Sort List of Objects by values from array;

Question

I have a List<Brand> categories; with 1000+ items.

For each item from the list I have id, to get that I use categories.getId();

And I have an array int[] sortedIdArr = {4,53,102,403,110,5,6,8,12};

I would like to sort my categories list and make order by id how it is in sortedIdArr. How can I implement that ?

private void sortBrandsById(List<Brand> categories) {
    Collections.sort(categories, new Comparator<Brand>() {
        public int compare(Brand o1, Brand o2) {

        }
    });    
}

Can I use Collections.sort ?

Check the index of both IDs in sortedIdArr and compare these indices. If the index of o1-ID is smaller than o2-ID, then it should return 1, -1 of smaller and 0 if equal. — Tom
– Tom, Commented Jan 27, 2016 at 12:55

Mena · Accepted Answer · 2016-01-27 14:28:23Z

Typically you would use Collections.sort, or the equivalent idioms in Java 8 if applicable, or a sorted Collection such as TreeSet.

However in this case you want to follow a pre-defined order, dictated by your sortedIdArr array.

One way to achieve that is to use a linked collection (e.g. a LinkedHashSet).

Then you iterate your sortedIdArr array, and search your List<Brand> for an object with the given ID.

If found, you add the Brand object with the given ID to your LinkedHashSet, which will retain the insertion order.

Note that if an ID is not found, your Set will not exactly "match" the array.

Self-enclosed example, using Java 8

package test;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Optional;
import java.util.Set;


public class Main {

    // simplified Brand pojo
    static class Brand {
        int id;
        public Brand(int id) {
            this.id = id;
        }
        public int getId() {
            return id;
        }

        // for output clarity
        @Override
        public String toString() {
            return String.format("Brand: %d", id);
        }
    }

    public static void main(String[] args) throws Exception {

        // simplified ID list
        int[] sortedIdArr = {4,53,102};
        // "randomly" ordered Brand list
        final List<Brand> categories = new ArrayList<Brand>() {  
            {
                add(new Brand(1));
                add(new Brand(102));
                add(new Brand(53));
                add(new Brand(4));
                add(new Brand(0));
            }
        };
        // destination: linked set
        Set<Brand> linked = new LinkedHashSet<Brand>();
        // streaming the ID array in order
        Arrays.stream(sortedIdArr)
            .forEach((i) -> {
                // retrieving a Brand with same ID as the current
                // from the "randomly" ordered list
                Optional<Brand> toAdd = categories.stream()
                .filter((b) -> b.getId() == i)
                .findFirst();
                // making sure there's one
                if (toAdd.isPresent()) {
                    // adding to linked set
                    linked.add(toAdd.get());
                }
            }
        );
        System.out.println(linked);
    }

}

Output

[Brand: 4, Brand: 53, Brand: 102]

Imperative idiom for older Java versions

for (int i: sortedIdArr) {
    for (Brand b: categories) {
        // assuming no nulls
        if (b.getId() == i) {
            linked.add(b);
            break;
        }
    }
}

Nice approach, but Is it real to avoid java 8 ? I`m using this code in android, and currently there is problems with java 8.
@VLeonovs if you're on Android you can forget the Java 8 idioms, yes :) But you can convert it to Java 6-styled approach relatively easily - see my edit.

Wael Sakhri · Accepted Answer · 2016-01-27 14:23:44Z

1

Yes you can use the Collections.sort()

To sort your Brand using id :

public int compare(Brand o1, Brand o2) {
   return o1.getId().compareTo(o2.getId());
}

To sort your Brand using the array of id sortedIdArr :

Implement the Comparator Class :

class C implements Comparator<A> {

    int[] idOrder;

    public C(int[] idOrder) {
        super();
        this.idOrder = idOrder;
    }

    @Override
    public int compare(A o1, A o2) {

        Integer indexofO1 = Arrays.binarySearch(idOrder, o1.getId());
        Integer indexofO2 = Arrays.binarySearch(idOrder, o2.getId());
        return indexofO1.compareTo(indexofO2);
    }
}

The key idea here is to inverse the process and compare using the index of the id instead of the id itself !

To use it :

Collections.sort(list, new C (idOrder));

Test Example :

int[] idOrder = new int [] {3,1,2};
List<A> list = new ArrayList<>();

list.add(new A(1));
list.add(new A(2));
list.add(new A(3));

System.out.println(list);
//Output : [A [id=1], A [id=2], A [id=3]]

Collections.sort(list, new C(idOrder));

System.out.println(list);
//Output : [A [id=3], A [id=1], A [id=2]]

edited Jan 27, 2016 at 14:23

answered Jan 27, 2016 at 12:51

Wael Sakhri

3971 silver badge8 bronze badges

2 Comments

STaefi Over a year ago

He wants to sort the category list by the order of ids which are in the sortedIdArr. Your approach sort the list by the Brand's id property in an ascending way. It has nothing to do with the order of ids listed in sortedIdArr.

Wael Sakhri Over a year ago

Thks gays .. i updated my answer to solve the sortedIdArr problem :) .. you can check it

STaefi · Accepted Answer · 2016-01-27 15:07:46Z

You can do it with Collections.sort(...) but I strongly recommend not to go for Comparator method in this situation with this number of items in your list.

You can have a loop on your List<Brand> categories and add them in a HashMap<Integer,Brand> named tempMap. Then use it for lookup them in the order of your sortedIdArr array. Change your sort method like this:

private void sortBrandsById(List<Brand> categories, int [] sortedIdArr) {
    HashMap<Integer,Brand> tempMap = new HashMap<Integer, Brand>(categories.size());
    for (int i = 0; i < categories.size(); i++) {
        tempMap.put(categories.get(i).getId(), categories.get(i));
    }
    categories = new ArrayList<Brand>(sortedIdArr.length);
    for (int i = 0; i < sortedIdArr.length; i++) {
        categories.add(tempMap.get(sortedIdArr[i]));
    }
}

This way of sorting is from order O(n) for creating the tempMap + O(n) for recreating the category list. Although, the O(1) is not guaranteed for HashMap's get/put operations, java's decent hashing implementation very good and reliable. So the total complexity can not be very greater than O(n). At least it is very better than O(n^2). Time complexity is very important and I don't think you can find a way better than something very near O(n) complexity.

Hope this would be helpful.

Sudheep Vallipoyil · Accepted Answer · 2016-01-27 13:46:35Z

0

You can use the following code(You need the Apache commons lang jar-otherwise you have to iterate through your array to find the index).

private static void sortBrandsById(List<Brand> categories,int[] array) {


            Collections.sort(categories, new Comparator<Brand>() {
                public int compare(Brand o1, Brand o2) {                            

    return ArrayUtils.indexOf(array,o1.getId())-ArrayUtils.indexOf(array,o2.getId());
                }
            });    
        }

If you are able to put your predefined sorting order in a list not in an array , it will be much easier.

answered Jan 27, 2016 at 13:46

Sudheep Vallipoyil

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