Java 8列表到地图的转换

时间:2019-02-10 23:06:37

标签: java collections java-8 java-stream

我将List Object转换为Map String,List Object时遇到问题。我正在寻找具有汽车中所有组件的键名称的Map,并且值由具有该组件的汽车表示

public class Car {
    private String model;
    private List<String> components;
    // getters and setters
}

我编写了一个解决方案,但正在寻找更好的流解决方案。

public Map<String, List<Car>> componentsInCar() {
    HashSet<String> components = new HashSet<>();
    cars.stream().forEach(x -> x.getComponents().stream().forEachOrdered(components::add));
    Map<String, List<Car>> mapCarsComponents  = new HashMap<>();
    for (String keys : components) {
        mapCarsComponents.put(keys,
                cars.stream().filter(c -> c.getComponents().contains(keys)).collect(Collectors.toList()));
    }
    return mapCarsComponents;
}

2 个答案:

答案 0 :(得分:2)

您也可以使用流来做到这一点,但是我发现它更具可读性:

public static Map<String, List<Car>> componentsInCar(List<Car> cars) {
    Map<String, List<Car>> result = new HashMap<>();
    cars.forEach(car -> {
        car.getComponents().forEach(comp -> {
            result.computeIfAbsent(comp, ignoreMe -> new ArrayList<>()).add(car);
        });
    });

    return result;
}

或使用流:

public static Map<String, List<Car>> componentsInCar(List<Car> cars) {
    return cars.stream()
               .flatMap(car -> car.getComponents().stream().distinct().map(comp -> new SimpleEntry<>(comp, car)))
               .collect(Collectors.groupingBy(
                   Entry::getKey,
                   Collectors.mapping(Entry::getValue, Collectors.toList())
               ));
}

答案 1 :(得分:0)

我知道这是一个Java问题,并且已经有一个Java答案。但是,我想补充一点,Kotlin是一种JVM语言,可以与Java完美地互操作,您可以非常轻松,干净地执行以下操作:

val carsByComponent = cars
    .flatMap { it.components }
    .distinct()
    .map { component -> component to cars.filter { car -> component in car.components } }
    .toMap()

或更简洁,但可读性更差:

val carsByComponent = cars
    .flatMap { car -> car.components.map { it to car } }
    .groupBy { it.first }
    .mapValues {it.value.map { it.second }}