Map<String, List<String>> map =list.stream().collect(Collectors.toMap(Person::getId, p->{ List<String> getNameList =newArrayList<>(); getNameList.add(p.getName());returngetNameList; }, (List<String> value1, List<String> value2) ->{ value1.addAll(value2);returnvalue1; } )) 3...
List 转 Set 可以通过使用HashSet或TreeSet来实现。HashSet是无序的,而TreeSet是有序的。 List<String> list = new ArrayList<>(); list.add("Alice"); list.add("Bob"); list.add("Charlie"); Set<String> set1 = new HashSet<>(list); // 使用 HashSet 转换 Set<String> set2 = new TreeSe...
15);User u3=newUser("laoBi",20);User u4=newUser("wangHao",20);list.add(u1);list.add(u2);list.add(u3);list.add(u4);staticclassUser{privateString name;privateint age;publicUser(String name,int age){this.name=name;this.age=age;}publicStringgetName(){returnname;}publicvoid...
Map<Integer, List<String>> personmap = ps.stream().collect(Collectors.toMap(Person::getId, s -> { List<String> names = new ArrayList<>(); if(s.getName()!=null){ names.add(s.getName()); } return names; }, ( v1, v2) -> { v1.addAll(v2); HashSet h = new HashSet(v1)...
1 将一个实体类的list集合转为map学生实体类：package test;public class Student {private Long id; private String age; public Long getId() { return id; } public void setId(Long id) { this.id = id; } public String getAge() { return age; } public void setAge(...
Map<String, Integer> map = list.stream() .collect(Collectors.toMap(Student::getName, Student::getAge)); 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 方式一存在的问题 Collectors.toMap()调用的方法如下: public static <T, K, U> Collector<T, ?, Map<K,U>> toMap( ...
private String name; private int age; //略其他属性和get、set方法 } 我们想把上述Student类型的List转换为Map,key是name,value是age,可以使用toMap()方法,如下代码所示: List<Student> studentList = new ArrayList<>(); //定如何根据Student的属性构建map的key和value,以及出现重复key时如何处理 Map<String,...
.toMap(ke -> String.valueOf(ke.getId()),v -> v.getName(),(k1,k2) -> k1));/*** 检验是否转map成功 ***///遍历方法一//获取map中的key键名的set集合Set<String> ketSet = map.keySet();// 得到迭代器Iterator<String> iterator = ketSet.iterator();//使用Lambda方式输出iterator.for...
();//方式一Map<String, String> stringMap = stuList.stream().collect(Collectors.toMap(v -> String.valueOf(v.getId()), v -> v.getName()));//方式二Map<Long, String> stringMap2 = stuList.stream().collect(Collectors.toMap(Stu::getId, Stu::getName));//转换成map的时候,可能出现key...
4 Map的Value值转换为ListMap<Integer,String> map = new HashMap<>();map.put(1,"AAAA");map.put(2,"BBBB");map.put(3,"CCCC");map.put(4,"DDDD");List<String> list = new ArrayList<String>(map.values());5 Array转换为SetString [] countries = {"AAAA", "BBBB", "CCCC", "DDDD"}...