Question

我有两个列表，我必须根据它们创建一个地图。首先，我在for循环中进行迭代，其次我想被流抛弃而不是收集来映射，但是我不知道如何在特定情况下使用Collectors.toMap。有可能吗？

我已经提出了解决方案，但没有使用流，但是我很好奇是否可以这样做，如果可以，怎么办？

public void findMatch(List<ObjectA> objectAList, List<ObjectB> objectBList) {
Map<ObjectB, ObjectA> objectBObjectAMap = new HashMap<>();
for (ObjectB objectB : objectBList) {
    if (isNull(objectB.getHandoverTime())) {
        objectBObjectAMap.putAll(
        objectAList
                .stream()
                .filter(objectA -> {
                    ObjectC objectC = objectB.getObjectC();
                    return objectA.getNumber().equals(objectC.getNumber())
                            && objectA.getQuality().equals(objectC.getQuality());
                })
                .collect(Collectors.toMap(???)));
        }
    }
}

Answer 1

您可以尝试使用flatMap：

Map<ObjectB, ObjectA> objectBObjectAMap =
    objectBList.stream()
               .filter(b -> isNull(b.getHandoverTime()))
               .flatMap(b -> objectAList.stream()
                                        .filter(a -> {
                                            ObjectC c = b.getObjectC();
                                            return a.getNumber().equals(c.getNumber()) &&
                                                   a.getQuality().equals(c.getQuality());
                                        })
                                        .map(a -> new SimpleEntry<>(b,a)))
               .collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

这是假设每个ObjectB实例都不会与多个ObjectA实例相关联。

Answer 2

我试图恢复您的问题，并创建了三个简单的案例

public class ObjectA {
    Long number;
    String quality;
    ...
}

public class ObjectB {
    Long number;
    String quality;
    Date handoverTime;
    ObjectC objectC;
    ...
}

public class ObjectC {
    Long number;
    String quality;
    ...
}

和主要功能，希望您能处理。

    ObjectA a1 = new ObjectA();
    a1.setNumber(1L);
    a1.setQuality("aaa1");

    ObjectA a2 = new ObjectA();
    a2.setNumber(2L);
    a2.setQuality("aaa2");

    List<ObjectA> aList = new ArrayList<>();
    aList.add(a1);
    aList.add(a2);

    ObjectB b1 = new ObjectB();
    b1.setNumber(3L);
    b1.setQuality("bbb1");
    //b1.setHandoverTime(new Date());
    ObjectC c1 = new ObjectC();
    c1.setNumber(1L);
    c1.setQuality("aaa1");
    b1.setObjectC(c1);

    ObjectB b2 = new ObjectB();
    b2.setNumber(4L);
    b2.setQuality("bbb2");
    //b2.setHandoverTime(new Date());
    ObjectC c2 = new ObjectC();
    c2.setNumber(2L);
    c2.setQuality("aaa2");
    b2.setObjectC(c2);

    List<ObjectB> bList = new ArrayList<>();
    bList.add(b1);
    bList.add(b2);

    Map<ObjectB, ObjectA> mapzz = findMatch(aList, bList);

    System.out.println(mapzz);

和以下方法

public static Map<ObjectB, ObjectA> findMatch(List<ObjectA> objectAList, List<ObjectB> objectBList) {
    List<ObjectA> checkPoint1 = new ArrayList<>();
    Map<ObjectB, ObjectA> mapzz = new HashMap<>();

    mapzz.putAll(
        objectBList.stream()
            .filter(objB -> isNull(objB.getHandoverTime()))
            .collect(Collectors.toMap(Function.identity(), 
                    objB -> objectAList
                        .stream()
                        .filter((a) -> a.getNumber().equals(objB.getObjectC().getNumber()) && a.getQuality().equals(objB.getObjectC().getQuality()))
                        .peek(checkPoint1::add)
                        .findAny().get()
                    ))
    );
    return mapzz;
}

结果如下：

{ObjectB {number = 3，quality = bbb1，handoverTime = null，objectC = ObjectC {number = 1，quality = aaa1}} = ObjectA {number = 1，quality = aaa1}，ObjectB {number = 4，quality = bbb2，handoverTime = null，objectC = ObjectC {number = 2，quality = aaa2}} = ObjectA {number = 2，quality = aaa2}}

希望对您有帮助。

在过滤到地图后收集到流的问题

2 个答案: