根据另一个列表中的属性从列表中提取不常见的元素

Question

My Structure

   A {
       String id;
       String bid;
   }

   B {
       String id;
   }

Given

List<A> aList = Arrays.asList(
   new A (1,2),
   new A (2,5),
   new A (3,9),
   new A (4,10),
   new A (5, 20),
   new A (6, 8),
   new A (7, 90)
   new A (8, 1)
);

List<B> bList = Arrays.asList(
   new B (2),
   new B (9),
   new B (10)
);

Now i want the elements of A which don't match with any B's element should be collected in another collection and these elements should be deleted from A collection itself.

Result

 List<A> aList = Arrays.asList(
       new A (1,2),
       new A (3,9),
       new A (4,10)
    );

    List<A> aListBin = Arrays.asList(
       new A (2,5),
       new A (5, 20),
       new A (6, 8),
       new A (7, 90)
       new A (8, 1)
    );

MY take

I can think of iterating A using iterator and for each element in A iterate through B and if found keep else keep adding to separate list and delete using iterator remove.

Is there a better way to do this using stream magic? Thanks

Answer 1

Collectors#partitionBy is your friend.

First, we'll extract the id s from the list of B s into a bare Set<Integer> , so we can use that for lookup:

Set<Integer> bSet = bList.stream()
    .map(b -> b.id)
    .collect(Collectors.toSet());

As mentioned by JB Nizet, a HashSet is fit for the job.

Then it's just as simple as this – we'll partition by a given predicate. The predicate is whether A.bid is contained in any B.id (which we stored into bSet for convenience).

Map<Boolean, List<A>> map = aList.stream()
    .collect(Collectors.partitioningBy(a -> bSet.contains(a.bid)));

Now map.get(true) contains all items contained in B , map.get(false) all others.

In order to replace aList , simply reassign aList :

aList = map.get(true);

Answer 2

You can't remove elements from a list created with Arrays.asList(). It returns a view of the array you pass as argument, so calling remove will throw UnsupportedOperationException.

Assuming you have ArrayList instead, I still don't think you can achieve this on a single step, and certainly not with "stream magic", since streams won't allow you to modify the original collection.

In two steps, something like:

List<A> newList = aList.stream()
    .filter(a -> !bList.contains(a.bid))
    .collect(Collectors.toList());
aList.removeAll(newList);

If performance is an issue, use Set or Map(with id as key) instead of List in order to perform the contains() and the removeAll() in O(1) and O(n) respectively.

Answer 3

Yes, you can use Java 8 Streams .

Here's the full example from your input:

import java.util.*;
import java.util.stream.*;
import static java.util.stream.Collectors.toList;

public class MyClass {
    public static void main(String args[]) {

        class A {
            public int id;
            public int bid;
            public A(int id, int bid) { this.id = id; this.bid = bid; }
            public String toString() { return "(" + id + "," + bid + ")"; }
        };

        class B {
            public int id;
            public B(int id) { this.id = id; }
            public String toString() { return "" + id; }
        };

        List<A> aList = Arrays.asList(
                new A (1,2),  // not removed
                new A (2,5),  // removed
                new A (3,9),  // not removed
                new A (4,10), // not removed
                new A (5, 20),// not removed
                new A (6, 8), // not removed
                new A (7, 90),// not removed
                new A (8, 1)// not removed
        );


        List<B> bList = Arrays.asList(
                new B (2),
                new B (9),
                new B (10)
        );


        List<A> aListBin = new ArrayList<>();
        aList.stream()
            .forEach( a -> {
                if (bList.stream().noneMatch(b -> b.id == a.bid )) {
                    aListBin.add(a);        
                }
            });

        aList = aList.stream()
        .filter( a -> bList.stream().anyMatch(b -> b.id == a.bid))
        .collect(toList());

        System.out.println("Alist-> " + aList);
        System.out.println("Blist-> " + bList);
        System.out.println("Removed-> " + aListBin);
    }
}

Output:

Alist-> [(1,2), (3,9), (4,10)]
Blist-> [2, 9, 10]
Removed-> [(2,5), (5,20), (6,8), (7,90), (8,1)]

Answer 4

You can use Collectors.partitioningBy . Is it better? Depends on your definition of better. It is much more concise code, however it is not as efficient as the simple iterator loop you described.

I cannot think of a more efficient way than the iterator route, except maybe using a string hashset for lookup of the B class id.

However, if you prefer concise code, here is the code using partitioningBy:

class A {
    int id;
    int bid;

    public A(int id, int bid){
        this.id = id;
        this.bid = bid;
    }

    public boolean containsBId(List<B> bList) {
        return bList.stream().anyMatch(b -> bid == b.id);
    }
}

class B {
    int id;

    public B(int id){
        this.id = id;
    }
}

class Main {

public static void main(String[] args) {
    List<A> aList = Arrays.asList(
        new A (1,2),
        new A (2,5),
        new A (3,9),
        new A (4,10),
        new A (5, 20),
        new A (6, 8),
        new A (7, 90),
        new A (8, 1)
    );
    List<B> bList = Arrays.asList(
        new B (2),
        new B (9),
        new B (10)
    );
    Map<Boolean, List<A>> split = aList.stream()
        .collect(Collectors.partitioningBy(a -> a.containsBId(bList)));

    aList = split.get(true);
    List<A> aListBin = split.get(false);
}

Answer 5

Since you are dealing with two different classes you can't compare them directly. So you need to reduce to the least common denominator which is the integer ID's.

   // start a stream of aList.
   List<A> aListBin = aList.stream()

   // Convert the bList to a collection of
   // of ID's so you can filter.       
   .filter(a -> !bList.stream()

         // get the b ID
         .map(b->b.id)

         // put it in a list         
        .collect(Collectors.toList())

         // test to see if that list of b's ID's
         // contains a's bID   
         .contains(a.bid))

    //if id doesn't contain it, then at to the list.
    .collect(Collectors.toList());

To finish up, remove the newly created list from the aList.

        aList.removeAll(aListBin);

They are displayed as follows:

        System.out.println("aListBin = " + aListBin);
        System.out.println("aList = " + aList);
        aListBin = [[2, 5], [5, 20], [6, 8], [7, 90], [8, 1]]
        aList = [[1, 2], [3, 9], [4, 10]]

Note:

• To reverse the contents of each final list, remove the bang(!) from the filter.
• I added toString methods in the classes to allow printing.