Java 8中Collection转为Map的方法

  • Z时代
  • 2024-01-10
  • 分类：综合

java

Java 8中java.util.stream.Collectors提供了几个方法可用于把Collection转为Map结构，本文记录了个人对其中三个的理解。

MethodReturn Type
groupingByMap<K, List<T>>
partitioningByMap<Boolean, List<T>>
toMapMap<K,U>

1. 环境

Java: jdk1.8.0_144

2. 特性说明

Student.java

public class Student {
    private String studentNo;
    private String name;
    private Boolean gender;
    private int age;
    public Student(String studentNo, String name, Boolean gender, int age) {
        this.studentNo = studentNo;
        this.name = name;
        this.gender = gender;
        this.age = age;
    }
    public String getStudentNo() {
        return studentNo;
    }
    public String getName() {
        return name;
    }
    public Boolean getGender() {
        return gender;
    }
    public int getAge() {
        return age;
    }
    @Override
    public String toString() {
        return String.format("Student [studentNo=%s, name=%s, gender=%s, age=%s]", studentNo, name, gender, age);
    }
}

fakeStudent()方法

private List<Student> fakeStudent() {
    List<Student> students = new ArrayList<>();
    students.add(new Student("1", "name1", false, 2));
    students.add(new Student("2", "name2", false, 2));
    students.add(new Student("3", "name2", null, 2));
    students.add(new Student("4", "name4", true, 2));
    students.add(new Student(null, "name5", true, 2));
    return students;
}

2.1. Collectors.groupingBy

public static <T, K> Collector<T, ?, Map<K, List<T>>> groupingBy(Function<? super T, ? extends K> classifier) {
    return groupingBy(classifier, toList());
}

a) 按Function的返回值把集合分组，并以之为Key，对应的列表为Value，返回Map

b) 若Key对应的列表为空时，返回的Map中将不包含该Key

c) 若Function的返回值为Null，抛出NullPointerException

@Test(expected = NullPointerException.class)
public void shouldThrowNPEWhenGroupingByNullKey() {
    fakeStudent().stream().collect(Collectors.groupingBy(Student::getStudentNo));
}

2.2. Collectors.partitioningBy

public static <T> Collector<T, ?, Map<Boolean, List<T>>> partitioningBy(Predicate<? super T> predicate) {
    return partitioningBy(predicate, toList());
}

a) 按Predicate的返回值把集合分为两组，符合条件的列表以true为Key，不符合的列表以false为Key

b) 若Predicate的返回值为Null，抛出NullPointerException

@Test(expected = NullPointerException.class)
public void shouldReturnMapWhenPartitioningByNullKey() {
    fakeStudent().stream().collect(Collectors.partitioningBy(Student::getGender));
}

2.3. Collectors.toMap

public static <T, K, U> Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper, Function<? super T, ? extends U> valueMapper) {
    return toMap(keyMapper, valueMapper, throwingMerger(), HashMap::new);
}

a) 以keyMapper的Function返回值为Key且以valueMapper的Function返回值为Value，形成Map

b) 若Key为Null，依然可以正确返回

@Test
public void shouldReturnMapWhenToMapNullKey() {
    Map<String, Student> map = fakeStudent().stream()
            .collect(Collectors.toMap(Student::getStudentNo, Function.identity()));
    assertEquals("{null=Student [studentNo=null, name=name5, gender=true, age=2], "
            + "1=Student [studentNo=1, name=name1, gender=false, age=2], "
            + "2=Student [studentNo=2, name=name2, gender=false, age=2], "
            + "3=Student [studentNo=3, name=name2, gender=null, age=2], "
            + "4=Student [studentNo=4, name=name4, gender=true, age=2]}", map.toString());
}

c) 若Key值出现重复，默认抛出IllegalStateException

@Test
public void shouldThrowIllegalStateExceptionWhenToMapDuplicateKey() {
    Map<String, Student> map = null;
    try {
        map = fakeStudent().stream().collect(Collectors.toMap(Student::getName, Function.identity()));
    } catch (Exception e) {
        assertTrue(e instanceof IllegalStateException);
        assertEquals("Duplicate key Student [studentNo=2, name=name2, gender=false, age=2]", e.getMessage());
    }
    assertNull(map);
}

若需要避免Duplicate Key的问题，可以有两个选择

  • 确定toMap的冲突策略，例如指定前者

@Test
public void shouldReturnMapWhenToMapDuplicateKey() {
    Map<String, Student> map = fakeStudent().stream()
            .collect(Collectors.toMap(Student::getName, Function.identity(), (student1, student2) -> student1));
    assertEquals("{name5=Student [studentNo=null, name=name5, gender=true, age=2], "
            + "name4=Student [studentNo=4, name=name4, gender=true, age=2], "
            + "name2=Student [studentNo=2, name=name2, gender=false, age=2], "
            + "name1=Student [studentNo=1, name=name1, gender=false, age=2]}", map.toString());
}

  • 放弃toMap方法，而利用collect

@Test
public void shouldReturnMapWhenCollectDuplicateKey() {
    Map<String, Student> map = fakeStudent().stream().collect(HashMap::new, (m, v) -> m.put(v.getName(), v),
            HashMap::putAll);
    assertEquals("{name5=Student [studentNo=null, name=name5, gender=true, age=2], "
            + "name4=Student [studentNo=4, name=name4, gender=true, age=2], "
            + "name2=Student [studentNo=3, name=name2, gender=null, age=2], "
            + "name1=Student [studentNo=1, name=name1, gender=false, age=2]}", map.toString());
}

d) 若Value为Null，则抛出NullPointerException

@Test(expected = NullPointerException.class)
public void shouldThrowNPEWhenToMapNullValue() {
    fakeStudent().stream().collect(Collectors.toMap(Student::getStudentNo, Student::getGender));
}

3. 结语

  • Collectors.groupingBy/Collectors.partitioningBy中心思想都是把原来集合以某种条件分组，分组条件不能为Null；只是Collectors.partitioningBy的分组条件是断言，且永远返回true/false对应的两组值，它们对应的Value可能是空列表，而Collectors.groupingBy的分组结果是空列表则会被抛弃

@Test
public void shouldReturnSameMapWhenGroupingByAndPartitioningBy() {
    List<Student> students = fakeStudent().stream().filter(student -> student.getGender() != null)
            .collect(Collectors.toList());
    Map<Boolean, List<Student>> groupingByMap = students.stream()
            .collect(Collectors.groupingBy(Student::getGender));
    Map<Boolean, List<Student>> partitioningByMap = students.stream()
            .collect(Collectors.partitioningBy(Student::getGender));
    assertEquals("{false=[Student [studentNo=1, name=name1, gender=false, age=2], "
            + "Student [studentNo=2, name=name2, gender=false, age=2]], "
            + "true=[Student [studentNo=4, name=name4, gender=true, age=2], "
            + "Student [studentNo=null, name=name5, gender=true, age=2]]}", groupingByMap.toString());
    assertEquals(groupingByMap.toString(), partitioningByMap.toString());
}
@Test
public void shouldReturnDifferentMapWhenGroupingByAndPartitioningBy() {
    Function<Student, Boolean> function = student -> student.getAge() > 3;
    List<Student> students = fakeStudent();
    Map<Boolean, List<Student>> groupingByMap = students.stream().collect(Collectors.groupingBy(function));
    Map<Boolean, List<Student>> partitioningByMap = students.stream()
            .collect(Collectors.partitioningBy(function::apply));
    assertEquals("{false=[Student [studentNo=1, name=name1, gender=false, age=2], "
            + "Student [studentNo=2, name=name2, gender=false, age=2], "
            + "Student [studentNo=3, name=name2, gender=null, age=2], "
            + "Student [studentNo=4, name=name4, gender=true, age=2], "
            + "Student [studentNo=null, name=name5, gender=true, age=2]]}", groupingByMap.toString());
    assertEquals(
            "{false=[Student [studentNo=1, name=name1, gender=false, age=2], "
                    + "Student [studentNo=2, name=name2, gender=false, age=2], "
                    + "Student [studentNo=3, name=name2, gender=null, age=2], "
                    + "Student [studentNo=4, name=name4, gender=true, age=2], "
                    + "Student [studentNo=null, name=name5, gender=true, age=2]], true=[]}",
            partitioningByMap.toString());
}

  • Collectors.toMap与Collectors.groupingBy/Collectors.partitioningBy不一样，它只负责把集合中的元素根据某种形式拆解为一个Map，该Map的key可以为Null但不允许重复，同时Map的Value不可以为Null

4. 参考资料

  • 附代码地址
    https://github.com/hivsuper/study/blob/master/study-java8/src/test/java/org/lxp/java8/map/CollectionToMapTest.java

  • https://stackoverflow.com/questions/27993604/whats-the-purpose-of-partitioningby

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