Java Streams - Sort if Comparator exists

Question

I have a class where optionally a Comparator can be specified.

---

Since the Comparator is optional, I have to evaluate its presence and execute the same stream code, either with sorted() or without:

if(comparator != null) {
    [...].stream().map()[...].sorted(comparator)[...];
} else {
    [...].stream().map()[...];
}

Question:
Is there a more elegant way to do this without the code duplication?

Note:
A default Comparator is not an option, I just want to keep the original order of the values I am streaming.

Besides, the elements are already mapped at the point of sorting, so I can not somehow reference the root list of the stream, as I do not have the original elements anymore.

Answer 1

You can do something like this:

Stream<Something> stream = [...].stream().map()[...]; // preliminary processing
if(comparator != null) {
    stream = stream.sorted(comparator); // optional sorting
}
stream... // resumed processing, which ends in some terminal operation (such as collect)

Answer 2

Another way would be to use Optional:

Stream<Whatever> stream = [...].stream().map()[...];

List<WhateverElse> result = Optional.ofNullable(comparator)
    .map(stream::sorted)
    .orElse(stream)
    .[...] // <-- go on with the stream pipeline
    .collect(Collectors.toList());

Answer 3

You could define a comparator of your type (I used E as a placeholder here) that will not change the order:

 Comparator<E> NO_SORTING = (one, other) -> 0;

If the comparator field is an Optional of Comparator, you can then use

.sorted(comparator.orElse(NO_SORTING))

Answer 4

If you don't mind use third party library StreamEx

StreamEx(source).[...].chain(s -> comparator == null ? s : s.sorted(comparator)).[...];