Java8新特性

分组（多字段分组）:

Map<String,List<Fruit>> fruitMap = fruitList.stream().collect(Collectors.groupingBy(Fruit::getName));
Map<String, List<CalbPrOutputSum>> map = outputSums.stream().collect(Collectors.groupingBy(t -> t.getModelNum()));
Map<String, List<WorkPlanVo>> collect = value.stream().collect(Collectors.groupingBy(a -> MessageFormat.format("{0}#{1}#{2}#{3}#{4}", a.getWorkshop(),
        a.getProjectCode(), a.getProdLineCode(), a.getModelCode(), a.getOutputOp())));

使用年龄进行升序排序:

List<StudentInfo> studentsSortName = studentList.stream().sorted(Comparator.comparing(StudentInfo::getAge)).collect(Collectors.toList());

使用年龄进行降序排序（使用reversed()方法）:

List<StudentInfo> studentsSortName = studentList.stream().sorted(Comparator.comparing(StudentInfo::getAge).reversed()).collect(Collectors.toList());

使用年龄进行降序排序，年龄相同再使用身高升序排序:

List<StudentInfo> studentsSortName = studentList.stream().sorted(Comparator.comparing(StudentInfo::getAge).reversed()
                        .thenComparing(StudentInfo::getHeight)).collect(Collectors.toList());

排序，改变原来的集合

personList.sort(Comparator.comparing(Person::getAge)); // 正序
personList.sort(Comparator.comparing(Person::getAge).reversed()); // 倒序

过滤

List<Employee> employees = new ArrayList<Employee>();
employees.addAll(Arrays.asList(e1, e2, e3, e4, e5, e6, e7, e8, e9, e10));
Optional<Employee> first = employees.stream().filter(t -> "M".equals(t.getGender())).findFirst();
List<Employee> collect = employees.stream().filter(t -> "M".equals(t.getGender())).collect(Collectors.toList());
List<Employee> collect1 = employees.stream().filter(t -> "M".equals(t.getGender()))
    .filter(t -> "Ricky".equals(t.getFirstName())).collect(Collectors.toList());

去重后的size

dtos.stream().distinct().count()

在使用Collectors.groupingBy分组时,如果分组的字段中有值为null,会抛出空指针异常

class Stu {
    String age;
    String sex;
}
@Test
void test1() {
    List<Stu> objects = new ArrayList<>();
    objects = Arrays.asList(new Stu("18", null), new Stu("19", "男"), new Stu("17", "女"));
    Map<String, List<Stu>> collect = objects.stream().collect(Collectors.groupingBy(x -> x.getSex()));
    System.out.println(collect);
}
//输出：java.lang.NullPointerException: element cannot be mapped to a null key

解决办法：

@Test
public void test4() {
    List<Stu> objects = new ArrayList<>();
    objects = Arrays.asList(new Stu("18", null), new Stu("19", "男"), new Stu("17", "女"));
    Map<Optional<String>, List<Stu>> collect = objects.stream().collect(Collectors.groupingBy(x -> Optional.ofNullable(x.getSex())));
    System.out.println(collect);
    List<Stu> finalList = new ArrayList<>();
    for (Map.Entry<Optional<String>, List<Stu>> optionalListEntry : collect.entrySet()) {
        if (optionalListEntry.getKey().isPresent()) {
            finalList.addAll(optionalListEntry.getValue());
        }
    }
    System.out.println(finalList);
}
// 输出： {Optional.empty=[Stu{age='18', sex='null'}], Optional[女]=[Stu{age='17', sex='女'}], Optional[男]=[Stu{age='19', sex='男'}]} 
// [Stu{age='17', sex='女'}, Stu{age='19', sex='男'}]

java8 Stream 取出两个list中相同部分

List l = (List)list.stream().filter(it->list2.contains(it)).collect(Collectors.toList());

获取年龄最大/最小的Person

Person maxAgePerson = personList.stream().max(Comparator.comparing(Person::getAge)).get();
long asLong = list.stream().mapToLong(User::getAge).max().getAsLong();
Person minAgePerson = personList.stream().min(Comparator.comparing(Person::getAge)).get();
long asLong1 = list.stream().mapToLong(User::getAge).min().getAsLong();

统计出年龄等于20的个数

long count = personList.stream().filter(person -> person.getAge() == 20).count();

获得年龄的平均值

double asDouble = personList.stream().mapToInt(person -> person.getAge()).average().getAsDouble();

获得年龄的求和

int sum = personList.stream().mapToInt(person -> person.getAge()).sum();
int sum = list.stream().reduce(0, (acc, value) -> acc + value);

// 将名称积累到一个树集
Set<String> set = people.stream().map(Person::getName).collect(Collectors.toCollection(TreeSet::new));
// 将元素转换为字符串，并用逗号分隔
String joined = things.stream()
    .map(Object::toString)
    .collect(Collectors.joining(", "));
// 计算员工工资总额
int total = employees.stream()
    .collect(Collectors.summingInt(Employee::getSalary)));
// 按部门计算工资总额
Map<Department, Integer> totalByDept
            = employees.stream()
    .collect(Collectors.groupingBy(Employee::getDepartment,
                                   Collectors.summingInt(Employee::getSalary)));
// 把学生分成及格和不及格
Map<Boolean, List<Student>> passingFailing =
    students.stream()
    .collect(Collectors.partitioningBy(s -> s.getGrade() >= PASS_THRESHOLD));
// 找出集合李重复元素个数大于2的
List<String> list = Arrays.asList("aaa","aaa","ccc","aaa","bbb","bbb","ddd","ddd","ddd","ddd");
List<String> collect = list.stream()
    .collect(Collectors.toMap(e-> {
        return e;
    }, e -> 1, (a, b) -> {
        return a + b;
    })) // 获得元素出现频率的 Map，键为元素，值为元素出现的次数
    .entrySet().stream() // Set<Entry>转换为Stream<Entry>
    .filter(entry -> entry.getValue() > 2) // 过滤出元素出现次数大于 1 的 entry
    .map(entry -> entry.getKey()) // 获得 entry 的键（重复元素）对应的 Stream
    .collect(Collectors.toList());

Java8新特性Stream之list转map及问题解决
1.Java8 使用 stream().map()提取List对象的某一列值及排重

@Test
void contextLoads0() {
    ArrayList<Product> products = new ArrayList<>();
    products.add(new Product(1,"name1","location1"));
    products.add(new Product(2,"name2","location2"));
    products.add(new Product(3,"name3","location3"));
    List<String> collect = products.stream().map(Product::getName).collect(Collectors.toList());
    //products.stream().map(t -> t.getName()).collect(Collectors.toList());
    System.out.println(collect);
}
输出：[name1, name2, name3]
排重：List<Integer> ageList = studentList.stream().map(StudentInfo::getAge).distinct().collect(Collectors.toList());

2. List集合转Map

@Test
void contextLoad1() {
    ArrayList<Product> products = new ArrayList<>();
    products.add(new Product(1,"name1","location1"));
    products.add(new Product(2,"name2","location2"));
    products.add(new Product(3,"name3","location3"));
    Map<Integer, String> collect = products.stream().collect(Collectors.toMap(Product::getId, Product::getName));
    System.out.println(collect);
}
输出：{1=name1, 2=name2, 3=name3}

3. key重复时用后面的value 覆盖前面的value

@Test
void contextLoad2() {
    ArrayList<Product> products = new ArrayList<>();
    products.add(new Product(1,"name1","location1"));
    products.add(new Product(1,"name2","location2"));
    products.add(new Product(3,"name3","location3"));
    Map<Integer, String> collect = products.stream().collect(Collectors.toMap(Product::getId, Product::getName,(k1,k2)->k2));
    System.out.println(collect);
}
输出：{1=name2, 3=name3}

4. 重复时将前面的value和后面的value逗号拼接起来

@Test
void contextLoad3() {
    ArrayList<Product> products = new ArrayList<>();
    products.add(new Product(1,"name1","location1"));
    products.add(new Product(1,"name2","location2"));
    products.add(new Product(3,"name3","location3"));
    Map<Integer, String> collect = products.stream()
            .collect(Collectors.toMap(Product::getId, Product::getName,(k1,k2)->k1 + "," + k2));
    System.out.println(collect);
}
输出：{1=name1,name2, 3=name3}

5.重复时将重复value的数据组成集合

@Test
void contextLoad4() {
    ArrayList<Product> products = new ArrayList<>();
    products.add(new Product(1,"name1","location1"));
    products.add(new Product(1,"name2","location2"));
    products.add(new Product(3,"name3","location3"));
    Map<Integer, List<String>> collect = products.stream().collect(Collectors.toMap(Product::getId,
            p -> {
                List<String> NameList = new ArrayList<>();
                NameList.add(p.getName());
                return NameList;
            },
            (List<String> value1, List<String> value2) -> {
                value1.addAll(value2);
                return value1;
            }
    ));
    System.out.println(collect);
}
输出：{1=[name1, name2], 3=[name3]}

6. value为null时会出现NullPointerException
解决办法：设置时加判断如果是null，则设置成一个特定值

Map<String,String> resultMap = list.stream().collect(Collectors.toMap(Student::getId, t -> t.getName()==null ? "":t.getName()));