You can skip the first match by adding skip(1) after the filter:

employees.stream()
         .filter(e -> e.getName().charAt(0) == 's')
         .skip(1)
         .findAny()
         .ifPresent(e -> System.out.println("Employee : " + e));

If you're on Java8 you can use a Stream:

RECOBeacon second = recoBeacons.stream().skip(1).findFirst().orElse(null);

The nice thing about this solution is that findFirst returns an Optional, so you don't have to do the hasNext checks like when using an iterator.

Also note that the Collection interface does not guarantee order, so getting the n-th element may yield unexpected results.

It is possible to get the last element with the method Stream::reduce. The following listing contains a minimal example for the general case:

Stream<T> stream = ...; // sequential or parallel stream
Optional<T> last = stream.reduce((first, second) -> second);

This implementations works for all ordered streams (including streams created from Lists). For unordered streams it is for obvious reasons unspecified which element will be returned.

The implementation works for both sequential and parallel streams. That might be surprising at first glance, and unfortunately the documentation doesn't state it explicitly. However, it is an important feature of streams, and I try to clarify it:

• The Javadoc for the method Stream::reduce states, that it "is not constrained to execute sequentially".
• The Javadoc also requires that the "accumulator function must be an associative, non-interfering, stateless function for combining two values", which is obviously the case for the lambda expression (first, second) -> second.
• The Javadoc for reduction operations states: "The streams classes have multiple forms of general reduction operations, called reduce() and collect() [..]" and "a properly constructed reduce operation is inherently parallelizable, so long as the function(s) used to process the elements are associative and stateless."

The documentation for the closely related Collectors is even more explicit: "To ensure that sequential and parallel executions produce equivalent results, the collector functions must satisfy an identity and an associativity constraints."

Back to the original question: The following code stores a reference to the last element in the variable last and throws an exception if the stream is empty. The complexity is linear in the length of the stream.

CArea last = data.careas
                 .stream()
                 .filter(c -> c.bbox.orientationHorizontal)
                 .reduce((first, second) -> second).get();

One of the prime motivations for the introduction of Java streams was to allow parallel operations. This led to a requirement that operations on Java streams such as map and filter be independent of the position of the item in the stream or the items around it. This has the advantage of making it easy to split streams for parallel processing. It has the disadvantage of making certain operations more complex.

So the simple answer is that there is no easy way to do things such as take every nth item or map each item to the sum of all previous items.

The most straightforward way to implement your requirement is to use the index of the list you are streaming from:

List<String> list = ...;
return IntStream.range(0, list.size())
    .filter(n -> n % 3 == 0)
    .mapToObj(list::get)
    .toList();

A more complicated solution would be to create a custom collector that collects every nth item into a list.

class EveryNth<C> {
    private final int nth;
    private final List<List<C>> lists = new ArrayList<>();
    private int next = 0;

    private EveryNth(int nth) {
        this.nth = nth;
        IntStream.range(0, nth).forEach(i -> lists.add(new ArrayList<>()));
    }

    private void accept(C item) {
        lists.get(next++ % nth).add(item);
    }

    private EveryNth<C> combine(EveryNth<C> other) {
        other.lists.forEach(l -> lists.get(next++ % nth).addAll(l));
        next += other.next;
        return this;
    }

    private List<C> getResult() {
        return lists.get(0);
    }

    public static Collector<Integer, ?, List<Integer>> collector(int nth) {
        return Collector.of(() -> new EveryNth(nth), 
            EveryNth::accept, EveryNth::combine, EveryNth::getResult));
}

This could be used as follows:

Stream.of("Anne", "Bill", "Chris", "Dean", "Eve", "Fred", "George")
    .parallel().collect(EveryNth.collector(3)).toList();

Which returns the result ["Anne", "Dean", "George"] as you would expect.

This is a very inefficient algorithm even with parallel processing. It splits all items it accepts into n lists and then just returns the first. Unfortunately it has to keep all items through the accumulation process because it's not until they are combined that it knows which list is the nth one.

Given the complexity and inefficiency of the collector solution I would definitely recommend sticking with the indices based solution above in preference to this if you can. If you aren't using a collection that supports get (e.g. you are passed a Stream rather than a List) then you will either need to collect the stream using Collectors.toList or use the EveryNth solution above.

You can build a custom Collector for this task.

Map<String, String> map = 
    Stream.of("a", "b", "err1", "c", "d", "err2", "e", "f", "g", "h", "err3", "i", "j")
          .collect(MappingErrors.collector());

with:

private static final class MappingErrors {

    private Map<String, String> map = new HashMap<>();

    private String first, second;

    public void accept(String str) {
        first = second;
        second = str;
        if (first != null && first.startsWith("err")) {
            map.put(first, second);
        }
    }

    public MappingErrors combine(MappingErrors other) {
        throw new UnsupportedOperationException("Parallel Stream not supported");
    }

    public Map<String, String> finish() {
        return map;
    }

    public static Collector<String, ?, Map<String, String>> collector() {
        return Collector.of(MappingErrors::new, MappingErrors::accept, MappingErrors::combine, MappingErrors::finish);
    }

}

In this collector, two running elements are kept. Each time a String is accepted, they are updated and if the first starts with "err", the two elements are added to a map.

Another solution is to use the StreamEx library which provides a pairMap method that applies a given function to the every adjacent pair of elements of this stream. In the following code, the operation returns a String array consisting of the first and second element of the pair if the first element starts with "err", null otherwise. null elements are then filtered out and the Stream is collected into a map.

Map<String, String> map = 
    StreamEx.of("a", "b", "err1", "c", "d", "err2", "e", "f", "g", "h", "err3", "i", "j")
            .pairMap((s1, s2) -> s1.startsWith("err") ? new String[] { s1, s2 } : null)
            .nonNull()
            .toMap(a -> a[0], a -> a[1]);

System.out.println(map);

No, this is not possible using streams, at least not easily. The stream API abstracts away from the order in which the elements are processed: the stream might be processed in parallel, or in reverse order. So "the next element" and "previous element" do not exist in the stream abstraction.

You should use the API best suited for the job: stream are excellent if you need to apply some operation to all elements of a collection and you are not interested in the order. If you need to process the elements in a certain order, you have to use iterators or maybe access the list elements through indices.

Create a custom Collector

public static <T> Collector<T, ?, T> toSingleton() {
    return Collectors.collectingAndThen(
            Collectors.toList(),
            list -> {
                if (list.size() != 1) {
                    throw new IllegalStateException();
                }
                return list.get(0);
            }
    );
}

We use Collectors.collectingAndThen to construct our desired Collector by

1. Collecting our objects in a List with the Collectors.toList() collector.
2. Applying an extra finisher at the end, that returns the single element — or throws an IllegalStateException if list.size != 1.

Used as:

User resultUser = users.stream()
        .filter(user -> user.getId() > 0)
        .collect(toSingleton());

You can then customize this Collector as much as you want, for example give the exception as argument in the constructor, tweak it to allow two values, and more.

An alternative — arguably less elegant — solution:

You can use a 'workaround' that involves peek() and an AtomicInteger, but really you shouldn't be using that.

What you could do instead is just collecting it in a List, like this:

LinkedList<User> users = new LinkedList<>();
users.add(new User(1, "User1"));
users.add(new User(2, "User2"));
users.add(new User(3, "User3"));
List<User> resultUserList = users.stream()
        .filter(user -> user.getId() == 1)
        .collect(Collectors.toList());
if (resultUserList.size() != 1) {
    throw new IllegalStateException();
}
User resultUser = resultUserList.get(0);

To get a range from a Stream<T>, you can use skip(long n) to first skip a set number of elements, and then you can call limit(long n) to only take a specific amount of items.

Consider a stream with 10 elements, then to get elements 3 to 7, you would normally call from a List:

list.subList(3, 7);

Now with a Stream, you need to first skip 3 items, and then take 7 - 3 = 4 items, so it becomes:

stream.skip(3).limit(4);

As a variant to @StuartMarks' solution to the second answer, I'll offer you the following solution which leaves the possibility to chain intact, it works similar to how @StuartMarks does it:

private <T> Collector<T, ?, Stream<T>> topPercentFromRangeCollector(Comparator<T> comparator, double from, double to) {
    return Collectors.collectingAndThen(
        Collectors.toList(),
        list -> list.stream()
            .sorted(comparator)
            .skip((long)(list.size() * from))
            .limit((long)(list.size() * (to - from)))
    );
}

and

IntStream.range(0, 100)
        .boxed()
        .collect(topPercentFromRangeCollector(Comparator.comparingInt(i -> i), 0.1d, 0.3d))
        .forEach(System.out::println);

This will print the elements 10 through 29.

It works by using a Collector<T, ?, Stream<T>> that takes in your elements from the stream, transforms them into a List<T>, then obtains a Stream<T>, sorts it and applies the (correct) bounds to it.