Java中对象的聚合列表
Java中是否有任何聚合器函数来执行以下聚合Java中对象的聚合列表,java,java-8,aggregate-functions,java-stream,Java,Java 8,Aggregate Functions,Java Stream,Java中是否有任何聚合器函数来执行以下聚合 Person { String name; String subject; String department; Long mark1; Long mark2; Long mark3; } 列表包含如下数据 Name |Subject |Department |Mark1 |Mark2 |Mark3 --------|-----------|-----------|-------|--
Person {
String name;
String subject;
String department;
Long mark1;
Long mark2;
Long mark3;
}
列表包含如下数据
Name |Subject |Department |Mark1 |Mark2 |Mark3
--------|-----------|-----------|-------|-------|-----
Clark |English |DEP1 |7 |8 |6
Michel |English |DEP1 |6 |4 |7
Dave |Maths |DEP2 |3 |5 |6
Mario |Maths |DEP1 |9 |7 |8
private static List<Person> getGrouped(List<Person> origList) {
Map<String, Person> grpMap = new HashMap<String, Person>();
for (Person person : origList) {
String key = person.getDepartment() + person.getSubject();
if (grpMap.containsKey(key)) {
Person grpdPerson = grpMap.get(key);
grpdPerson.setMark1(grpdPerson.getMark1() + person.getMark1());
grpdPerson.setMark2(grpdPerson.getMark2() + person.getMark2());
grpdPerson.setMark3(grpdPerson.getMark3() + person.getMark3());
} else {
grpMap.put(key, person);
}
}
return new ArrayList<Person>(grpMap.values());
}
姓名|科目|部门|标记1 |标记2 |标记3
--------|-----------|-----------|-------|-------|-----
克拉克|英语| DEP1 | 7 | 8 | 6
米歇尔|英语| DEP1 | 6 | 4 | 7
戴夫|数学| DEP2 | 3 | 5 | 6
马里奥|数学| DEP1 | 9 | 7 | 8
聚合条件为Subject&Dep。结果对象需要
Subject |Department |Mark1 |Mark2 |Mark3
----------- |-----------|-------|-------|-----
English |DEP1 |13 |12 |13
Maths |DEP2 |3 |5 |6
Maths |DEP1 |9 |7 |8
科目|部门|标记1 |标记2 |标记3
----------- |-----------|-------|-------|-----
英语| DEP1 | 13 | 12 | 13
数学| DEP2 | 3 | 5 | 6
数学| DEP1 | 9 | 7 | 8
可以通过手动迭代列表并创建聚合列表来实现此聚合。示例如下
Name |Subject |Department |Mark1 |Mark2 |Mark3
--------|-----------|-----------|-------|-------|-----
Clark |English |DEP1 |7 |8 |6
Michel |English |DEP1 |6 |4 |7
Dave |Maths |DEP2 |3 |5 |6
Mario |Maths |DEP1 |9 |7 |8
private static List<Person> getGrouped(List<Person> origList) {
Map<String, Person> grpMap = new HashMap<String, Person>();
for (Person person : origList) {
String key = person.getDepartment() + person.getSubject();
if (grpMap.containsKey(key)) {
Person grpdPerson = grpMap.get(key);
grpdPerson.setMark1(grpdPerson.getMark1() + person.getMark1());
grpdPerson.setMark2(grpdPerson.getMark2() + person.getMark2());
grpdPerson.setMark3(grpdPerson.getMark3() + person.getMark3());
} else {
grpMap.put(key, person);
}
}
return new ArrayList<Person>(grpMap.values());
}
私有静态列表getGrouped(List origList){
Map grpMap=newhashmap();
for(个人:origList){
字符串键=person.getDepartment()+person.getSubject();
if(grpMap.CONTANSKEY(关键)){
Person grpdPerson=grpMap.get(键);
grpdPerson.setMark1(grpdPerson.getMark1()+person.getMark1());
grpdPerson.setMark2(grpdPerson.getMark2()+person.getMark2());
grpdPerson.setMark3(grpdPerson.getMark3()+person.getMark3());
}否则{
grpMap.put(钥匙、人员);
}
}
返回新的ArrayList(grpMap.values());
}
但是,我们是否可以利用Java 8的任何聚合功能或特性?您可以使用。聚合标记1的样本如下所示
public class Test {
static class Person {
Person(String name, String subject, String department, Long mark1, Long mark2, Long mark3) {
this.name = name;
this.subject = subject;
this.department = department;
this.mark1 = mark1;
this.mark2 = mark2;
this.mark3= mark3;
}
String name;
String subject;
String department;
Long mark1;
Long mark2;
Long mark3;
String group() {
return subject+department;
}
Long getMark1() {
return mark1;
}
}
public static void main(String[] args)
{
List<Person> list = new ArrayList<Test.Person>();
list.add(new Test.Person("Clark","English","DEP1",7l,8l,6l));
list.add(new Test.Person("Michel","English","DEP1",6l,4l,7l));
list.add(new Test.Person("Dave","Maths","DEP2",3l,5l,6l));
list.add(new Test.Person("Mario","Maths","DEP1",9l,7l,8l));
Map<String, Long> groups = list.stream().collect(Collectors.groupingBy(Person::group, Collectors.reducing(
0l, Person::getMark1, Long::sum)));
//Or alternatively as suggested by Holger
Map<String, Long> groupsNew = list.stream().collect(Collectors.groupingBy(Person::group, Collectors.summingLong(Person::getMark1)));
System.out.println(groups);
}
}
公共类测试{
静态类人{
人员(字符串名称、字符串主题、字符串部门、长标记1、长标记2、长标记3){
this.name=名称;
this.subject=主语;
这个部门=部门;
this.mark1=mark1;
this.mark2=mark2;
这个.mark3=mark3;
}
字符串名;
字符串主题;
弦乐部;
长标记1;
长标记2;
长标记3;
字符串组(){
返回主题+部门;
}
Long getMark1(){
返回标记1;
}
}
公共静态void main(字符串[]args)
{
列表=新的ArrayList();
添加(新测试人员(“克拉克”、“英语”、“DEP1”、7l、8l、6l”);
添加(新的测试人员(“Michel”、“English”、“DEP1”、6l、4l、7l”);
添加(新的测试人员(“Dave”、“数学”、“DEP2”、3l、5l、6l”);
添加(新的测试人员(“马里奥”、“数学”、“DEP1”、9l、7l、8l”);
映射组=list.stream().collect(Collectors.groupingBy(Person::group,Collectors.com)(
0l,Person::getMark1,Long::sum));
//或者按照霍尔格的建议
Map groupsNew=list.stream().collect(Collectors.groupingBy(Person::group,Collectors.summingLong(Person::getMark1));
系统输出打印项次(组);
}
}
仍在考虑通过单个函数生成输出。将在完成后更新。使用JDK中的标准收集器,可以这样做(假设创建了
Tuple3
类):
在这种情况下,您可能还希望使用toMap
收集器使用另一个变量。逻辑保持不变,映射值的函数将创建一个映射,其中包含部门作为键,学生的成绩作为值。合并功能将负责添加或更新映射
Map<String, Map<String, Tuple3<Long, Long, Long>>> res3 =
persons.stream()
.collect(toMap(p -> p.subject,
p -> {
Map<String, Tuple3<Long, Long, Long>> value = new HashMap<>();
value.put(p.department, new Tuple3<>(p.mark1, p.mark2, p.mark3));
return value;
},
(v1, v2) -> {
v2.forEach((k, v) -> v1.merge(k, v, (t1, t2) -> new Tuple3<>(t1.e1 + t2.e1, t1.e2 + t2.e2, t1.e3 + t2.e3)));
return v1;
}
));
Map res3=
人流()
.收集(toMap)(p->p.subject,
p->{
Map value=newhashmap();
value.put(p.department,新元组3(p.mark1,p.mark2,p.mark3));
返回值;
},
(v1,v2)->{
v2.forEach((k,v)->v1.merge(k,v,(t1,t2)->新元组3(t1.e1+t2.e1,t1.e2+t2.e2,t1.e3+t2.e3));
返回v1;
}
));
当然,您可以对这些解决方案的“美”提出疑问,也许您想引入一个自定义收集器或自定义类,以使意图更加明确。使用来自自定义密钥类的方法,我的建议如下:
Map<DepSubject, Grades> map = persons.stream().
collect(Collectors.groupingBy(x -> new DepSubject(x.department, x.subject),
Collectors.reducing(
new Grades(0, 0, 0),
y -> new Grades(y.mark1, y.mark2, y.mark3),
(x, y) -> new Grades(x.m1 + y.m1, x.m2 + y.m2, x.m3 + y.m3)
)));
也可以将结果收集到列表中。这样,自定义类DepSubject
和Grades
仅用于中间操作:
List<Person> list = persons.stream().
collect(Collectors.collectingAndThen(
Collectors.groupingBy(x -> new DepSubject(x.department, x.subject),
Collectors.reducing(
new Grades(0, 0, 0),
y -> new Grades(y.mark1, y.mark2, y.mark3),
(x, y) -> new Grades(x.m1 + y.m1, x.m2 + y.m2, x.m3 + y.m3)
)),
map -> map.entrySet().stream()
.map(e -> new Person(null, e.getKey().subject, e.getKey().department, e.getValue().m1, e.getValue().m2, e.getValue().m3))
.collect(Collectors.toList())
));
List List=persons.stream()。
收集(收集器。收集然后(
收集者。分组方式(x->new DepSubject(x.department,x.subject),
还原剂(
新职系(0,0,0),,
y->新等级(y.mark1、y.mark2、y.mark3),
(x,y)->新等级(x.m1+y.m1,x.m2+y.m2,x.m3+y.m3)
)),
map->map.entrySet().stream()
.map(e->newperson(null,e.getKey().subject,e.getKey().department,e.getValue().m1,e.getValue().m2,e.getValue().m3))
.collect(收集器.toList())
));
您还可以将groupingBy逻辑提取到函数中:
private static <T> List<Person> groupBy(List<Person> persons, Function<Person,T> function, BiFunction<T,Grades,Person> biFunction) {
return persons.stream().
collect(Collectors.collectingAndThen(
Collectors.groupingBy(function,
Collectors.reducing(
new Grades(0, 0, 0),
y -> new Grades(y.mark1, y.mark2, y.mark3),
(x, y) -> new Grades(x.m1 + y.m1, x.m2 + y.m2, x.m3 + y.m3)
)),
map -> map.entrySet().stream()
.map(e -> biFunction.apply(e.getKey(),e.getValue()))
.collect(Collectors.toList())
));
}
private static List groupBy(列出人员、函数、双函数){
return persons.stream()。
收集(收集器。收集然后(
收集器。分组方式(函数,
还原剂(
新职系(0,0,0),,
y->新年级(y.ma)
private static <T> List<Person> groupBy(List<Person> persons, Function<Person,T> function, BiFunction<T,Grades,Person> biFunction) {
return persons.stream().
collect(Collectors.collectingAndThen(
Collectors.groupingBy(function,
Collectors.reducing(
new Grades(0, 0, 0),
y -> new Grades(y.mark1, y.mark2, y.mark3),
(x, y) -> new Grades(x.m1 + y.m1, x.m2 + y.m2, x.m3 + y.m3)
)),
map -> map.entrySet().stream()
.map(e -> biFunction.apply(e.getKey(),e.getValue()))
.collect(Collectors.toList())
));
}
List<Person> list = groupBy(persons,
x -> new DepSubject(x.department, x.subject),
(depSubject,grades) -> new Person(null, depSubject.subject, depSubject.department, grades.m1, grades.m2, grades.m3));
List<Person> list2 = groupBy(persons,
Person::getSubject,
(subject,grades) -> new Person(null,subject, null, grades.m1, grades.m2, grades.m3));