按属性将对象列表分组:Java


97

我需要使用特定对象的属性(位置)对对象(学生)的列表进行分组,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;

    }
}

请建议我一种干净的方法。


2
以位置为键,学生列表为值的哈希图。
Omoro 2014年

按位置排序会解决您的问题,还是还有其他问题?
督军2014年

尝试使用比较器并按位置排序。
pshemek 2014年

1
@Warlord是的,但是如果我需要得到更多信息,可以
走得更远,

@Omoro请您通过代码给我一个提示,我不太熟悉Hashmaps
Dilukshan Mahendra

Answers:


130

这会将学生对象添加到HashMapwith locationID作为键。

HashMap<Integer, List<Student>> hashMap = new HashMap<Integer, List<Student>>();

迭代在这个代码,并添加学生到HashMap

if (!hashMap.containsKey(locationId)) {
    List<Student> list = new ArrayList<Student>();
    list.add(student);

    hashMap.put(locationId, list);
} else {
    hashMap.get(locationId).add(student);
}

如果要让所有学生都具有特定的位置详细信息,则可以使用以下方法:

hashMap.get(locationId);

这将为您提供具有相同位置ID的所有学生。


4
您声明了位置对象列表,然后在下一行中将Student对象添加到上一个列表中,这应该引发错误。
OJVM

当hashMap.contanisKey()返回false时,hashMap.get()返回null。如果您调用第一个hashMap.get()并将结果存储在本地变量中,并检查该本地变量是否为空,则可以将调用保存到containsKey()方法中
Esteve

246

在Java 8中:

Map<String, List<Student>> studlistGrouped =
    studlist.stream().collect(Collectors.groupingBy(w -> w.stud_location));

那是因为在Student课堂上stud_location被指定为友好。只有Student一流的和在同一个包中定义的任何类Student可以访问stud_location。如果您放置public String stud_location;而不是String stud_location;,这应该可以工作。或者,您可以定义一个吸气功能。在更多信息cs.princeton.edu/courses/archive/spr96/cs333/java/tutorial/java/...
Eranga鹤山

32
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);
    }

}

8

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]
    }
}

通过坚持与问题相同的示例可以改善此答案。而且结果与问题中要求的期望输出不匹配。
Pim Hazebroek,

5

Java 8分组按Collector

可能已经晚了,但是我想分享一个改进的想法来解决这个问题。这与@Vitalii Fedorenko的回答基本相同,但比较方便。

您可以Collectors.groupingBy()通过将分组逻辑作为函数参数使用来使用,您将获得带有关键参数映射的拆分列表。请注意,Optional当提供的列表为null

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。对于这里的用例

Map<String, List<Student>> map = groupBy(studlist, Student::getLocation);

也许您还想研究一下Java 8分组指南by Collector


4

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);
}

//打印

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);
    }
}

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,如果将排序后的数据保留为分组数据,例如,如果您具有相同的重复列数据,则排序机制对它们进行排序,以将相同的数据保留在一侧,然后查找其他数据,这些数据是不同的数据。间接将其视为相同数据的分组。

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);
        }
    }
}

上面的ProductBean列表的输出是在GROUP BY条件下完成的,如果您看到的输入数据是给给Collections.sort的ProductBean列表给出的输入数据(列表,所需列的Comparator对象),则将根据您的比较器实现进行排序您将可以在下面的输出中看到GROUPED数据。希望这可以帮助...

    ********在分组输入数据之前,这种方式******
    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 Nov 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 Nov Nov 17 09:31:01 IST 2014]
    ProductBean [P3 60 Mon Nov 17 09:31:01 IST 2014]

    ********按价格分组后**
    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 Nov 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]


1
嗨,请不要多次发布相同的答案,也请不要发布原始代码,而无需对原始代码的工作原理以及它如何解决上述问题进行解释。
2014年

抱歉,伙计,在粘贴代码时有些错误,因为它可能已经多次出现。我已经编辑了我发布的内容的说明。希望现在看起来很好???
拉维·贝利

我遗漏了一些东西,或者此代码正在排序而不是按字段分组?我看到产品按ID排序,然后按价格排序
Funder

0

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());
        }
    });

假设您在学生班上也有自己的位置。


3
为什么要排序?问题是对元素进行分组!
桑卡尔普(Sankalp)'18

0

您可以这样做:

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);

键将是位置和学生的值列表。因此,稍后您可以使用以下方法获得一组学生:

studlist = map.get("New York");

0

您可以使用guavaMultimaps

@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 })

它打印:

[qianzi:[com.ctcf.message.Persion(qianzi,100),com.ctcf.message.Persion(qianzi,88)],zhijia:[com.ctcf.message.Persion(zhijia,99)]


0
Function<Student, List<Object>> compositKey = std ->
                Arrays.asList(std.stud_location());
        studentList.stream().collect(Collectors.groupingBy(compositKey, Collectors.toList()));

如果要为group by添加多个对象,则只需在compositKey方法中用逗号分隔即可添加对象:

Function<Student, List<Object>> compositKey = std ->
                Arrays.asList(std.stud_location(),std.stud_name());
        studentList.stream().collect(Collectors.groupingBy(compositKey, Collectors.toList()));

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
}
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