问题

import java.util.List;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        List<ObjectOne> objectOneList = ...; // Initialize your ObjectOne list
        List<ObjectTwo> objectTwoList = ...; // Initialize your ObjectTwo list

        // Find and filter by field not on the other list using Java Stream API
        List<ObjectOne> filteredObjectOneList = objectOneList.stream()
                .filter(objectOne -> objectTwoList.stream()
                        .noneMatch(objectTwo -> objectOne.getId() == objectTwo.getXId()
                                && objectOne.getName().equals(objectTwo.getText())))
                .collect(Collectors.toList());
    }
}

class ObjectOne {
    private int id;
    private String name;

    public int getId() {
        return id;
    }

    public String getName() {
        return name;
    }
}

class ObjectTwo {
    private int xId;
    private String text;

    public int getXId() {
        return xId;
    }

    public String getText() {
        return text;
    }
}

Please note that the code provided is a Java example of how you can achieve the desired filtering using the Java Stream API. You would need to replace the placeholder comments (...) with actual list instances and make sure that the ObjectOne and ObjectTwo classes are defined with the necessary methods and fields.

List<ObjectOne>      List<ObjectTwo>
field | value        field | value
{id=5, name="aaa"}   {xId=4, text="aaa"}
{id=6, name="bbb"}   {xId=6, text="bbb"}
{id=7, name="ccc"}   {xId=5, text="ccc"}

CollectionUtils.filter(objectOne, s -> (
(ObjectOne) s).getId() == anyObject.getXId()
&&  (ObjectOne) s).getName() == anyObject.getText());

objectOne = {id=5, name="aaa"} , {id=7, name="ccc"}

答案1

得分: 1

你可以根据 ObjectTwo 列表创建一个 ObjectOne 列表，方法如下：

List<ObjectOne> objectOne = listTwo.stream()
        .map(x -> new ObjectOne(x.getxId(), x.getText()))
        .collect(Collectors.toList());

然后你可以使用 retainAll 方法来找到共同的元素：

listOne.retainAll(objectOne);

如果你不想修改 ObjectOne 列表，那么你可以从 listOne 创建第二个列表：

List<ObjectOne> listOne2 = new ArrayList<>(listOne);
listOne2.retainAll(objectOne);

注意，这个解决方案需要在 ObjectOne 类中使用 hashcode 和 equals 方法。

List<ObjectOne> objectOne = listTwo.stream()
        .map(x -> new ObjectOne(x.getxId(), x.getText()))
        .collect(Collectors.toList());

listOne.retainAll(objectOne);

List<ObjectOne> listOne2 = new ArrayList<>(listOne);
listOne2.retainAll(objectOne);

答案2

得分: 1

noneMatch 在这里对你有所帮助。

objectOnes.stream()
          .filter(x -> objectTwos.stream()
                                 .noneMatch(y -> y.text.equals(x.name) && y.xId == x.id))
          .collect(Collectors.toList());

objectOnes.stream()
          .filter(x -> objectTwos.stream()
                                 .noneMatch(y -> y.text.equals(x.name) && y.xId == x.id))
          .collect(Collectors.toList());

答案3

得分: 0

我不知道如何只使用一个流来完成这个操作，但至少我得到了一个处理两个流的解决方案。

List<ObjectOne> list1 = new ArrayList<>();
List<ObjectTwo> list2 = new ArrayList<>();
list1.stream()
        .filter(o1 -> isInObjectTwoList(list2, o1))
        .collect(Collectors.toList());

private boolean isInObjectTwoList(List<ObjectTwo> objectTwoList, ObjectOne o1) {
    return objectTwoList.stream()
            .filter(o2 -> o2.getText().equals(o1.getValue()))
            .findAny()
            .isPresent();
}

List<ObjectOne> list1 = new ArrayList<>();
List<ObjectTwo> list2 = new ArrayList<>();
list1.stream()
        .filter(o1 -> isInObjectTwoList(list2, o1))
        .collect(Collectors.toList());

private boolean isInObjectTwoList(List<ObjectTwo> objectTwoList, ObjectOne o1) {
    return objectTwoList.stream()
            .filter(o2 -> o2.getText().equals(o1.getValue()))
            .findAny()
            .isPresent();
}