如何将列表中具有相同值的两个字段的对象组合成一个精简列表？

Question

class Item
{
    private String name;
    private String parentName;
    private Integer someValue;
    private Double anotherValue;

    public Item() {}

    //...elided getters/setters...
}

//PSEUDO CODE (Not JavaScript, but using JSON is easier to follow)
List<Item> items = [
    {
        "name": "Joe",
        "parentName": "Frank",
        "someValue": 10,
        "anotherValue": 15.0
    },
    {
        "name": "Joe",
        "parentName": "Frank",
        "someValue": 40,
        "anotherValue": 0.5
    },
    {
        "name": "Joe",
        "parentName": "Jack",
        "someValue": 10,
        "anotherValue": 10.0
    },
    {
        "name": "Jeff",
        "parentName": "Frank",
        "someValue": 10,
        "anotherValue": 10.0
    }
];

List<Item> items = [
    {
        "name": "Joe",
        "parentName": "Frank",
        "someValue": 50,
        "anotherValue": 15.5
    },
    {
        "name": "Joe",
        "parentName": "Jack",
        "someValue": 10,
        "anotherValue": 10.0
    },
    {
        "name": "Jeff",
        "parentName": "Frank",
        "someValue": 10,
        "anotherValue": 10.0
    }
];

null

在Java8流中如何做到这一点？

我开始把它们放进桶里（见下文），但不确定从这里去哪里：

items
    .stream()
    .collect(
        Collectors.groupingBy(
            item -> new ArrayList<String>( Arrays.asList( item.getName(), item.getParentName() ) ) 
        )
    );

Answer 1

collect(groupingBy())方法返回一个映射。

如果定义以下类：

class Tuple {
    String name;
    String parentName;
}

然后可以获得以下地图：

Map<Tuple, List<Item>> groupedItems = items.stream().collect(
            groupingBy(item -> new Tuple(item.getName(), item.getParentName())));

List<Item> finalItems = groupedItems.entrySet().stream().map(entry -> 
               new Item(entry.getKey().name, 
                        entry.getKey().parentName,
                        entry.getValue().stream().mapToInt(
                                  item -> item.someValue).sum(),
                        entry.getValue().stream().mapToDouble(
                                  item -> item.anotherValue).sum()))
               .collect(Collectors.toList());