流过滤以获得最佳匹配

Question

My aim is to filter for a best match. In my example I have a list of persons, which I want to filter by surname and firstname.

The matching prescendence would be:

1. both surname and firstname match, return first match
2. only surname matches, return first match
3. none match, throw some exception

My code so far:

final List<Person> persons = Arrays.asList(
  new Person("Doe", "John"),
  new Person("Doe", "Jane"),
  new Person("Munster", "Herman");

Person person = persons.stream().filter(p -> p.getSurname().equals("Doe")).???

Answer 1

Assuming Person implements equals and hashCode:

Person personToFind = new Person("Doe", "Jane");

Person person = persons.stream()
    .filter(p -> p.equals(personToFind))
    .findFirst()
    .orElseGet(() -> 
        persons.stream()
            .filter(p -> p.getSurname().equals(personToFind.getSurname()))
            .findFirst()
            .orElseThrow(() -> new RuntimeException("Could not find person ..."))
    );

Answer 2

You can use

Person person = persons.stream()
        .filter(p -> p.getSurName().equals("Doe"))
        .max(Comparator.comparing(p -> p.getFirstName().equals("Jane")))
        .orElse(null);

It will only consider elements having the right surname and return the best element of them, which is the one with a matching first name. Otherwise, the first matching element is returned.

As already mentioned in a comment , a for loop could be more efficient if there is a best element, as it can short circuit. If there is no best element with matching surname and first name, all element have to be checked in all implementations.

Answer 3

I would propose this:

Optional<Person> bestMatch = persons.stream()
            .filter(p -> "Doe".equals(p.getSurname()))
            .reduce((person, person2) -> {
                if ("John".equals(person.getFirstName())) {
                    return person;
                } else if ("John".equals(person2.getFirstName())) {
                    return person2;
                }
                return person;
            });
Person result = bestMatch.orElseThrow(IllegalArgumentException::new);

Answer 4

The right tool is .findFirst() . You can also use .limit(1) .