Sorting an ArrayList of objects using a custom sor

Question

I am looking to implement a sort feature for my address book application.

I want to sort an ArrayList<Contact> contactArray. Contact is a class which contains four fields: name, home number, mobile number and address. I want to sort on name.

How can I write a custom sort function to do this?

Answer 1

You shoud use the Arrays.sort function. The containing classes should implement Comparable.

Answer 2

This page tells you all you need to know about sorting collections, such as ArrayList.

Basically you need to

• make your Contact class implement the Comparable interface by
  • creating a method public int compareTo(Contact anotherContact) within it.
• Once you do this, you can just call Collections.sort(myContactList);,
  • where myContactList is ArrayList<Contact> (or any other collection of Contact).

There's another way as well, involving creating a Comparator class, and you can read about that from the linked page as well.

Example:

public class Contact implements Comparable<Contact> {

    ....

    //return -1 for less than, 0 for equals, and 1 for more than
    public compareTo(Contact anotherContact) {
        int result = 0;
        result = getName().compareTo(anotherContact.getName());
        if (result != 0)
        {
            return result;
        }
        result = getNunmber().compareTo(anotherContact.getNumber());
        if (result != 0)
        {
            return result;
        }
        ...
    }
}

Answer 3

I did it by the following way. number and name are two arraylist. I have to sort name .If any change happen to name arralist order then the number arraylist also change its order.

public void sortval(){

        String tempname="",tempnum="";

         if (name.size()>1) // check if the number of orders is larger than 1
            {
                for (int x=0; x<name.size(); x++) // bubble sort outer loop
                {
                    for (int i=0; i < name.size()-x-1; i++) {
                        if (name.get(i).compareTo(name.get(i+1)) > 0)
                        {

                            tempname = name.get(i);

                            tempnum=number.get(i);


                           name.set(i,name.get(i+1) );
                           name.set(i+1, tempname);

                            number.set(i,number.get(i+1) );
                            number.set(i+1, tempnum);


                        }
                    }
                }
            }



}

Answer 4

Here's a tutorial about ordering objects:

• The Java Tutorials - Collections - Object Ordering

Although I will give some examples, I would recommend to read it anyway.

---

There are various way to sort an ArrayList. If you want to define a natural (default) ordering, then you need to let Contact implement Comparable. Assuming that you want to sort by default on name, then do (nullchecks omitted for simplicity):

public class Contact implements Comparable<Contact> {

    private String name;
    private String phone;
    private Address address;

    public int compareTo(Contact other) {
        return name.compareTo(other.name);
    }

    // Add/generate getters/setters and other boilerplate.
}

so that you can just do

List<Contact> contacts = new ArrayList<Contact>();
// Fill it.

Collections.sort(contacts);

---

If you want to define an external controllable ordering (which overrides the natural ordering), then you need to create a Comparator:

List<Contact> contacts = new ArrayList<Contact>();
// Fill it.

// Now sort by address instead of name (default).
Collections.sort(contacts, new Comparator<Contact>() {
    public int compare(Contact one, Contact other) {
        return one.getAddress().compareTo(other.getAddress());
    }
}); 

---

You can even define the Comparators in the Contact itself so that you can reuse them instead of recreating them everytime:

public class Contact {

    private String name;
    private String phone;
    private Address address;

    // ...

    public static Comparator<Contact> COMPARE_BY_PHONE = new Comparator<Contact>() {
        public int compare(Contact one, Contact other) {
            return one.phone.compareTo(other.phone);
        }
    };

    public static Comparator<Contact> COMPARE_BY_ADDRESS = new Comparator<Contact>() {
        public int compare(Contact one, Contact other) {
            return one.address.compareTo(other.address);
        }
    };

}

which can be used as follows:

List<Contact> contacts = new ArrayList<Contact>();
// Fill it.

// Sort by address.
Collections.sort(contacts, Contact.COMPARE_BY_ADDRESS);

// Sort later by phone.
Collections.sort(contacts, Contact.COMPARE_BY_PHONE);

---

And to cream the top off, you could consider to use a generic javabean comparator:

public class BeanComparator implements Comparator<Object> {

    private String getter;

    public BeanComparator(String field) {
        this.getter = "get" + field.substring(0, 1).toUpperCase() + field.substring(1);
    }

    public int compare(Object o1, Object o2) {
        try {
            if (o1 != null && o2 != null) {
                o1 = o1.getClass().getMethod(getter, new Class[0]).invoke(o1, new Object[0]);
                o2 = o2.getClass().getMethod(getter, new Class[0]).invoke(o2, new Object[0]);
            }
        } catch (Exception e) {
            // If this exception occurs, then it is usually a fault of the developer.
            throw new RuntimeException("Cannot compare " + o1 + " with " + o2 + " on " + getter, e);
        }

        return (o1 == null) ? -1 : ((o2 == null) ? 1 : ((Comparable<Object>) o1).compareTo(o2));
    }

}

which you can use as follows:

// Sort on "phone" field of the Contact bean.
Collections.sort(contacts, new BeanComparator("phone"));

(as you see in the code, possibly null fields are already covered to avoid NPE's during sort)

Answer 5

The Collections.sort is a good sort implementation. If you don't have The comparable implemented for Contact, you will need to pass in a Comparator implementation

Of note:

The sorting algorithm is a modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist). This algorithm offers guaranteed n log(n) performance. The specified list must be modifiable, but need not be resizable. This implementation dumps the specified list into an array, sorts the array, and iterates over the list resetting each element from the corresponding position in the array. This avoids the n2 log(n) performance that would result from attempting to sort a linked list in place.

The merge sort is probably better than most search algorithm you can do.

Answer 6

BalusC and bguiz have already given very complete answers on how to use Java's built-in Comparators.

I just want to add that google-collections has an Ordering class which is more "powerful" than the standard Comparators. It might be worth checking out. You can do cool things such as compounding Orderings, reversing them, ordering depending on a function's result for your objects...

Here is a blog post that mentions some of its benefits.