Creating subset of a Set in Java

Question

I have a LinkedHashSet, i.e an ordered set. I'm trying to find a function to just return a subset of the set, i.e the first 20 elements of the set. I know I can do it by creating a new set and then populating using an iteration of the first set but I was hoping for something more succinct.

Also took a look at Google's Guava libraries, but couldn't see what I wanted.

Solution 1

In Guava:

Set<Integer> subset = ImmutableSet.copyOf(Iterables.limit(set, 20));

Note that Iterables.limit() is evaluated lazily, so only one extra collection is created.

Solution 2

In Java 8 you can do

mySet.stream()
   .skip(start) // the offset
   .limit(count) // how many items you want
   .collect(Collectors.toSet());

Solution 3

A solution using streams and collectors:

Set<Integer> subSet = set.stream()
    // .skip(10) // Use this to get elements later in the stream
    .limit(20)
    .collect(toCollection(LinkedHashSet::new));
    // You could also collect to something else 
    // with another collector like this: 
    // .collect(toList());

This assumes the following import:

import static java.util.stream.Collectors.toCollection;

Solution 4

You could do this:

Set<Integer> set = new LinkedHashSet<>();
for (int i = 0; i < 50; i++) {
   set.add(i);
}

List<Integer> list = new ArrayList<>(set);
Set<Integer> subSet = new LinkedHashSet<>(list.subList(0, 20));

Solution 5

You can either use at first a SortedSet as the subSet method exists on it.

You can also add the content of your set to a List and use the subList method on it. But it depends on the amount of data stored in your Set as you would not want to duplicate an enormous volume of data.

Otherwise you should stay with the iteration over the Set as it will be more efficient.

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Comments

[Paul Taylor]

THanks that look neat, except for the fact you have to create both another Set and an ArrayList but I can live with that.

[Paul Taylor]

Thanks I had an inkling there would be a way to do this in guava.

[Dejell]

what if I want get from index 20-40?

[dtrunk]

To avoid a IndexOutOfBoundsException use Math.min(list.size(), 20) as 2nd parameter of subList in case of a dynamically filled list.

[dnellis74]

You mean limit() not skip() I think.

[Farid]

@Dejell As you have already figured out by now, there is no index in Set unless its implementation has one e.g. SortedSet