Android-java- How to sort a list of objects by a certain value within the object

Question

Im trying to sort through an arraylist of objects by a particular value within the object. What would be the best approach to do such a thing. Should I use Collections.sort() with some kind of comparator?

Im trying to sort a list of objects by a float value they hold in one of the variables.

EDIT: This is what I have so far:

public class CustomComparator implements Comparator<Marker> {
    @Override
    public int compare(Mark o1, Mark o2) {
        return o1.getDistance().compareTo(o2.getDistance());
    }
}

the error states: Cannot invoke compareTo(double) on the primitive type double.

Is it because a comparator cant return anything other than a certain type?

"Should I use Collections.sort() with some kind of comparator? " Yes, sounds like a good idea — Kennet, Feb 02 '12 at 09:31
I dont know if it matters but the number of objects in the list will be as high as 80. Thats why Im kind of confused about using a comparator if that is the way to go because it only compares two values at once. — James andresakis, Feb 02 '12 at 09:33
That's how the sorting works. First add one item to a list. When adding next; should this go before or after current in the list. When adding third item compare to first item in list, if after then compare to next item. And so on. — Kennet, Feb 02 '12 at 09:41

Atif Mahmood · Answer 1 · 2018-11-20T04:26:42.443

427

Follow this code to sort any ArrayList

Collections.sort(myList, new Comparator<EmployeeClass>(){
    public int compare(EmployeeClass obj1, EmployeeClass obj2) {
        // ## Ascending order
        return obj1.firstName.compareToIgnoreCase(obj2.firstName); // To compare string values
        // return Integer.valueOf(obj1.empId).compareTo(Integer.valueOf(obj2.empId)); // To compare integer values

        // ## Descending order
        // return obj2.firstName.compareToIgnoreCase(obj1.firstName); // To compare string values
        // return Integer.valueOf(obj2.empId).compareTo(Integer.valueOf(obj1.empId)); // To compare integer values
        }
    });

edited Nov 20 '18 at 04:26

answered Dec 11 '12 at 13:36

Atif Mahmood

8,882
2
41
44

This example works for comparing strings. For sorting by comparing integers look at Pranav's answer below. – Ambran Aug 25 '16 at 13:41
@Themelis it's java function – Atif Mahmood Jul 24 '19 at 15:51
4

int compare can be replaced by: return Integer.compare(obj1.empId, obj2.empId); – Ali Zarei Oct 13 '19 at 17:40

score 104 · Accepted Answer · edited May 23 '17 at 12:26

You should use Comparable instead of a Comparator if a default sort is what your looking for.

See here, this may be of some help - When should a class be Comparable and/or Comparator?

Try this -

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class TestSort {

    public static void main(String args[]){

        ToSort toSort1 = new ToSort(new Float(3), "3");
        ToSort toSort2 = new ToSort(new Float(6), "6");
        ToSort toSort3 = new ToSort(new Float(9), "9");
        ToSort toSort4 = new ToSort(new Float(1), "1");
        ToSort toSort5 = new ToSort(new Float(5), "5");
        ToSort toSort6 = new ToSort(new Float(0), "0");
        ToSort toSort7 = new ToSort(new Float(3), "3");
        ToSort toSort8 = new ToSort(new Float(-3), "-3");

        List<ToSort> sortList = new ArrayList<ToSort>();
        sortList.add(toSort1);
        sortList.add(toSort2);
        sortList.add(toSort3);
        sortList.add(toSort4);
        sortList.add(toSort5);
        sortList.add(toSort6);
        sortList.add(toSort7);
        sortList.add(toSort8);

        Collections.sort(sortList);

        for(ToSort toSort : sortList){
            System.out.println(toSort.toString());
        }
    }

}

public class ToSort implements Comparable<ToSort> {

    private Float val;
    private String id;

    public ToSort(Float val, String id){
        this.val = val;
        this.id = id;
    }

    @Override
    public int compareTo(ToSort f) {

        if (val.floatValue() > f.val.floatValue()) {
            return 1;
        }
        else if (val.floatValue() <  f.val.floatValue()) {
            return -1;
        }
        else {
            return 0;
        }

    }

    @Override
    public String toString(){
        return this.id;
    }
}

Hey thanks for the link :) I go back and forth from one language to the next so Im always missing something :p you know how it goes......jack of all trades but a master of none lol — James andresakis, Feb 02 '12 at 21:02

score 46 · Answer 3 · edited Jan 22 '16 at 13:32

I think this will help you better

Person p = new Person("Bruce", "Willis");
Person p1  = new Person("Tom", "Hanks");
Person p2 = new Person("Nicolas", "Cage");
Person p3 = new Person("John", "Travolta");

ArrayList<Person> list = new ArrayList<Person>();
list.add(p);
list.add(p1);
list.add(p2);
list.add(p3);

Collections.sort(list, new Comparator() {
    @Override
    public int compare(Object o1, Object o2) {
        Person p1 = (Person) o1;
        Person p2 = (Person) o2;
        return p1.getFirstName().compareToIgnoreCase(p2.getFirstName());
    }
});

score 33 · Answer 4 · answered Sep 18 '15 at 15:24

33

Now no need to Boxing (i.e no need to Creating OBJECT using new Operator use valueOf insted with compareTo of Collections.Sort..)

1)For Ascending order

Collections.sort(temp, new Comparator<XYZBean>() 
{
     @Override
     public int compare(XYZBean lhs, XYZBean rhs) {

       return Integer.valueOf(lhs.getDistance()).compareTo(rhs.getDistance());
      }
 });

1)For Deascending order

Collections.sort(temp, new Comparator<XYZBean>() 
{
     @Override
     public int compare(XYZBean lhs, XYZBean rhs) {

       return Integer.valueOf(rhs.getDistance()).compareTo(lhs.getDistance());
      }
 });

answered Sep 18 '15 at 15:24

Pranav

4,172
3
30
31

Someone up above reccommended your answer for comparing integer values. However, and even I am gonna try your style anyway, I wonder: why you convert to Integer the first value, but not the second one? Will it work like this anyway? The `compareTo` already has an "under the hood" conversion of the parameter, or this was just a mistake? – SebasSBM Apr 14 '23 at 06:03
After trying it myself, I think I know why: maybe the input for the Comparator are just `int`, which by itself doesn't have the `compareTo` method... so, if I didn't overlook anything, it was just because of that, right? So, after applying your code, Android Studio just reccommended me to replace it with `Integer.compare( rhs.getDistance(), lhs.getDistance() )`... doesn't it look cleaner this way? I still have to test this though... – SebasSBM Apr 14 '23 at 06:22
But well, this answer is 8 years old already... perhaps this suggestion of mine was not an option with the latest API levels of those days... – SebasSBM Apr 14 '23 at 06:29

score 6 · Answer 5 · answered Dec 29 '18 at 15:20

6

It's very easy for Kotlin!

listToBeSorted.sortBy { it.distance }

answered Dec 29 '18 at 15:20

Musab Gultekin

353
3
7

score 2 · Answer 6 · edited Mar 06 '17 at 08:44

You can compare two String by using this.

Collections.sort(contactsList, new Comparator<ContactsData>() {

                    @Override
                    public int compare(ContactsData lhs, ContactsData rhs) {

                        char l = Character.toUpperCase(lhs.name.charAt(0));

                        if (l < 'A' || l > 'Z')

                            l += 'Z';

                        char r = Character.toUpperCase(rhs.name.charAt(0));

                        if (r < 'A' || r > 'Z')

                            r += 'Z';

                        String s1 = l + lhs.name.substring(1);

                        String s2 = r + rhs.name.substring(1);

                        return s1.compareTo(s2);

                    }

                });

And Now make a ContactData Class.

public class ContactsData {

public String name;
public String id;
public String email;
public String avatar; 
public String connection_type;
public String thumb;
public String small;
public String first_name;
public String last_name;
public String no_of_user;
public int grpIndex;

public ContactsData(String name, String id, String email, String avatar, String connection_type)
{
    this.name = name;
    this.id = id;
    this.email = email;
    this.avatar = avatar;
    this.connection_type = connection_type;

}
}

Here contactsList is :

public static ArrayList<ContactsData> contactsList = new ArrayList<ContactsData>();

this places names starting from numbers (i.e 1gbfilename) on bottom. how they can be placed at top? — User, Mar 05 '21 at 11:03

score 2 · Answer 7 · answered May 31 '20 at 08:45

For Kotlin, you can use this function

fun sortList(list: List<YourCustomPOJOClass?>) {

    //descending
    Collections.sort(
        list
    ) { o1, o2 -> Integer.valueOf(o1!!.intValueXYZ!!).compareTo(o2!!.intValueXYZ!!) }

//    //ascending
//    Collections.sort(
//        list
//    ) { o1, o2 -> Integer.valueOf(o2!!.intValueXYZ!!).compareTo(o1!!.intValueXYZ!!) }
}

and just call it in your activity or fragment by

sortList(list)

score 2 · Answer 8 · answered Sep 09 '21 at 18:05

Person Model Class

 class Person {
        int id;
        String name;
        String fatherName;

        public Person(int id, String name, String fatherName) {
            this.id = id;
            this.name = name;
            this.fatherName = fatherName;
        }
        
    }

Sorting

class SORT_BY_ID implements Comparator<Person> {
        public int compare(Person a, Person b)
        {
            return a.id - b.id;
        }
    }

Usage

 ArrayList<Person> personArrayList = new ArrayList<Person>();
            personArrayList.add(new Person(111, "bb", "oakla"));
            personArrayList.add(new Person(131, "aa", "fast"));
            personArrayList.add(new Person(121, "cccc", "paokla"));

            System.out.println("Before Sorted Array list of PErson");
            for (int i=0; i<personArrayList.size(); i++)
                System.out.println(personArrayList.get(i));

            Collections.sort(personArrayList, new SORT_BY_ID());

            System.out.println("After Sorted ");
            for (int i=0; i<personArrayList.size(); i++)
                System.out.println(personArrayList.get(i));

same answer is already given. You can also see different different approaches too. — Snigdhajyoti, Sep 09 '21 at 20:52

score 1 · Answer 9 · edited Jan 22 '16 at 13:34

public class DateComparator implements Comparator<Marker> {
    @Override
    public int compare(Mark lhs, Mark rhs) {
        Double distance = Double.valueOf(lhs.getDistance());
        Double distance1 = Double.valueOf(rhs.getDistance());
        if (distance.compareTo(distance1) < 0) {
            return -1;
        } else if (distance.compareTo(distance1) > 0) {
            return 1;
        } else {
            return 0;
        }
    }
}

ArrayList(Marker) arraylist;

How To use:

Collections.sort(arraylist, new DateComparator());

score 1 · Answer 10 · edited Jan 22 '16 at 13:35

Model Class:

public class ToDoModel implements Comparable<ToDoModel> {
    private String id;
    private Date taskDate;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public Date getTaskDate() {
        return taskDate;
    }

    public void setTaskDate(Date taskDate) {
        this.taskDate = taskDate;
    }

    @Override
    public int compareTo(ToDoModel another) {
        return getTaskDate().compareTo(another.getTaskDate());  
    }
}

Now set data in ArrayList

for (int i = 0; i < your_array_length; i++) {
    ToDoModel tm = new ToDoModel();
    tm.setId(your_id);
    tm.setTaskDate(your_date);
    mArrayList.add(tm);
}

Now Sort ArrayList

Collections.sort(toDoList);

Summary: It will sort your data datewise

score 1 · Answer 11 · answered Feb 02 '12 at 09:32

1

"Android-java" is here by no means different than "normal java", so yes Collections.sort() would be a good approach.

answered Feb 02 '12 at 09:32

Heiko Rupp

30,426
13
82
119

1

But how do I make it sort by a value within the object. Thats what Im stuck on. – James andresakis Feb 02 '12 at 09:38

Hridoy Krisna Das · Answer 12 · 2022-02-05T11:31:23.483

0

I am using this code:

Collections.sort(groupMemberList, new Comparator<GroupMember>() {
    @Override
    public int compare(GroupMember o1, GroupMember o2) {
        return Long.valueOf(o2.getStatus()).compareTo(Long.valueOf(o1.getStatus()));
}
    @Override
    public boolean equals(Object obj) {
        return false;
    }

});

edited Feb 05 '22 at 11:31

answered Jan 31 '22 at 05:48

Hridoy Krisna Das

159
7

score 0 · Answer 13 · answered Jan 09 '23 at 13:06

When sorting a view model you can sort very easily, even with multiple elements if you create a comparator.

// Create the comparator
Comparator<NewsViewModel> sortByDateTime = Comparator.comparing(NewsViewModel::getNewsDate)
                .thenComparing(NewsViewModel::getNewsTime);

// Sort on date, then on time
Collections.sort(newsList, sortByDateTime);

// And even reverse the list afterwards
Collections.reverse(newsList);

score 0 · Answer 14 · answered Feb 02 '12 at 09:35

0

Either make a Comparator that can compare your objects, or if they are all instances of the same class, you can make that class implement Comparable. You can then use Collections.sort() to do the actual sorting.

answered Feb 02 '12 at 09:35

Jave

31,598
14
77
90

I went ahead and implemented Comparable in my class but and created a method to call sort on the list when i need to have it sorted but how do I sort it by a value with in the object? – James andresakis Feb 02 '12 at 09:42
The `compareTo()`-method is where you do the comparison. A little googling gave several detailed examples on how to use it, here is one of them: http://www.javadeveloper.co.in/java-example/java-comparable-example.html – Jave Feb 02 '12 at 09:45
I set up a method similar to the example however I get an error stating I cant use compareTo on a cast double. It seems like for what ever reason, what Im doing it doesnt like. Cannot invoke compareTo(double) on the primitive type double. Ill add my code up above to show what I mean – James andresakis Feb 02 '12 at 20:56

score -1 · Answer 15 · edited Sep 19 '13 at 07:08

I have a listview which shows the Information about the all clients I am sorting the clients name using this custom comparator class. They are having some extra lerret apart from english letters which i am managing with this setStrength(Collator.SECONDARY)

 public class CustomNameComparator implements Comparator<ClientInfo> {
        @Override

    public int compare(ClientInfo o1, ClientInfo o2) { 

        Locale locale=Locale.getDefault();
        Collator collator = Collator.getInstance(locale);
        collator.setStrength(Collator.SECONDARY);
        return collator.compare(o1.title, o2.title);

    }
}


PRIMARY strength: Typically, this is used to denote differences between base characters (for example, "a" < "b"). It is the strongest difference. For example, dictionaries are divided into different sections by base character. 
SECONDARY strength: Accents in the characters are considered secondary differences (for example, "as" < "às" < "at"). Other differences between letters can also be considered secondary differences, depending on the language. A secondary difference is ignored when there is a primary difference anywhere in the strings. 
TERTIARY strength: Upper and lower case differences in characters are distinguished at tertiary strength (for example, "ao" < "Ao" < "aò"). In addition, a variant of a letter differs from the base form on the tertiary strength (such as "A" and "Ⓐ"). Another example is the difference between large and small Kana. A tertiary difference is ignored when there is a primary or secondary difference anywhere in the strings. 
IDENTICAL strength: When all other strengths are equal, the IDENTICAL strength is used as a tiebreaker. The Unicode code point values of the NFD form of each string are compared, just in case there is no difference. For example, Hebrew cantellation marks are only distinguished at this strength. This strength should be used sparingly, as only code point value differences between two strings are an extremely rare occurrence. Using this strength substantially decreases the performance for both comparison and collation key generation APIs. This strength also increases the size of the collation key. 

**Here is a another way to make a rule base sorting if u need it just sharing**

/*      String rules="< å,Å< ä,Ä< a,A< b,B< c,C< d,D< é< e,E< f,F< g,G< h,H< ï< i,I"+"< j,J< k,K< l,L< m,M< n,N< ö,Ö< o,O< p,P< q,Q< r,R"+"< s,S< t,T< ü< u,U< v,V< w,W< x,X< y,Y< z,Z";
        RuleBasedCollator rbc = null;
        try {
            rbc = new RuleBasedCollator(rules);
        } catch (ParseException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        String myTitles[]={o1.title,o2.title};
        Collections.sort(Arrays.asList(myTitles), rbc);*/

Android-java- How to sort a list of objects by a certain value within the object

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