Java 8 Distinct by property

An alternative would be to place the persons in a map using the name as a key:

persons.collect(Collectors.toMap(Person::getName, p -> p, (p, q) -> p)).values();

Note that the Person that is kept, in case of a duplicate name, will be the first encontered.

Consider distinct to be a stateful filter. Here is a function that returns a predicate that maintains state about what it's seen previously, and that returns whether the given element was seen for the first time:

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

Then you can write:

persons.stream().filter(distinctByKey(Person::getName))

Note that if the stream is ordered and is run in parallel, this will preserve an arbitrary element from among the duplicates, instead of the first one, as distinct() does.

(This is essentially the same as my answer to this question: Java Lambda Stream Distinct() on arbitrary key?)

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();
    }
}