JavaScript數組排序通過總結相同的元素

Question

當嘗試對JavaScript中的數組項進行排序和分組時，我遇到了一些麻煩。這裏是採樣輸入：

var arr = [ 
{merchantName: '', branchName: 'e', branchAddress: '', total: 10.5}, 
]; 

，我想實現的輸出：

var arr = [ 
{merchantName: '', branchName: '', branchAddress: '', total: 10.5}, 
]; 

我想把它由BRANCHNAME排序，例如總結總爲同一BRANCHNAME然後在同時綁定所有其他屬性，如MERCHANTNAME和branchAddress與它在一起，這樣我可以訪問他們喜歡的：

for(var i = 0; i < arr.length; i++){ 
      console.log(arr[i].merchantName + ' ' + arr[i].branchName + ' ' + arr[i].branchAddress + ' ' + arr[i].total); 
     } 

其實我對如何甚至開始它不知道。任何想法如何實現它？

感謝先進！

Answer 1

因此，這裏是我會怎麼做：

1. 組陣列成hashmap基礎上，branchName財產 - 計算總伴隨於此。

2. 從hashmap取出數組和對它們進行排序

參見下面演示：使用

var arr = [ 
 
{merchantName: 'Giant', branchName: 'Giant Marine', branchAddress: 'Terrace 56 Branch Blk 56 Marine Terrace #01-259/261 Singapore 440056', total: 10.5}, 
 
{merchantName: 'Ntuc', branchName: 'Ntuc Zhongshan Mall Branch', branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', total: 12.149999999999999}, 
 
{merchantName: 'Giant', branchName: 'Giant Kim Keat 260 Branch', branchAddress: ' Blk 260 Kim Keat Avenue #01-01 Singapore 310260', total: 5.1}, 
 
{merchantName: 'Ntuc', branchName: 'Ntuc Scotts Square Branch', branchAddress: ' Scotts Square #B1-03 To 07 & #B1-10, 6 Scotts Road, 228209', total: 4}, 
 
{merchantName: 'Ntuc', branchName: 'Ntuc Zhongshan Mall Branch', branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', total: 4}, 
 
{merchantName: 'Ntuc', branchName: 'Ntuc Zhongshan Mall Branch', branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', total: 8} 
 
]; 
 

 
// create a hashmap 
 
var hash = arr.reduce(function(p,c){ 
 
    if(!p[c.branchName]) 
 
    p[c.branchName] = c; 
 
    else 
 
    p[c.branchName].total += c.total; 
 
    return p; 
 
}, Object.create(null)) 
 

 
// now extract the result and sort them 
 
var result = Object.keys(hash).map(function(e){ 
 
    return hash[e]; 
 
}).sort(function(a,b){ 
 
    return a.branchName - b.branchName; 
 
}); 
 

 
console.log(result);

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

Answer 2

看起來你想要做2件事情：按照branchName排序，然後爲每個branchName輸出一個值（去除重複結果/ group-by branchName）。

有幾個潛在的問題。 1）上面顯示的示例輸出沒有按branchName排序，即使您聲明瞭a）您希望它按branchName排序，並且b）這將是一個示例。其次，輸出並不是完全確定性的 - 特別是它似乎只輸出了第一個匹配的記錄格式branchName，因此總屬性的值（在同一個分支名稱的記錄中變化）就是顯示的內容。所以......假設你A）要求結果排序，B）不關心「總」屬性的值，這可以通過以下方式輕鬆完成：

I）對數組進行排序。舉例來說，請參見https://gist.github.com/umidjons/9614157：只需編寫一個比較分支名稱值的比較函數即可。並且， II）循環遍歷結果，只要branchName從前一個值更改時輸出第一條記錄。

Answer 3

溶液reduce()

var arr = [{ 
 
    merchantName: 'Giant', 
 
    branchName: 'Giant Marine', 
 
    branchAddress: 'Terrace 56 Branch Blk 56 Marine Terrace #01-259/261 Singapore 440056', 
 
    total: 10.5 
 
    }, 
 
    { 
 
    merchantName: 'Ntuc', 
 
    branchName: 'Ntuc Zhongshan Mall Branch', 
 
    branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', 
 
    total: 12.149999999999999 
 
    }, 
 
    { 
 
    merchantName: 'Giant', 
 
    branchName: 'Giant Kim Keat 260 Branch', 
 
    branchAddress: ' Blk 260 Kim Keat Avenue #01-01 Singapore 310260', 
 
    total: 5.1 
 
    }, 
 
    { 
 
    merchantName: 'Ntuc', 
 
    branchName: 'Ntuc Scotts Square Branch', 
 
    branchAddress: ' Scotts Square #B1-03 To 07 & #B1-10, 6 Scotts Road, 228209', 
 
    total: 4 
 
    }, 
 
    { 
 
    merchantName: 'Ntuc', 
 
    branchName: 'Ntuc Zhongshan Mall Branch', 
 
    branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', 
 
    total: 4 
 
    }, 
 
    { 
 
    merchantName: 'Ntuc', 
 
    branchName: 'Ntuc Zhongshan Mall Branch', 
 
    branchAddress: ' Zhongshan Mall Balestier #02-01, 20 Ah Hood Road, 329984', 
 
    total: 8 
 
    } 
 
]; 
 

 
var newArr = arr.reduce(function(items, item) { 
 

 
    var existing = items.find(function(i) { 
 
    return i.branchName === item.branchName; 
 
    }); 
 
    
 
    if (existing) { 
 
    existing.total += item.total; 
 
    } else { 
 
    items.push(item); 
 
    } 
 
    
 
    return items; 
 
}, []); 
 

 
console.log(newArr);