如何使用java8流对TreeSet的列表进行排序

Question

My list contains sets like [1,3,5][2,6,4] etc, all of the same size. I tried doing this but it doesn't seem to work.

List<TreeSet<T>> block;
    for(TreeSet<T> t : block){
        block.stream().sorted((n,m)->n.compareTo(m)).collect(Collectors.toSet());

    }

The end result I want is [1,2,3][4,5,6] .

I could try to add all the elements in an ArrayList and sort that out then make a new List of TreeSet 's. But is there is some kind of one liner?

UPDATE:

List<T> list=new ArrayList<T>();
    for(TreeSet<T> t : block){

        for(T t1 : t)
        {
            list.add(t1);   

        }
    }

    list=list.stream().sorted((n,m)->n.compareTo(m)).collect(Collectors.toList());

This works but could this be simplified?

Answer 1

@Eugene's answer is sweet, because Guava is sweet. But if you happen to not have Guava in your classpath, here's another way:

List<Set<Integer>> list = block.stream()
    .flatMap(Set::stream)
    .sorted()
    .collect(partitioning(3));

First, I'm flatmapping all the sets into one stream, then I'm sorting all the elements and finally, I'm collecting the whole sorted stream to a list of sets. For this, I'm invoking a helper method that uses a custom collector:

private static <T> Collector<T, ?, List<Set<T>>> partitioning(int size) {
    class Acc {
        int count = 0;
        List<Set<T>> list = new ArrayList<>();

        void add(T elem) {
            int index = count++ / size;
            if (index == list.size()) list.add(new LinkedHashSet<>());
            list.get(index).add(elem);
        }

        Acc merge(Acc another) {
            another.list.stream().flatMap(Set::stream).forEach(this::add);
            return this;
        }
    }
    return Collector.of(Acc::new, Acc::add, Acc::merge, acc -> acc.list);
}

The method receives the size of each partition and uses the Acc local class as the mutable structure to be used by the collector. Inside the Acc class, I'm using a List that will contain LinkedHashSet instances, which will hold the elements of the stream.

The Acc class keeps the count of all the elements that have been already collected. In the add method, I calculate the index of the list and increment this count, and if there was no set in that position of the list, I append a new empty LinkedHashSet to it. Then, I add the element to the set.

As I'm calling sorted() on the stream to sort its elements before collecting, I need to use data structures that preserve insertion order. This is why I'm using ArrayList for the outer list and LinkedHashSet for the inner sets.

The merge method is to be used by parallel streams, to merge two previously accumulated Acc instances. I'm just adding all the elements of the received Acc instance to this Acc instance, by delegating to the add method.

Finally, I'm using Collector.of to create a collector based on the methods of the Acc class. The last argument is a finisher function, which just returns the Acc instance's list.

Answer 2

If you have guava on the classpath this is a breeze:

        block
            .stream()
            .flatMap(Set::stream)
            .collect(Collectors.toCollection(TreeSet::new));

    Iterable<List<Integer>> result = Iterables.partition(sorted, 3);

Answer 3

Adding another answer since this would be bigger than a comment. It's really what the accepted answer has done, but with a "smarter" combiner that does not have to stream all the time again.

 private static <T> Collector<T, ?, List<Set<T>>> partitioning(int size) {
    class Acc {
        int count = 0;

        List<List<T>> list = new ArrayList<>();

        void add(T elem) {
            int index = count++ / size;
            if (index == list.size()) {
                list.add(new ArrayList<>());
            }
            list.get(index).add(elem);
        }

        Acc merge(Acc right) {

            List<T> lastLeftList = list.get(list.size() - 1);
            List<T> firstRightList = right.list.get(0);
            int lastLeftSize = lastLeftList.size();
            int firstRightSize = firstRightList.size();

            // they have both the same size, simply addAll will work
            if (lastLeftSize + firstRightSize == 2 * size) {
                System.out.println("Perfect!");
                list.addAll(right.list);
                return this;
            }

            // last and first from each chunk are merged "perfectly"
            if (lastLeftSize + firstRightSize == size) {
                System.out.println("Almost perfect");
                int x = 0;
                while (x < firstRightSize) {
                    lastLeftList.add(firstRightList.remove(x));
                    --firstRightSize;
                }
                right.list.remove(0);
                list.addAll(right.list);
                return this;
            }

            right.list.stream().flatMap(List::stream).forEach(this::add);
            return this;
        }

        public List<Set<T>> finisher() {
            return list.stream().map(LinkedHashSet::new).collect(Collectors.toList());
        }

    }
    return Collector.of(Acc::new, Acc::add, Acc::merge, Acc::finisher);
}