帶{}的數組jave數據對象，我需要忽略它

Question

我有一個數組，其中我需要總結每個鍵的所有值並獲得總數。然後我用每個鍵和總數創建一個新的數組。但是當我收到空數據時，我遇到了一個問題，並且我需要忽略數組對象內的{}。我遇到了一些問題這樣做的，解決這個，我試圖驗證，如果我的數據比{}

var data = [{"data":{"since":1500896035232,"passerby":212,"live":159,"accesspoints":37,"associated":122,"knownvendor":139,"knownvendorinternal":109,"knownvendorcommon":107,"proberequest":75}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}}]; 


var array=[]; 
var labels = [9]; 
var total = [9]; 

//iniciar a zero todos os elementos do array 

for(var x=0; x<9;x++){ 
total[x] = 0; 
} 

var live=0, associated=0, ap=0, others=0; 
var cas=0, official=0, common=0; 

console.log("live data ",data[2]); 

for(var i=0; i<data.length;i++){ 
    if(data[i]!={}){ 
     for(var j=1;j<9;j++){ 
      switch(j){ 
       case 1: 
        labels[j-1] = "passerby"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        break; 
       case 2: 
        labels[j-1]="live"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        live=total[j-1]; 
        break; 
       case 3: 
        labels[j-1]="accesspoints"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        ap=total[j-1]; 
        break; 
       case 4: 
        labels[j-1]="associated"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        associated=total[j-1];    
        break; 
       case 5: 
        labels[j-1]="knownvendor"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        official=total[j-1]; 

        break; 
       case 6: 
        labels[j-1]="knownvendorinternal"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        cas=total[j-1];    
        break; 
       case 7: 
        labels[j-1]="knownvendorcommon"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        common=total[j-1]; 
        break; 
       case 8: 
        labels[j-1]="proberequest"; 
        total[j-1] += data[i]['data'][labels[j-1]]; 
        break; 
      } 
     } 
    } 
} 

for(var i=0; i<labels.length; i++){ 
    var obj = { 
     "label": labels[i], 
     "value": total[i] 
    }; 
    array.push(obj); 
} 

others1 = live - ap - associated; 

var othersObj = { 
    "label": "others", 
    "value": others1 
}; 

array.push(othersObj); 

console.log("array ---> ", array); 
console.log("data size ---> ", data.length); 


console.log("-------------------------------"); 


console.log("live ---> ", live); 
console.log("ap ---> ", ap); 
console.log("associated ---> ", associated); 
console.log("others ---> ", others1); 

console.log("-------------------------------"); 

others2 = live-cas-official; 

console.log("cas --> ",cas); 
console.log("official --> ",official); 
console.log("common --> ",common); 
console.log("others ---> ", others2); 

日誌

live data { data: {} } 
array ---> [ { label: 'passerby', value: NaN }, 
    { label: 'live', value: NaN }, 
    { label: 'accesspoints', value: NaN }, 
    { label: 'associated', value: NaN }, 
    { label: 'knownvendor', value: NaN }, 
    { label: 'knownvendorinternal', value: NaN }, 
    { label: 'knownvendorcommon', value: NaN }, 
    { label: 'proberequest', value: NaN }, 
    { label: 'others', value: NaN } ] 
data size ---> 20 
------------------------------- 
live ---> NaN 
ap ---> NaN 
associated ---> NaN 
others ---> NaN 
------------------------------- 
cas --> NaN 
official --> NaN 
common --> NaN 
others ---> NaN 

Answer 1

只需使用reduce()通過鑰匙來總結的值不同，然後映射結果對象的鍵以獲得所需的數據結構。

let data = [{"data":{"since":1500896035232,"passerby":212,"live":159,"accesspoints":37,"associated":122,"knownvendor":139,"knownvendorinternal":109,"knownvendorcommon":107,"proberequest":75}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}},{"data":{}}]; 
 
let r = data.reduce((a,b) => { 
 
    Object.keys(b.data).forEach(v => { 
 
    if (v === 'since') return; 
 
    a[v] = a[v] ? a[v] : 0; 
 
    a[v] += parseInt(b.data[v]) || 0; 
 
    }); 
 
    return a; 
 
}, {}); 
 
let res = Object.keys(r).map(v => ({ 
 
    label: v, 
 
    value: r[v] 
 
})); 
 
console.log(res);