Take n random elements from a List<E>?

user568866 picture user568866 · Jan 15, 2011 · Viewed 47.5k times · Source

How can I take n random elements from an ArrayList<E>? Ideally, I'd like to be able to make successive calls to the take() method to get another x elements, without replacement.

Answer

BalusC picture BalusC · Jan 15, 2011

Two main ways.

  1. Use Random#nextInt(int):

    List<Foo> list = createItSomehow();
    Random random = new Random();
    Foo foo = list.get(random.nextInt(list.size()));
    

    It's however not guaranteed that successive n calls returns unique elements.

  2. Use Collections#shuffle():

    List<Foo> list = createItSomehow();
    Collections.shuffle(list);
    Foo foo = list.get(0);
    

    It enables you to get n unique elements by an incremented index (assuming that the list itself contains unique elements).


In case you're wondering if there's a Java 8 Stream approach; no, there isn't a built-in one. There's no such thing as Comparator#randomOrder() in standard API (yet?). You could try something like below while still satisfying the strict Comparator contract (although the distribution is pretty terrible):

List<Foo> list = createItSomehow();
int random = new Random().nextInt();
Foo foo = list.stream().sorted(Comparator.comparingInt(o -> System.identityHashCode(o) ^ random)).findFirst().get();

Better use Collections#shuffle() instead.