我需要使用特定对象的属性（Location）对对象列表（Student）进行分组，代码如下所示，
public class Grouping {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
List<Student> studlist = new ArrayList<Student>();
studlist.add(new Student("1726", "John", "New York"));
studlist.add(new Student("4321", "Max", "California"));
studlist.add(new Student("2234", "Andrew", "Los Angeles"));
studlist.add(new Student("5223", "Michael", "New York"));
studlist.add(new Student("7765", "Sam", "California"));
studlist.add(new Student("3442", "Mark", "New York"));
//Code to group students by location
/* Output should be Like below
ID : 1726 Name : John Location : New York
ID : 5223 Name : Michael Location : New York
ID : 4321 Name : Max Location : California
ID : 7765 Name : Sam Location : California
*/
for (Student student : studlist) {
System.out.println("ID : "+student.stud_id+"\t"+"Name : "+student.stud_name+"\t"+"Location : "+student.stud_location);
}
}
}
class Student {
String stud_id;
String stud_name;
String stud_location;
Student(String sid, String sname, String slocation) {
this.stud_id = sid;
this.stud_name = sname;
this.stud_location = slocation;
}
}
请建议我一个干净的方法。
答案 0 :(得分：198)
在Java 8中：
Map<String, List<Student>> studlistGrouped =
studlist.stream().collect(Collectors.groupingBy(w -> w.stud_location));
答案 1 :(得分：100)
这会将学生对象添加到以locationID作为关键字的HashMap。
HashMap<Integer, List<Location>> hashMap = new HashMap<Integer, List<Location>>();
对此代码进行迭代并将学生添加到HashMap：
if (!hashMap.containsKey(locationId)) {
List<Location> list = new ArrayList<Location>();
list.add(student);
hashMap.put(locationId, list);
} else {
hashMap.get(locationId).add(student);
}
如果您希望所有学生都有特定的位置详细信息，那么您可以使用：
hashMap.get(locationId);
将为您提供具有相同位置ID的所有学生。
答案 2 :(得分：28)
Map<String, List<Student>> map = new HashMap<String, List<Student>>();
for (Student student : studlist) {
String key = student.stud_location;
if(map.containsKey(key)){
List<Student> list = map.get(key);
list.add(student);
}else{
List<Student> list = new ArrayList<Student>();
list.add(student);
map.put(key, list);
}
}
答案 3 :(得分：6)
使用 Java 8
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import java.util.stream.Stream;
class Student {
String stud_id;
String stud_name;
String stud_location;
public String getStud_id() {
return stud_id;
}
public String getStud_name() {
return stud_name;
}
public String getStud_location() {
return stud_location;
}
Student(String sid, String sname, String slocation) {
this.stud_id = sid;
this.stud_name = sname;
this.stud_location = slocation;
}
}
class Temp
{
public static void main(String args[])
{
Stream<Student> studs =
Stream.of(new Student("1726", "John", "New York"),
new Student("4321", "Max", "California"),
new Student("2234", "Max", "Los Angeles"),
new Student("7765", "Sam", "California"));
Map<String, Map<Object, List<Student>>> map= studs.collect(Collectors.groupingBy(Student::getStud_name,Collectors.groupingBy(Student::getStud_location)));
System.out.println(map);//print by name and then location
}
}
结果将是：
{
Max={
Los Angeles=[Student@214c265e],
California=[Student@448139f0]
},
John={
New York=[Student@7cca494b]
},
Sam={
California=[Student@7ba4f24f]
}
}
答案 4 :(得分：4)
使用Comparator在Java中实现SQL GROUP BY功能，比较器将比较您的列数据并对其进行排序。基本上，如果您保持排序数据看起来像分组数据，例如，如果您有相同的重复列数据，则排序机制对它们进行排序，将数据保持在一侧，然后查找其他数据，这些数据是不同的数据。这间接被视为相同数据的GROUPING。
public class GroupByFeatureInJava {
public static void main(String[] args) {
ProductBean p1 = new ProductBean("P1", 20, new Date());
ProductBean p2 = new ProductBean("P1", 30, new Date());
ProductBean p3 = new ProductBean("P2", 20, new Date());
ProductBean p4 = new ProductBean("P1", 20, new Date());
ProductBean p5 = new ProductBean("P3", 60, new Date());
ProductBean p6 = new ProductBean("P1", 20, new Date());
List<ProductBean> list = new ArrayList<ProductBean>();
list.add(p1);
list.add(p2);
list.add(p3);
list.add(p4);
list.add(p5);
list.add(p6);
for (Iterator iterator = list.iterator(); iterator.hasNext();) {
ProductBean bean = (ProductBean) iterator.next();
System.out.println(bean);
}
System.out.println("******** AFTER GROUP BY PRODUCT_ID ******");
Collections.sort(list, new ProductBean().new CompareByProductID());
for (Iterator iterator = list.iterator(); iterator.hasNext();) {
ProductBean bean = (ProductBean) iterator.next();
System.out.println(bean);
}
System.out.println("******** AFTER GROUP BY PRICE ******");
Collections.sort(list, new ProductBean().new CompareByProductPrice());
for (Iterator iterator = list.iterator(); iterator.hasNext();) {
ProductBean bean = (ProductBean) iterator.next();
System.out.println(bean);
}
}
}
class ProductBean {
String productId;
int price;
Date date;
@Override
public String toString() {
return "ProductBean [" + productId + " " + price + " " + date + "]";
}
ProductBean() {
}
ProductBean(String productId, int price, Date date) {
this.productId = productId;
this.price = price;
this.date = date;
}
class CompareByProductID implements Comparator<ProductBean> {
public int compare(ProductBean p1, ProductBean p2) {
if (p1.productId.compareTo(p2.productId) > 0) {
return 1;
}
if (p1.productId.compareTo(p2.productId) < 0) {
return -1;
}
// at this point all a.b,c,d are equal... so return "equal"
return 0;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
return super.equals(obj);
}
}
class CompareByProductPrice implements Comparator<ProductBean> {
@Override
public int compare(ProductBean p1, ProductBean p2) {
// this mean the first column is tied in thee two rows
if (p1.price > p2.price) {
return 1;
}
if (p1.price < p2.price) {
return -1;
}
return 0;
}
public boolean equals(Object obj) {
// TODO Auto-generated method stub
return super.equals(obj);
}
}
class CompareByCreateDate implements Comparator<ProductBean> {
@Override
public int compare(ProductBean p1, ProductBean p2) {
if (p1.date.after(p2.date)) {
return 1;
}
if (p1.date.before(p2.date)) {
return -1;
}
return 0;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
return super.equals(obj);
}
}
}
输出在这里为上面的ProductBean列表完成GROUP BY标准，这里如果你看到给出了ProductBean列表的输入数据到Collections.sort（列表，Comparator的对象为你需要的列）这将基于您的比较器实现，您将能够在下面的输出中看到GROUPED数据。希望这会有所帮助...
******** BEFORE GROUPING INPUT DATA LOOKS THIS WAY ******
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 30 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P2 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P3 60 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
******** AFTER GROUP BY PRODUCT_ID ******
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 30 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P2 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P3 60 Mon Nov 17 09:31:01 IST 2014]
******** AFTER GROUP BY PRICE ******
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P2 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 20 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P1 30 Mon Nov 17 09:31:01 IST 2014]
ProductBean [P3 60 Mon Nov 17 09:31:01 IST 2014]
答案 5 :(得分：3)
您可以使用以下内容：
Map<String, List<Student>> groupedStudents = new HashMap<String, List<Student>>();
for (Student student: studlist) {
String key = student.stud_location;
if (groupedStudents.get(key) == null) {
groupedStudents.put(key, new ArrayList<Student>());
}
groupedStudents.get(key).add(student);
}
Set<String> groupedStudentsKeySet = groupedCustomer.keySet();
for (String location: groupedStudentsKeySet) {
List<Student> stdnts = groupedStudents.get(location);
for (Student student : stdnts) {
System.out.println("ID : "+student.stud_id+"\t"+"Name : "+student.stud_name+"\t"+"Location : "+student.stud_location);
}
}
答案 6 :(得分：1)
您可以这样排序：
Collections.sort(studlist, new Comparator<Student>() {
@Override
public int compare(Student o1, Student o2) {
return o1.getStud_location().compareTo(o2.getStud_location());
}
});
假设你的学生班上也有位置的吸气剂。
答案 7 :(得分：0)
你可以这样做：
Map<String, List<Student>> map = new HashMap<String, List<Student>>();
List<Student> studlist = new ArrayList<Student>();
studlist.add(new Student("1726", "John", "New York"));
map.put("New York", studlist);
键将是学生的位置和值列表。所以稍后你可以通过使用以下方式获得一组学生：
studlist = map.get("New York");
答案 8 :(得分：0)
您可以使用guava＆＃39; s Multimaps
@Canonical
class Persion {
String name
Integer age
}
List<Persion> list = [
new Persion("qianzi", 100),
new Persion("qianzi", 99),
new Persion("zhijia", 99)
]
println Multimaps.index(list, { Persion p -> return p.name })
打印：
[qianzi：[com.ctcf.message.Persion（qianzi，100），com.ctcf.message.Persion（qianzi，88）]，zhijia：[com.ctcf.message.Persion（zhijia，99）] ]
答案 9 :(得分：0)
Function<Student, List<Object>> compositKey = std ->
Arrays.asList(std.stud_location());
studentList.stream().collect(Collectors.groupingBy(compositKey, Collectors.toList()));
如果要为组添加多个对象，只需在compositKey方法中添加以逗号分隔的对象：
Function<Student, List<Object>> compositKey = std ->
Arrays.asList(std.stud_location(),std.stud_name());
studentList.stream().collect(Collectors.groupingBy(compositKey, Collectors.toList()));
答案 10 :(得分：0)
public class Test9 {
static class Student {
String stud_id;
String stud_name;
String stud_location;
public Student(String stud_id, String stud_name, String stud_location) {
super();
this.stud_id = stud_id;
this.stud_name = stud_name;
this.stud_location = stud_location;
}
public String getStud_id() {
return stud_id;
}
public void setStud_id(String stud_id) {
this.stud_id = stud_id;
}
public String getStud_name() {
return stud_name;
}
public void setStud_name(String stud_name) {
this.stud_name = stud_name;
}
public String getStud_location() {
return stud_location;
}
public void setStud_location(String stud_location) {
this.stud_location = stud_location;
}
@Override
public String toString() {
return " [stud_id=" + stud_id + ", stud_name=" + stud_name + "]";
}
}
public static void main(String[] args) {
List<Student> list = new ArrayList<Student>();
list.add(new Student("1726", "John Easton", "Lancaster"));
list.add(new Student("4321", "Max Carrados", "London"));
list.add(new Student("2234", "Andrew Lewis", "Lancaster"));
list.add(new Student("5223", "Michael Benson", "Leeds"));
list.add(new Student("5225", "Sanath Jayasuriya", "Leeds"));
list.add(new Student("7765", "Samuael Vatican", "California"));
list.add(new Student("3442", "Mark Farley", "Ladykirk"));
list.add(new Student("3443", "Alex Stuart", "Ladykirk"));
list.add(new Student("4321", "Michael Stuart", "California"));
Map<String, List<Student>> map1 =
list
.stream()
.sorted(Comparator.comparing(Student::getStud_id)
.thenComparing(Student::getStud_name)
.thenComparing(Student::getStud_location)
)
.collect(Collectors.groupingBy(
ch -> ch.stud_location
));
System.out.println(map1);
/*
Output :
{Ladykirk=[ [stud_id=3442, stud_name=Mark Farley],
[stud_id=3443, stud_name=Alex Stuart]],
Leeds=[ [stud_id=5223, stud_name=Michael Benson],
[stud_id=5225, stud_name=Sanath Jayasuriya]],
London=[ [stud_id=4321, stud_name=Max Carrados]],
Lancaster=[ [stud_id=1726, stud_name=John Easton],
[stud_id=2234, stud_name=Andrew Lewis]],
California=[ [stud_id=4321, stud_name=Michael Stuart],
[stud_id=7765, stud_name=Samuael Vatican]]}
*/
}// main
}
答案 11 :(得分：0)
可能已经晚了，但我想分享一个改进的解决方案。这与@Vitalii Fedorenko的回答基本相同，但比较方便。
因此，您只需通过将分组逻辑作为函数参数来使用Collectors.groupingBy()，您将获得带有关键参数映射的拆分列表。请注意，在提供的列表为Optional
null有助于避免不必要的NPE
public static <E, K> Map<K, List<E>> groupBy(List<E> list, Function<E, K> keyFunction) {
return Optional.ofNullable(list)
.orElseGet(ArrayList::new)
.stream()
.collect(Collectors.groupingBy(keyFunction));
}
现在，您可以与此分组。对于问题
中的用例 Map<String, List<Student>> map = groupBy(studlist, Student::getLocation);