Is Java 8 findFirst().isPresent() more efficient than count() > 0?

Question

Given I have a stream Stream stream = list.stream().filter(some predicate) where the list is very large, is it more efficient to check whether the stream

Answer 1

If all you want to know, is whether there is a match, you should use
list.stream().anyMatch(some predicate), not only because it is more efficient, but also, because it is the correct idiom for expressing you intention.

As said by others, anyMatch is short-circuiting, which implies that it will stop at the first match, whereas count will, as the name suggests, count all matches before returning. Depending on the stream contents, this can make a huge performance difference. But note, that you can make count similarly efficient, by using list.stream().filter(some predicate).limit(1).count() > 0

Then, it will also stop after the first occurrence, but, as said, anyMatch is still the preferred way to express that you are interested in whether there is any match. Things change, when the task is to find out whether there are at least n matches. Then, .limit(n).count() > n-1 (or >= n) becomes the natural idiom.

Note that findFirst() differs from the other solution because its answer depends on the ordering. So if you only want to know whether there is a match, you should use findAny() instead. Still, there is a theoretical difference due to the requirement of returning the matching value, compared to just telling whether there is a match, like anyMatch does, though currently that difference lies only in the construction of an Optional instance, hence is negligible.

But since you are programming against an API to encode your intention, you should not use find… when you only want to know whether there is a match. anyMatch clearly expresses your intention and might have an even higher benefit in future implementations or more complex scenarios.