如何在 java 中组合 2 个具有相同属性值的对象数组列表

Question

我有 2 个数组列表，如下所示。 一个数组包含用id 、 name和city映射的学生信息。 另一个数组包含用id, marks and grade 映射的标记详细信息. The goal is to compare each of these elements against each other and combine them based on the common field id in both the . The goal is to compare each of these elements against each other and combine them based on the common field id in both the 。

student_details=[
        {
            "id": 1,
            "name": "sam",
            "city": "Chennai"
           
        },
        {
            "id": 2,
            "name": "peter",
            "city": "Chennai"
            
        }
    ]
student_grades=[
        {
            "id": 1,
            "marks": 95,
            "grade": "A"
           
        },
        {
            "id": 2,
            "marks":63,
            "grade": "B"
            
        }
    ]

结果应如下所示。

[
        {
            "id": 1,
            "name": "sam",
            "city": "Chennai",
            "marks": 95,
            "grade": "A"
           
        },
        {
            "id": 2,
            "name": "peter",
            "city": "Chennai",
            "marks":63,
            "grade": "B"
            
        }
    ]

我使用下面的代码在 javascript 中实现了相同的功能。 如何在 java 中实现这个？

async mergeData(data1, data2, key) {
    const result = data1.map(a =>
      Object.assign({}, a,
        data2.find(b => b[key] === a[key])
      )
    )
    return result
  }

Answer 1

如果您的数据包含的属性始终相同，那么最简单的方法是创建一个 ArrayList，然后在将所有条目添加到 ArrayList 后调用 toArrayMethod，否则您必须考虑 id 不存在的不同情况match 这将需要大量代码来解释不同的数组大小。

然后，您只需在给定的 arrays 上迭代 2 个嵌套 for 循环，如果 id 匹配，则使用 ArrayList 的 add 方法将新条目添加到 Z57A97A39435CFDFED96E03F2A3BC2。

The other question is what object type your data actually has, because I don't think you can create object pairs like of the nature of "attribute_name" = value in java, so your input is probably some json you're reading in. You could use gson or other libraries to map your json to 2 java classes, then use another 3rd class for the combined data.

Another way would be to map it yourself with a json handling library to a java class or some datastructure like a HashMap. 如果您不知道 java class 将来会有什么属性，那么 HashMap 最有意义。

Another way would be to map the object from json to another json with a json handling library, with 2 for loops iterating over your json arrays.

Answer 2

public static void main(String... args) {
    List<Map<String, String>> studentDetails = List.of(
            Map.of("id", "1", "name", "sam", "city", "Chennai"),
            Map.of("id", "2", "name", "peter", "city", "Chennai"));
    List<Map<String, String>> studentGrades = List.of(
            Map.of("id", "1", "marks", "95", "grade", "A"),
            Map.of("id", "2", "marks", "63", "grade", "B"));

    List<Map<String, String>> res = merge(studentDetails, studentGrades);
}

public static List<Map<String, String>> merge(
        List<Map<String, String>> studentDetails,
        List<Map<String, String>> studentGrades) {
    Map<String, Map<String, String>> map = new HashMap<>();
    Consumer<Map<String, String>> add = studentData -> {
        String id = studentData.get("id");
        map.computeIfAbsent(id, key -> new HashMap<>());
        map.get(id).putAll(studentData);
    };

    studentDetails.forEach(add);
    studentGrades.forEach(add);

    return new ArrayList<>(map.values());
}

Answer 3

您可以使用如下流和记录，如下所示：

package pl.maro;

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

public class MyUtils {

    public static void main(String[] args) {
        var studentDetails = List.of(
            new StudentDetail(1, "sam", "Chennai"),
            new StudentDetail(2, "peter", "Chennai")
        );

        var studentGrades = List.of(
                new StudentGrade(1, 95, 'A'),
                new StudentGrade(2, 63, 'B')
        );

        List<StudentDao> studentDaos = yourMethod(studentDetails, studentGrades);
        studentDaos.forEach(System.out::println);
    }

    private static List<StudentDao> yourMethod(List<StudentDetail> studentDetails, List<StudentGrade> studentGrades) {
        var detailMap = studentDetails.stream().collect(Collectors.toMap(k -> k.id, v -> v));
        var gradeMap = studentGrades.stream().collect(Collectors.toMap(k-> k.id, v->v));

        return Stream.of(detailMap.keySet(), gradeMap.keySet())
                .flatMap(Collection::stream)
                .collect(Collectors.toSet())
                .stream()
                .filter(id -> detailMap.containsKey(id) && gradeMap.containsKey(id))
                .map(id -> createStudentDao(detailMap.get(id), gradeMap.get(id)))
                .toList();
    }

    private static StudentDao createStudentDao(StudentDetail studentDetail, StudentGrade studentGrade) {
        return new StudentDao(studentDetail.id, studentDetail.name, studentDetail.surname, studentGrade.marks, studentGrade.grade);
    }

    record StudentDetail(int id, String name, String surname) {}
    record StudentGrade(int id, int marks, char grade) {}
    record StudentDao(int id, String name, String surname, int marks, char grade) {
        @Override
        public String toString() {
            return "StudentDao{" +
                    "id=" + id +
                    ", name='" + name + '\'' +
                    ", surname='" + surname + '\'' +
                    ", marks=" + marks +
                    ", grade=" + grade +
                    '}';
        }
    }
}