处理对象操作数组

Question

下面我有一个对象数组

var data = [{
  "time": "1572024707.4763825",
  "rssi": "32",
  "id": "77777"
}, {
  "time": "1572024709.0991757",
  "rssi": "32",
  "id": "77777"
}, {
  "time": "1572024704.4570136",
  "rssi": "32",
  "id": "555555"
}, {
  "time": "1572024708.3903246",
  "rssi": "32",
  "id": "77777"
}, {
  "time": "1572024699.7132683",
  "rssi": "32",
  "id": "66666"
}]

如何重组它以删除最旧时间的重复 id

我尝试从数组中提取所有唯一 ID，以便循环遍历数据数组，但随后代码开始变得太长。

  data.forEach(item => {

    IDs.push(item.id);
  });

  var unqIDs = [...new Set(IDs)];
  console.log(unqIDs);

输出应该是这样的

outPutShouldBe = [{
  "time": "1572024699.7132683",
  "rssi": "32",
  "id": "66666"
},{
  "time": "1572024709.0991757",
  "rssi": "32",
  "id": "77777"
},  {"time": "1572024704.4570136",
  "rssi": "32",
  "id": "555555"
}
]

Answer 1

创建一个对象，将 ID 映射到具有该 ID 的最早时间的项目：

var keydata = {};
data.forEach(item=>{ 
    var p = keydata[item.id]; 
    if ( !p || p.time>item.time ) { 
        keydata[item.id] = item; 
    }});

现在收集该对象中的值：

var newdata = [];
for ( var k in keydata ) { 
    newdata.push(keydata[k]);
}

或更优雅（感谢@TulioF。）：

var newdata = Object.values(keydata)

Answer 2

使用 forEach() find() filter() 和 filter() 来决定返回哪个元素

var data = [{"time": "1572024707.4763825","rssi": "32","id": "77777"},{"time": "1572024709.0991757","rssi": "32","id": "77777"}, {"time": "1572024704.4570136","rssi": "32","id": "555555"}, {"time": "1572024708.3903246","rssi": "32","id": "77777"}, {"time": "1572024699.7132683","rssi": "32","id": "66666"}]

let resultsArray = []

data.forEach(obj=>{
  const foundObj = resultsArray.find(data => data.id === obj.id)

  if(foundObj && new Date(foundObj.time) > new Date(obj.time)){
     const filteredArray = resultsArray.filter(data => data.id === obj.id)
     resultsArray = [...filteredArray , foundObj]
  } else if (!foundObj){
     resultsArray.push(obj)
  }
})

console.log(resultsArray)

Answer 3

您可以将一个对象作为哈希表并直接获取值。

var data = [{ time: "1572024707.4763825", rssi: "32", id: "77777" }, { time: "1572024709.0991757", rssi: "32", id: "77777" }, { time: "1572024704.4570136", rssi: "32", id: "555555" }, { time: "1572024708.3903246", rssi: "32", id: "77777" }, { time: "1572024699.7132683", rssi: "32", id: "66666" }],
    result = Object.values(data.reduce((r, o) => {
        if (!r[o.id] || +r[o.id].time > +o.time) r[o.id] = o;
        return r;
    }, {}));

console.log(result)

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Answer 4

使用lodash根据您的需要（desc，asc）对数组进行降序或升序排序并获取第零个对象。尝试这样的事情。 filter 和 orderBy

var data = [{
  "time": "1572024707.4763825",
  "rssi": "32",
  "id": "77777"
}, ....];

let idsSet = new Set();

data.map(item=> idsSet.add(item.id));

let idsArr = Array.from(idsSet);

let newArr = [];
idsArr.map(id=>{
let tempArray = data.filter(item => item.id === id);
return newArr.push((_.orderBy(tempArray, ['time'],['desc']))[0]);
} )

console.log(newArr);

控制台输出

[ {
  "time": "1572024709.0991757",
  "rssi": "32",
  "id": "77777"
}, {
  "time": "1572024704.4570136",
  "rssi": "32",
  "id": "555555"
},  {
  "time": "1572024699.7132683",
  "rssi": "32",
  "id": "66666"
}];

Answer 5

在这里你可以做这样的事情：

let existMap = {};
data.filter(val => {
  if((val.id in existMap) && (val.time>existMap[val.id])) return;
  else{
    existMap[val.id] = val.time;
    return true;
  }
})
console.log(result)

可以根据需要更改条件。只是想为您的问题提供参考。