Patan Patan - 2 months ago 78
Java Question

Remove duplicates from a list of objects based on property in Java 8

I am trying to remove duplicates from a List of objects based on some property.

can we do it in a simple way using java 8

List<Employee> employee


Can we remove duplicates from it based on
id
property of employee. I have seen posts removing duplicate strings form arraylist of string.

Answer

You can get a stream from the List and put in in the TreeSet from which you provide a custom comparator that compares id uniquely.

Then if you really need a list you can put then back this collection into an ArrayList.

import static java.util.Comparator.comparingInt;
import static java.util.stream.Collectors.collectingAndThen;
import static java.util.stream.Collectors.toCollection;

...
List<Employee> unique = employee.stream()
                                .collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparingInt(Employee::getId))),
                                                           ArrayList::new));

Given the example:

List<Employee> employee = Arrays.asList(new Employee(1, "John"), new Employee(1, "Bob"), new Employee(2, "Alice"));

It will output:

[Employee{id=1, name='John'}, Employee{id=2, name='Alice'}]

Another idea could be to use a wrapper that wraps an employee and have the equals and hashcode method based with its id:

class WrapperEmployee {
    private Employee e;

    public WrapperEmployee(Employee e) {
        this.e = e;
    }

    public Employee unwrap() {
        return this.e;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        WrapperEmployee that = (WrapperEmployee) o;
        return Objects.equals(e.getId(), that.e.getId());
    }

    @Override
    public int hashCode() {
        return Objects.hash(e.getId());
    }
}

Then you wrap each instance, call distinct(), unwrap them and collect the result in a list.

List<Employee> unique = employee.stream()
                                .map(WrapperEmployee::new)
                                .distinct()
                                .map(WrapperEmployee::unwrap)
                                .collect(toList());

In fact, I think you can make this wrapper generic by providing a function that will do the comparison:

class Wrapper<T, U> {
    private T t;
    private Function<T, U> equalityFunction;

    public Wrapper(T t, Function<T, U> equalityFunction) {
        this.t = t;
        this.equalityFunction = equalityFunction;
    }

    public T unwrap() {
        return this.t;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        @SuppressWarnings("unchecked")
        Wrapper<T, U> that = (Wrapper<T, U>) o;
        return Objects.equals(equalityFunction.apply(this.t), that.equalityFunction.apply(that.t));
    }

    @Override
    public int hashCode() {
        return Objects.hash(equalityFunction.apply(this.t));
    }
}

and the mapping will be:

.map(e -> new Wrapper<>(e, Employee::getId))