How to split a list in two by a condition

Short answer

partitioningBy() creates a Map with two keys: true for elements that match the condition and false for the rest:

void main() {
    var numbers = List.of(1, 2, 3, 4, 5, 6, 7);
 
    Map<Boolean, List<Integer>> parts = numbers.stream()
            .collect(Collectors.partitioningBy(n -> n % 2 == 0));
    IO.println(parts.get(true));
    IO.println(parts.get(false));
}

Output:

[2, 4, 6]
[1, 3, 5, 7]

With counting

The second argument processes each part — for example, counts the elements:

record Student(String name, int score) {}
 
void main() {
    var students = List.of(new Student("Anna", 85), new Student("Oleg", 40), new Student("Ivan", 62));
 
    var passed = students.stream()
            .collect(Collectors.partitioningBy(s -> s.score() >= 50, Collectors.counting()));
    IO.println(passed);
}

Output:

{false=1, true=2}

Both groups always exist

Unlike groupingBy(), both parts are present even when one of them is empty:

void main() {
    var numbers = List.of(2, 4, 6);
    IO.println(numbers.stream().collect(Collectors.partitioningBy(n -> n > 10)));
}

Output:

{false=[2, 4, 6], true=[]}

Common mistake

Filtering the list twice — once by the condition and once by its opposite. That is two passes and a duplicated condition that is easy to change in only one place. partitioningBy() does it all in one pass.

Thanks for reading!

Short answer

With counting

Both groups always exist

Common mistake

Functions used in this recipe

Thanks for reading!