupedData}

​

var modbusData = [
  { Type: "rtu", IdDevice: "1", Time: "11:01:00", Data: "1,12,23" },
  { Type: "rtu", IdDevice: "1", Time: "11:02:00", Data: "5,10,21" },
];

function groupBy(objectArray, property) {
  return objectArray.reduce(function (acc, obj) {
    let key = obj[property]
    if (!acc[key]) {
      acc[key] = []
    }
    acc[key].push(obj)
    return acc
  }, {})
}

let groupedData = groupBy(modbusData, 'Type')

let alltypes = { rtu: [], tcp: [] }
let result = {...alltypes, ...groupedData}
console.log(result)

​