Java8 filter and return if only element

Question

How can I filter a list using java8 streams and return the found element if it is the only element in the filtered list, otherwise(if there are more which meet the condition, or there is no result that meets the condition) return for example an Optional.empty()

I would need something like this:

Suppose I have a:

List<String> list = Arrays.asList("Apple","Banana","Peach");

then I want:

Optional<String> string = list.stream()
    .filter(item -> item.startsWith("A"))
    .findOne();

I know I can do it by:

boolean singleElement = list.stream()
        .filter(item -> item.startsWith("A"))
        .count() == 1;

String string = null;

if(singleElement){
  string = list.stream().filter(item -> item.startsWith("A")).iterator().next();
}

but I was wondering if I can do it in a single stream?

Is there any single stream solution?

Answer 1

Not very pretty, but you could limit the stream to 2 elements, collect those in a List, and see if that list has exactly one element. This still has more than one line, and has some overhead for creating the list, but that overhead is limited (to a list of size 2) and it does not have to iterate the stream twice, either.

List<String> tmp = list.stream()
    .filter(item -> item.startsWith("A"))
    .limit(2)
    .collect(Collectors.toList());
Optional<String> res = tmp.size() == 1 ? Optional.of(tmp.get(0)) : Optional.empty();

(My other idea was to use reduce((s1, s2) -> null) after limit(2) and reduce any two matches to null, but instead of returning an Optional.empty this will just raise an Exception, i.e. it does not work, but maybe this triggers some better (working) ideas.)

---

Update: It seems like while reduce raises an Exceptions, Collectors.reducing does not, and instead returns an Optional.empty as desired, so this also works, as shown in this answer to a very similar question. Still, I'd add limit(2) to make it stop early:

Optional<String> res = list.stream()
    .filter(item -> item.startsWith("A"))
    .limit(2)
    .collect(Collectors.reducing((s1, s2) -> null));

(If you like this last part, please upvote the original answer.)

Answer 2

You could use google Guava library's MoreCollectors.onlyElement as below:

    List<String> list = Arrays.asList("Apple", "Banana", "Peach");
    String string = null;
    try {
        string = list.stream()
                .filter(item -> item.startsWith("A"))
                .collect(MoreCollectors.onlyElement());
    } catch (NoSuchElementException | IllegalArgumentException iae) {
        System.out.println("zero or more than one elements found.");
    }
    Optional<String> res = string == null ? Optional.empty() : Optional.of(string);

Notice it throws NoSuchElementException if there is no element and it throws IllegalArgumentException if there are more than one elements.

Answer 3

I don't know if this counts as a single operation to you, but you can do :

Arrays.asList("Apple", "Banana", "Peach")
            .stream()
            .collect(Collectors.collectingAndThen(
                    Collectors.partitioningBy(
                            x -> x.startsWith("A")),
                    map -> {
                        List<String> list = map.get(Boolean.TRUE);
                        return list.size() == 1 ? Optional.of(list.get(0)) : Optional.empty();
                    }));