Java 8 - Sort list of integers with nulls

Question

Lets say you have a list of integers containing null vaues

List<Integer> integers = Arrays.asList(null, 9, 11, 7, 5, null);

Now if you want to get max value, you can try

OptionalInt max = integers.stream()
        .mapToInt(Integer::valueOf)
        .max();

This would work if the list did not contain nulls. In our case there will be a NullPointerException. After some research I can see such solution

Optional<Integer> max = integers.stream()
        .max(Comparator.nullsFirst(Comparator.naturalOrder()));

It works if you consider nulls less then non-nulls. But if you want them to be greater and change nullsFirst to nullsLast

Optional<Integer> max = integers.stream()
        .max(Comparator.nullsLast(Comparator.naturalOrder()));

then you will meet NullPointerException again.

I am interesting - why nullsLast doesn't work in this case considering that its doc states that it is null-friendly? And what is the best way to sort list of Integers containing nulls?

Thanks for your ideas!

Consider changing the name of this question. It has nothing to do with sorting, has it? — Erk, Jul 24 '18 at 19:19

score 4 · Answer 1 · answered Oct 04 '17 at 09:34

That's not about Comparator per-se:

List<Integer> result = integers.stream()
            .sorted(Comparator.nullsLast(Comparator.naturalOrder()))
            .collect(Collectors.toList());
System.out.println(result); // [5, 7, 9, 11, null, null]

You are putting your nulls to the end and that would be max; calling max on a null throws that NullPointerException.

Alexis C. · Accepted Answer · 2017-10-04T09:41:09.793

The documentation of Stream#max states:

Throws: NullPointerException - if the maximum element is null

As you provided a comparator that will consider null as a maximum element, then that is why it throws the exception.

This is what the stacktrace indicates when it tries to return Optional.of(state); with state being null in your example, and Optional.of throws a NPE if the value being passed is null.

Comparator.nullsLast works fine, for example:

Integer max = 
    Collections.max(integers, Comparator.nullsLast(Comparator.naturalOrder()));

will indeed give you null.

score 2 · Answer 3 · answered Oct 04 '17 at 09:45

If you consider null to be larger than any non-null value, the maximum will be, of course, null when list contains any null.

Since Stream.max already documents that it will throw a NullPointerException when the result is null, this is expected behavior. The design rationale is similar to findFirst(), which has been discussed in this Q&A.

Note that even

Optional<Integer> max = integers.stream()
    .max(Comparator.nullsFirst(Comparator.naturalOrder()));

can still fail, i.e. if all values are null. But there is no point in using a custom Comparator here, as the solution is straight-forward:

OptionalInt max = integers.stream()
    .filter(Objects::nonNull)
    .mapToInt(Integer::valueOf)
    .max();

When you get an empty OptionalInt, you may still test integers.isEmpty() to determine whether all elements were null or there were no elements at all.

Java 8 - Sort list of integers with nulls

3 Answers3