Java 8 Distinct by property

Question

In Java 8 how can I filter a collection using the Stream API by checking the distinctness of a property of each object?

For example I have a list of

Answer 1

You can wrap the person objects into another class, that only compares the names of the persons. Afterward, you unwrap the wrapped objects to get a person stream again. The stream operations might look as follows:

persons.stream()
    .map(Wrapper::new)
    .distinct()
    .map(Wrapper::unwrap)
    ...;

The class Wrapper might look as follows:

class Wrapper {
    private final Person person;
    public Wrapper(Person person) {
        this.person = person;
    }
    public Person unwrap() {
        return person;
    }
    public boolean equals(Object other) {
        if (other instanceof Wrapper) {
            return ((Wrapper) other).person.getName().equals(person.getName());
        } else {
            return false;
        }
    }
    public int hashCode() {
        return person.getName().hashCode();
    }
}

Answer 2

Set<YourPropertyType> set = new HashSet<>();
list
        .stream()
        .filter(it -> set.add(it.getYourProperty()))
        .forEach(it -> ...);

Answer 3

The Most simple code you can write:

    persons.stream().map(x-> x.getName()).distinct().collect(Collectors.toList());

Answer 4

I recommend using Vavr, if you can. With this library you can do the following:

io.vavr.collection.List.ofAll(persons)
                       .distinctBy(Person::getName)
                       .toJavaSet() // or any another Java 8 Collection

Answer 5

Extending Stuart Marks's answer, this can be done in a shorter way and without a concurrent map (if you don't need parallel streams):

public static <T> Predicate<T> distinctByKey(Function<? super T, ?> keyExtractor) {
    final Set<Object> seen = new HashSet<>();
    return t -> seen.add(keyExtractor.apply(t));
}

Then call:

persons.stream().filter(distinctByKey(p -> p.getName());

Answer 6

Maybe will be useful for somebody. I had a little bit another requirement. Having list of objects A from 3rd party remove all which have same A.b field for same A.id (multiple A object with same A.id in list). Stream partition answer by Tagir Valeev inspired me to use custom Collector which returns Map<A.id, List<A>>. Simple flatMap will do the rest.

 public static <T, K, K2> Collector<T, ?, Map<K, List<T>>> groupingDistinctBy(Function<T, K> keyFunction, Function<T, K2> distinctFunction) {
    return groupingBy(keyFunction, Collector.of((Supplier<Map<K2, T>>) HashMap::new,
            (map, error) -> map.putIfAbsent(distinctFunction.apply(error), error),
            (left, right) -> {
                left.putAll(right);
                return left;
            }, map -> new ArrayList<>(map.values()),
            Collector.Characteristics.UNORDERED)); }