Get last element of Stream/List in a one-liner

If you have a Collection (or more general an Iterable) you can use Google Guava's

Iterables.getLast(myIterable)

as handy oneliner.

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."

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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();

Guava has dedicated method for this case:

Stream<T> stream = ...;
Optional<T> lastItem = Streams.findLast(stream);

It's equivalent to stream.reduce((a, b) -> b) but creators claim it has much better performance.

From documentation:

This method's runtime will be between O(log n) and O(n), performing better on efficiently splittable streams.

It's worth to mention that if stream is unordered this method behaves like findAny().