在angular2中找到具有相同名称的对象数组的值的总和

时间:2018-07-10 13:58:15

标签: javascript

我有一个要求,我需要在具有相同名称的不同对象中找到amt的值之和。 以下是代码段

traveler = [
  {  description: 'Senior', Amount: 50},
   {  description: 'Senior', Amount: 50},
   {  description: 'Adult', Amount: 75},
   {  description: 'Child', Amount: 35},
   {  description: 'Infant', Amount: 25 },
];

在这里我要计算具有相同描述的不同对象中的金额总计。 例如:对象0和1包含相同的描述'Senior',因此总AMT为100

如何在angular2中实现这一目标?

我应该使用内部forloop还是有更好的方法? 请帮助我

6 个答案:

答案 0 :(得分:1)

您可以使用@Override public void setZoomLayout(String imageUri) { View view = getLayoutInflater().inflate(R.layout.layout_image_zoom, null); ImageView zoomImage = view.findViewById(R.id.ivImageZoom); zoomImage.setClickable(false); zoomImage.setFocusable(false); GlideApp.with(getActivity()) .load(imageUri) .into(zoomImage); dialogBuilder = new Dialog(getActivity(), -1); dialogBuilder.requestWindowFeature(Window.FEATURE_NO_TITLE); dialogBuilder.getWindow().setBackgroundDrawable( new ColorDrawable(Color.TRANSPARENT)); dialogBuilder.setContentView(view); dialogBuilder.show(); 将数组分组为一个对象。

reduce

答案 1 :(得分:1)

您可以使用Map来收集值,并使用分组结果呈现对象数组。

var traveler = [{ description: 'Senior', amount: 50 }, { description: 'Senior', amount: 50 }, { description: 'Adult', amount: 75 }, { description: 'Child', amount: 35 },  { description: 'Infant', amount: 25 }],
    grouped = Array.from(
        traveler.reduce(
            (m, { description, amount }) => m.set(description, (m.get(description) || 0) + amount),
            new Map
        ).entries(),
        ([description, amount]) => ({ description, amount })
    );
    
console.log(grouped);
.as-console-wrapper { max-height: 100% !important; top: 0; }

答案 2 :(得分:0)

我认为您应该使用array#reduce方法执行以下操作:

let traveler = [
       {  description: 'Senior', Amount: 50},
       {  description: 'Senior', Amount: 50},
       {  description: 'Adult', Amount: 75},
       {  description: 'Child', Amount: 35},
       {  description: 'Infant', Amount: 25 },
    ];
    
function sumFinder(description) {
 return traveler.reduce((sum, e) => {
   (e.description == description) ? (sum += e.Amount) : (sum += 0)
   return sum;
 }, 0);
}

console.log(sumFinder('Senior'));

答案 3 :(得分:0)

使用Array.reduce()获得以下输出:

var traveler = [
   {  description: 'Senior', Amount: 50},
   {  description: 'Senior', Amount: 50},
   {  description: 'Adult', Amount: 75},
   {  description: 'Child', Amount: 35},
   {  description: 'Infant', Amount: 25 },
];
var res = traveler.reduce((acc, obj)=>{
  var existItem = acc.find(item => item.description === obj.description);
  if(existItem){
    existItem.Amount += obj.Amount;
    return acc;
  } 
  acc.push(obj);
  return acc;
}, []);
console.log(res);

答案 4 :(得分:0)

您可以使用rxjs来完成此操作:

travelers = [
      {  description: 'Senior', amount: 50},
      {  description: 'Senior', amount: 50},
      {  description: 'Adult', amount: 75},
      {  description: 'Child', amount: 35},
      {  description: 'Infant', amount: 25 }
    ];
//emit each person
const source = Rx.Observable.from(travelers);
//group by description
const example = source
  .groupBy(traveler => traveler.description)
  //return each item in group as array
  .mergeMap(group => group.toArray())
const reducer = (accumulator, currentValue) => {return accumulator + currentValue.amount};
const subscribe = example.subscribe(
  val => 
    console.log(
      {
        description: val[0].description, 
        amount: val.reduce( reducer, 0 )
      }
    )
);

此代码的输出是您想要的:

[object Object] {
  amount: 100,
  description: "Senior"
}
[object Object] {
  amount: 75,
  description: "Adult"
}
[object Object] {
  amount: 35,
  description: "Child"
}
[object Object] {
  amount: 25,
  description: "Infant"
}

最后,这是您可以使用的JSBin: http://jsbin.com/kacoqulike/1/edit?js,console

答案 5 :(得分:0)

使用Array.prototype.reduce和ES6对象分配解构的干净简单的解决方案:

const traveler = [{"description":"Senior","Amount":50},{"description":"Senior","Amount":50},{"description":"Adult","Amount":75},{"description":"Child","Amount":35},{"description":"Infant","Amount":25}]

const result = traveler.reduce((all, {description: d, Amount: a}) => {

    all[d] = (all[d] || 0) + a;
    return all;

}, {});

console.log(result);