final orders = data["Orders"] as List;
final mapped = orders.fold<Map<String,Map<String,dynamic>>>({},(p,v) {
  final name = v["productName"];

  if (p.containsKey(name)) {
    p[name]["NumberOfItems"] += int.parse(v["NumberOfItems"]);
  } else {
    p[name] = {
      ...v,"NumberOfItems": int.parse(v["NumberOfItems"])
    };
  }

  return p;
});

final newData = {
  ...data,"Orders": mapped.values,};

print(newData);

结果是：

  

{订单：（{productname：Apple，NumberOfItems：8}，{productName：Orange，NumberOfItems：6}，{productName：Egg，NumberOfItems：14}）}

注意：此代码有2个循环，表示速度较慢。 伊戈尔·克拉霍霍尔丁（Igor Kharakhordin）回答的比较聪明，但是对于那些问这个问题的人来说可能很难。（因为他一次要做两件事。）基本上我是在做同一件事。

String string = await rootBundle.loadString("asset/data/Orders.json");
  Map orders = jsonDecode(string);
  List orderList = orders["Orders"];

  Map<String,int> sums = {};

  for(int i = 0 ; i < orderList.length; i++){
    dynamic item = orderList[i];
    if(sums.containsKey(item["productName"])){
      sums[item["productName"]] += int.parse(item["NumberOfItems"]);
    }
    else{
      sums[item["productName"]] = int.parse(item["NumberOfItems"]);
    }
  }

  List sumList = [];
  sums.forEach((key,value)=> 
      sumList.add({
      "productName":key,"NumberOfItems":value.toString()
    })
  );

  Map result = {
    "Orders":sumList
  };

  print(jsonEncode(result));

结果

{
    "Orders": [
        {
            "productName": "Apple","NumberOfItems": "8"
        },{
            "productName": "Orange","NumberOfItems": "6"
        },{
            "productName": "Egg","NumberOfItems": "14"
        }
    ]
}

private static class GetContacts extends AsyncTask<String,Void,String> { ProgressDialog dialog; ArrayList<Actors> actors = new ArrayList<>(); ... @Override protected String doInBackground(String... sText1) { ... for (int i = 0; i < actors.length(); i++) { ... // activityReference.get().actorsList.add(actor); <-- remove this. actors.add(actor); } ... } protected void onPostExecute(String result) { super.onPostExecute(result); Zoznam activity = activityReference.get(); if (activity == null || activity.isFinishing()) return; dialog.dismiss(); activityReference.get().actorsList.add(actors); <-- add this activity.adapter.notifyDataSetChanged(); } } 方法允许您通过指定的分组键对序列值进行分组。根据您的情况，可以从groupBy的值获取此密钥。
输入序列将被延迟转换为新序列，其中每个组均可使用该键。
然后，您可以选择此新序列（使用productName方法）并生成所需的结果。
同时，在形成结果时，您可以变换每个分组的元素（或对它们进行任何计算，例如，对分组元素的值求和）。

select

结果：

{Orders: [{productName: Apple,NumberOfItems: 8},{productName: Orange,NumberOfItems: 6},{productName: Egg,NumberOfItems: 14}]}

另一个更复杂的示例（同时实现起来也并不困难）：

import 'package:enumerable/enumerable.dart';

void main(List<String> args) {
  final q = orders['Orders'].groupBy((e) => e['productName']).select((g) => {
        'productName': g.key,'NumberOfItems': g.sum$1((e) => int.parse(e['NumberOfItems']))
      });
  print({'Orders': q.toList()});
}

final orders = {
  "Orders": [
    {"productName": "Apple","NumberOfItems": "5"},{"productName": "Orange","NumberOfItems": "2"},{"productName": "Egg",{"productName": "Apple","NumberOfItems": "3"},"NumberOfItems": "4"},"NumberOfItems": "9"},]
};

{Orders: [{productName: Apple,NumberOfItems: 8,Amount: 77.25},NumberOfItems: 6,Amount: 45.8},NumberOfItems: 14,Amount: 46.55}]}