如何使用id javascript對group對象求和並得到最大值

Question

var myArray = [
  {"Emiten_ID":'SMBR',"Lot":500,"Price":2500},     
  {"Emiten_ID":'SMBR',"Lot":300,"Price":2200},     
  {"Emiten_ID":'ELSA',"Lot":500,"Price":1000},
  {"Emiten_ID":'SMBR',"Lot":100,"Price":3000},     
  {"Emiten_ID":'BI',"Lot":300,"Price":500},     
  {"Emiten_ID":'AAI',"Lot":200,"Price":1300},
  {"Emiten_ID":'BTB',"Lot":700,"Price":2900},     
  {"Emiten_ID":'BI',"Lot":150,"Price":200},     
  {"Emiten_ID":'AAI',"Lot":200,"Price":600},
];

我想要這樣的結果，在那里我從批次和價格的最大價值得到總和

var Result= [
  {"Emiten_ID":'ELSA',"Lot":500,"Price":1000},
  {"Emiten_ID":'SMBR',"Lot":900,"Price":3000},     
  {"Emiten_ID":'BI',"Lot":450,"Price":500},     
  {"Emiten_ID":'BTB',"Lot":700,"Price":2900},     
  {"Emiten_ID":'AAI',"Lot":400,"Price":1300},
];

Answer 1

您可以使用Array#forEach和一個對象作為哈希表， Emiten_ID對其進行Emiten_ID 。

 var myArray = [{ "Emiten_ID": 'SMBR', "Lot": 500, "Price": 2500 }, { "Emiten_ID": 'SMBR', "Lot": 300, "Price": 2200 }, { "Emiten_ID": 'ELSA', "Lot": 500, "Price": 1000 }, { "Emiten_ID": 'SMBR', "Lot": 100, "Price": 3000 }, { "Emiten_ID": 'BI', "Lot": 300, "Price": 500 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 1300 }, { "Emiten_ID": 'BTB', "Lot": 700, "Price": 2900 }, { "Emiten_ID": 'BI', "Lot": 150, "Price": 200 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 600 }, ], result = []; myArray.forEach(function (a) { if (!this[a.Emiten_ID]) { this[a.Emiten_ID] = { Emiten_ID: a.Emiten_ID, Lot: 0, Price: 0 }; result.push(this[a.Emiten_ID]); } this[a.Emiten_ID].Lot += a.Lot; this[a.Emiten_ID].Price = Math.max(this[a.Emiten_ID].Price, a.Price); }, Object.create(null)); console.log(result); 

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

ES6帶有hash的閉包而不使用this 。

 var myArray = [{ "Emiten_ID": 'SMBR', "Lot": 500, "Price": 2500 }, { "Emiten_ID": 'SMBR', "Lot": 300, "Price": 2200 }, { "Emiten_ID": 'ELSA', "Lot": 500, "Price": 1000 }, { "Emiten_ID": 'SMBR', "Lot": 100, "Price": 3000 }, { "Emiten_ID": 'BI', "Lot": 300, "Price": 500 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 1300 }, { "Emiten_ID": 'BTB', "Lot": 700, "Price": 2900 }, { "Emiten_ID": 'BI', "Lot": 150, "Price": 200 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 600 }, ], result = []; myArray.forEach((hash => a => { if (!hash[a.Emiten_ID]) { hash[a.Emiten_ID] = { Emiten_ID: a.Emiten_ID, Lot: 0, Price: 0 }; result.push(hash[a.Emiten_ID]); } hash[a.Emiten_ID].Lot += a.Lot; hash[a.Emiten_ID].Price = Math.max(hash[a.Emiten_ID].Price, a.Price); })(Object.create(null))); console.log(result); 

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

Answer 2

您可以使用linq.js庫：

 var myArray = [ {"Emiten_ID":'SMBR',"Lot":500,"Price":2500}, {"Emiten_ID":'SMBR',"Lot":300,"Price":2200}, {"Emiten_ID":'ELSA',"Lot":500,"Price":1000}, {"Emiten_ID":'SMBR',"Lot":100,"Price":3000}, {"Emiten_ID":'BI',"Lot":300,"Price":500}, {"Emiten_ID":'AAI',"Lot":200,"Price":1300}, {"Emiten_ID":'BTB',"Lot":700,"Price":2900}, {"Emiten_ID":'BI',"Lot":150,"Price":200}, {"Emiten_ID":'AAI',"Lot":200,"Price":600}, ]; var answer = Enumerable.From(myArray).GroupBy("x => x.Emiten_ID", "x => {Lot: x.Lot, Price: x.Price}").Select("x => {Emiten_ID:x.Key(), Lot:x.Sum(y=>y.Lot), Price:x.Max(y=>y.Price)}").ToArray(); answer.forEach(x => console.log(x)); 

 <script data-require="jquery@3.0.0" data-semver="3.0.0" src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.0.0/jquery.js"></script> <script data-require="linq.js@2.2.0+2" data-semver="2.2.0+2" src="//cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.js"></script> <script data-require="linq.js@2.2.0+2" data-semver="2.2.0+2" src="//cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/jquery.linq.js"></script> 

Answer 3

使用ECMAScript 2015（或polyfill）中提供的Map類：

var temp = new Map();

for (var item of myArray) {
    var e = temp.get(item.Emiten_ID)
    if (e) {
        e.Lot += item.Lot;
        e.Price = Math.max( e.Price, item.Price );
    } else {
        temp.set( item.Emiten_ID, 
            { Emiten_ID: item.Emiten_ID, Lot:item.Lot, Price:item.Price })
    }
}

var result = Array.from(temp.values());

console.log(result)

Answer 4

這是很長的路要走，希望有更簡單的方法做到這一點。

var myArray = [
  {"Emiten_ID":'SMBR',"Lot":500,"Price":2500},     
  {"Emiten_ID":'SMBR',"Lot":300,"Price":2200},     
  {"Emiten_ID":'ELSA',"Lot":500,"Price":1000},
  {"Emiten_ID":'SMBR',"Lot":100,"Price":3000},     
  {"Emiten_ID":'BI',"Lot":300,"Price":500},     
  {"Emiten_ID":'AAI',"Lot":200,"Price":1300},
  {"Emiten_ID":'BTB',"Lot":700,"Price":2900},     
  {"Emiten_ID":'BI',"Lot":150,"Price":200},     
  {"Emiten_ID":'AAI',"Lot":200,"Price":600},
];

var sortedIds = [];
var Result = [];
function sortArray(){

    for(var i = 0; i < myArray.length; i++){
        if(myArray.indexOf(myArray[i].Emiten_ID) < 0){
            sortedIds.push(myArray[i].Emiten_ID);
            Result.push({
                "Emiten_ID" : myArray[i].Emiten_ID,
                "Lot"       : sumLot(myArray[i].Emiten_ID),
                "Price"     : maxPrice(myArray[i].Emiten_ID);
            });


        }
    }   

    //out put
    console.log(Result);
}

function sumLot(id){
    var sum = 0;
    for(var i = 0; i < myArray.length; i++){
        if(myArray[i].Emiten_ID == id){
            sum += myArray[i].lot;
        }
    }   
    return sum;
}

function maxPrice(id){
    var max = 0;
    for(var i = 0; i < myArray.length; i++){
        if(myArray[i].Price > max){
            max = myArray[i].Price;
        }
    }   
    return max;
}

Answer 5

 var myArray = [ {"Emiten_ID":'SMBR',"Lot":500,"Price":2500}, {"Emiten_ID":'SMBR',"Lot":300,"Price":2200}, {"Emiten_ID":'ELSA',"Lot":500,"Price":1000}, {"Emiten_ID":'SMBR',"Lot":100,"Price":3000}, {"Emiten_ID":'BI',"Lot":300,"Price":500}, {"Emiten_ID":'AAI',"Lot":200,"Price":1300}, {"Emiten_ID":'BTB',"Lot":700,"Price":2900}, {"Emiten_ID":'BI',"Lot":150,"Price":200}, {"Emiten_ID":'AAI',"Lot":200,"Price":600}, ]; myArray = myArray.sort(function(a,b){ if(a.Emiten_ID > b.Emiten_ID) return 1; else if(a.Emiten_ID < b.Emiten_ID) return -1; else{ return a.Price - b.Price; } }); var result = [myArray[0]]; for(var i = 1 ; i < myArray.length ; i ++){ var obj = myArray[i]; var res = result[result.length - 1]; if(obj.Emiten_ID == res.Emiten_ID){ res.Lot += obj.Lot; res.Price = Math.max(res.Price,obj.Price); }else{ result.push(obj); } } console.log(result); 

Answer 6

使用lodash的解決方案：

_.chain(myArray).groupBy('Emiten_ID').map(emiten => ({
  "Emiten_ID":emiten[0]['Emiten_ID'],
  "Lot": _.sumBy(emiten, 'Lot'),
  "Price": _.maxBy(emiten, 'Price')['Price']
})).value()

Answer 7

• 迭代所有array元素
• 從新數組中查找each-element的索引
• 如果element不存在，請將其推入array
• 如果元素存在，則將Lot的值相加，如果Price值更高，則覆蓋更高的值

 var myArray = [{ "Emiten_ID": 'SMBR', "Lot": 500, "Price": 2500 }, { "Emiten_ID": 'SMBR', "Lot": 300, "Price": 2200 }, { "Emiten_ID": 'ELSA', "Lot": 500, "Price": 1000 }, { "Emiten_ID": 'SMBR', "Lot": 100, "Price": 3000 }, { "Emiten_ID": 'BI', "Lot": 300, "Price": 500 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 1300 }, { "Emiten_ID": 'BTB', "Lot": 700, "Price": 2900 }, { "Emiten_ID": 'BI', "Lot": 150, "Price": 200 }, { "Emiten_ID": 'AAI', "Lot": 200, "Price": 600 }]; var newArr = []; myArray.forEach(function (el) { var findIndex = newArr.findIndex(function (item) { return item.Emiten_ID === el.Emiten_ID; }); if (findIndex === -1) { newArr.push(el); } else if (el.Price > newArr[findIndex].Price) { newArr[findIndex].Price = el.Price; newArr[findIndex].Lot += el.Lot; } else { newArr[findIndex].Lot += el.Lot; } }); console.log(JSON.stringify(newArr, null, 4)); 

Answer 8

讓我們嘗試分解這個。 我們將通過編寫一個通用的效用函數，我們將調用start combine ，其結合使用在哈希叫每個屬性指定的函數對象的屬性combiners ：

function combine(array, combiners) {
  const result = {};
  for (prop of Object.keys(combiners)) {
    result[prop] = combiners[prop](...array.map(elt => elt[prop]));
  }
  return result;
}

使用此示例：

combine(
  [{a: 1, b: 10}, {a: 42, b: 80}],
  {a: sum, b: Math.max}
)

這將導致

{a: 43, b: 80}

當然，為了完成這項工作，我們必須定義sum ：

function sum(...vals) { return vals.reduce(add); }

其中add就是

function add(a, b) { return a + b; }

接下來，我們將按Emiten_ID屬性對輸入進行Emiten_ID 。 您可以使用_.groupBy的_.groupBy ，或者自己編寫（見下文）。

const groups = _.groupBy(myArray, 'Emiten_ID`);

這將導致看起來像

{ SMBR: [
    { Emiten_ID: 'SMBR', "Lot": 500, "Price": 2500},     
    { Emiten_ID: 'SMBR', "Lot": 300, "Price": 2200}
  ],
  ELSA: [
  ]
}

完成此准備工作后，只需使用我們的combine實用程序映射groups每個值，就可以輕松獲得結果：

const Result = Object.keys(groups).map(key =>
  combine(
    groups[key], 
    {
      Emiten_ID: identity, 
      Lot: sum, 
      Price: Math.max}));

identity是公正的

function identity(id) { return id; }

如果您希望在映射對象屬性的combine中抽象出概念，您可以再次使用Underscore中的一些實用程序，或者自己編寫它：

function mapObject(obj, fn) {
  const result = {};

  for (prop of obj) result[prop] = fn(obj[prop], prop);
  return result;
}

使用此示例：

mapObject({a: 2, b: 3}, x => x * x)
// yields {a: 4, b: 9}

現在你可以更簡單地編寫combine了

function combine(array, combiners) {
  return mapObject(combiners, function(combiner, prop) {
    return combiner(...array.map(elt => elt[prop]));
  };
}

如果你不想使用_.groupBy的_.groupBy ，這里是一個本土版本：

function groupBy(array, prop) {
  var result = {};
  array.forEach(elt => (result[elt[prop]] = result[elt[prop]] || []).push(elt));
  return result;
}