從一個數組創建2個數組ecmascript 2015

Question

我有一個數組，我需要能夠從它得到2個數組。我需要這兩個數組作爲圖表。圖表組件需要一個標籤數組和一個總數組（每個標籤）。

我還在學習新的數組方法，的forEach，地圖等於是用我裸，這是數組我有：

const income = [ 
    { 
     "date": 1482233886000, 
     "name": "Company A", 
     "type": "Salary", 
     "description": "Salary jan. 2017", 
     "amount": 50.6 
    }, 
    { 
     "date": 1482406686000, 
     "name": "Company B", 
     "type": "Salary", 
     "description": "Salary jan. 2017", 
     "amount": 40.23 
    }, 
    { 
     "date": 1485171486000, 
     "name": "Company A", 
     "type": "Salary", 
     "description": "Salary feb. 2017", 
     "amount": 200.2 
    }, 
    { 
     "date": 1485344286000, 
     "name": "Company B", 
     "type": "Salary", 
     "description": "Salary feb. 2017", 
     "amount": 100.4 
    }, 
    { 
     "date": 1490655892000, 
     "name": "Company A", 
     "type": "Salary", 
     "description": "Salary mar. 2017", 
     "amount": 20 
    }, 
    { 
     "date": 1493335850000, 
     "name": "Company A", 
     "type": "Salary", 
     "description": "Salary apr. 2017", 
     "amount": 15 
    }, 
    { 
     "date": 1493335850000, 
     "name": "Company B", 
     "type": "Salary", 
     "description": "Salary apr. 2017", 
     "amount": 10 
    } 
    ]; 

爲了讓我這樣做是爲了得到獨一無二的標籤標籤（說明）：

[...new Set(income.map(value => value.description))] 
=> ["Salary jan. 2017", "Salary feb. 2017", "Salary mar. 2017", "Salary apr. 2017"] 

這工作，但現在我想每一個描述如總量：

//[90.83, 300.6, 20, 25] 

如果可能的話，是否有人可以幫助我使用最後一個陣列和ES6風格。我想減少，但我無法得到任何的。所以我希望一個主人可以教我; D。

Answer 1

您可以縮小到一個Map中，得到的keys（）和values（）將是您的數組，並按描述進行分組和彙總。

const income = [{"date": 1482233886000,"name": "Company A","type": "Salary","description": "Salary jan. 2017","amount": 50.6},{"date": 1482406686000,"name": "Company B","type": "Salary","description": "Salary jan. 2017","amount": 40.23},{"date": 1485171486000,"name": "Company A","type": "Salary","description": "Salary feb. 2017","amount": 200.2},{"date": 1485344286000,"name": "Company B","type": "Salary","description": "Salary feb. 2017","amount": 100.4},{"date": 1490655892000,"name": "Company A","type": "Salary","description": "Salary mar. 2017","amount": 20},{"date": 1493335850000,"name": "Company A","type": "Salary","description": "Salary apr. 2017","amount": 15},{"date": 1493335850000,"name": "Company B","type": "Salary","description": "Salary apr. 2017","amount": 10}]; 
 

 
let res = income.reduce((a, b) => { 
 
    a.set(b.description, (a.get(b.description) || 0) + b.amount); 
 
    return a; 
 
}, new Map()); 
 

 
console.log([...res.values()]); // the summed amount 
 
console.log([...res.keys()]); // the description property

編輯的評論

要添加到您使用set（）映射中的項 - 在上面的例子中，我使用的說明財產作爲密鑰，並且添加金額。在這部分代碼：

(a.get(b.description) || 0) 

基本上只是檢查，如果創建的地圖擁有像迭代描述的關鍵，如果是的話，它需要在現有量補充。 ||短路，所以如果沒有現有量（a.get（b.description）未定義），則取而代之。英語不是我的第一語言 - 這裏是什麼，我的意思是一個簡單的例子：

let m = new Map(); 
 
m.set("foo", 1); 
 
let a = m.get("foo"); 
 
console.log(a); // 1 - obvious 
 
let b = m.get("bar"); 
 
console.log(b); // undefined 
 

 
console.log(b || 0); // 0, as b is undefined, 0 is taken 
 
// in the original answer, this would be the same as 
 
if (! b) b = 0; 
 
console.log(b); 0 
 

 
let c = (b || 0) + 1; // 0 + 1 because b is undefined; 
 
console.log("c: ", c); 
 

 
let d = (a || 0) + 1; // 1 + 1 because a holds 1; 
 
console.log("d:", d);

Answer 2

另一種方法：

• 濾波器income可變擺脫重複說明
的
• 獲取已過濾數組的每個元素並遍歷income數組以獲取具有相同的條目值然後總結它們的amount屬性
• 將每個元素的總數推入新建的arr數組。

const income=[{date:1482233886e3,name:"Company A",type:"Salary",description:"Salary jan. 2017",amount:50.6},{date:1482406686e3,name:"Company B",type:"Salary",description:"Salary jan. 2017",amount:40.23},{date:1485171486e3,name:"Company A",type:"Salary",description:"Salary feb. 2017",amount:200.2},{date:1485344286e3,name:"Company B",type:"Salary",description:"Salary feb. 2017",amount:100.4},{date:1490655892e3,name:"Company A",type:"Salary",description:"Salary mar. 2017",amount:20},{date:149333585e4,name:"Company A",type:"Salary",description:"Salary apr. 2017",amount:15},{date:149333585e4,name:"Company B",type:"Salary",description:"Salary apr. 2017",amount:10}]; 
 

 
var arr = []; 
 
var sorted = [...new Set(income.map(value => value.description))]; 
 
sorted.forEach(function(v){ 
 
    var obj = {}; 
 
    var sum = 0; 
 
    income.forEach(function(c){ 
 
    if (c.description == v){ 
 
     sum += c.amount; 
 
    } 
 
    }); 
 
    obj[v] = sum; 
 
    arr.push(obj); 
 
}); 
 

 
console.log(arr);