使用流将Java列表转换为map,其中map中的每个值共享相同的属性

Question

我试图将list转换成一个map,这样对于list元素的某个属性,map的键是元素ID,而值是具有相同属性的元素的其他元素ID。使用for循环非常简单，但我尝试使用专门的Java流来实现。主要目标是可读性，而不是性能。

class Apple {
    int _appleId; // different for each apple
    int _colorId; // same for some apples

    Apple(int appleId, int colorId) {
        _appleId = appleId;
        _colorId = colorId;
    }

    int getAppleId() { return _appleId; }
    int getColorId() { return _colorId; }
}

public static void main(String[] args) {
    Apple apple1 = new Apple(1, 7);
    Apple apple2 = new Apple(2, 7);
    Apple apple3 = new Apple(3, 7);
    Apple apple4 = new Apple(4, 8);
    Apple apple5 = new Apple(5, 8);
    Apple apple6 = new Apple(6, 9);

    List<Apple> apples = List.of(apple1, apple2, apple3, apple4,
        apple5, apple6);

    // applesByColor: {7=[1, 2, 3], 8=[4, 5], 9=[6]}
    Map<Integer, Set<Integer>> applesByColor = apples.stream().collect(
        Collectors.groupingBy(Apple::getColorId,
            Collectors.mapping(Apple::getAppleId, Collectors.toSet())));

    // applesWithSameColor:
    // {1=[1, 2, 3], 2=[1, 2, 3], 3=[1, 2, 3], 4=[4, 5], 5=[4, 5], 6=[6]}
    Map<Integer, Set<Integer>> applesWithSameColor = new HashMap<>();
    for (var entry : applesByColor.entrySet()) {
        for (var appleId : entry.getValue()) {
            applesWithSameColor
                .computeIfAbsent(appleId, k -> new HashSet<>())
                .addAll(entry.getValue());
        }
    }
}

best:   {1=[2, 3], 2=[1, 3], 3=[1, 2], 4=[5], 5=[4]}
better: {1=[2, 3], 2=[1, 3], 3=[1, 2], 4=[5], 5=[4], 6=[]}
good:   {1=[1, 2, 3], 2=[1, 2, 3], 3=[1, 2, 3], 4=[4, 5], 5=[4, 5], 6=[6]}

Answer 1

您可以只从AppleSbyColorMap中对值进行流式处理，然后在Set中为每个值创建条目，然后将它们收集到Map中

Map<Integer, Set<Integer>> applesWithSameColor = applesByColor.values().stream()
            .flatMap(set -> set.stream().map(val -> Map.entry(val, set)))
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

在重复键的情况下，也可以使用合并功能，以避免冲突

collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (v1, v2) ->v1)));