How to convert a Collection to List?

Question

I am using TreeBidiMap from the Apache Collections library. I want to sort this on the values which are doubles.

My method is to retrieve a Collection of the values using:

Collection coll = themap.values();

Which naturally works fine.

Main Question: I now want to know how I can convert/cast (not sure which is correct) coll into a List so it can be sorted?

I then intend to iterate over the sorted List object, which should be in order and get the appropriate keys from the TreeBidiMap (themap) using themap.getKey(iterator.next()) where the iterator will be over the list of doubles.

Answer 1

Java 8 onwards...

You can convert Collection to any collection (i.e, List, Set, and Queue) using Streams and Collectors.toCollection().

Consider the following example map

Map<Integer, Double> map = Map.of(
    1, 1015.45,
    2, 8956.31,
    3, 1234.86,
    4, 2348.26,
    5, 7351.03
);

to ArrayList

List<Double> arrayList = map.values()
                            .stream()
                            .collect(
                                Collectors.toCollection(ArrayList::new)
                            );

Output: [7351.03, 2348.26, 1234.86, 8956.31, 1015.45]

to Sorted ArrayList (Ascending order)

List<Double> arrayListSortedAsc = map.values()
                                        .stream()
                                        .sorted()
                                        .collect(
                                            Collectors.toCollection(ArrayList::new)
                                        );

Output: [1015.45, 1234.86, 2348.26, 7351.03, 8956.31]

to Sorted ArrayList (Descending order)

List<Double> arrayListSortedDesc = map.values()
                                        .stream()
                                        .sorted(
                                            (a, b) -> b.compareTo(a)
                                        )
                                        .collect(
                                            Collectors.toCollection(ArrayList::new)
                                        );

Output: [8956.31, 7351.03, 2348.26, 1234.86, 1015.45]

to LinkedList

List<Double> linkedList = map.values()
                                .stream()
                                .collect(
                                    Collectors.toCollection(LinkedList::new)
                                );

Output: [7351.03, 2348.26, 1234.86, 8956.31, 1015.45]

to HashSet

Set<Double> hashSet = map.values()
                            .stream()
                            .collect(
                                Collectors.toCollection(HashSet::new)
                            );

Output: [2348.26, 8956.31, 1015.45, 1234.86, 7351.03]

to PriorityQueue

PriorityQueue<Double> priorityQueue = map.values()
                                            .stream()
                                            .collect(
                                                Collectors.toCollection(PriorityQueue::new)
                                            );

Output: [1015.45, 1234.86, 2348.26, 8956.31, 7351.03]

Reference

Java - Package java.util.stream

Java - Package java.util

Answer 2

Use streams:

someCollection.stream().collect(Collectors.toList())

Answer 3

Java 10 introduced List#copyOf which returns unmodifiable List while preserving the order:

List<Integer> list = List.copyOf(coll);

Answer 4

What you request is quite a costy operation, make sure you don't need to do it often (e.g in a cycle).

If you need it to stay sorted and you update it frequently, you can create a custom collection. For example, I came up with one that has your TreeBidiMap and TreeMultiset under the hood. Implement only what you need and care about data integrity.

class MyCustomCollection implements Map<K, V> {
    TreeBidiMap<K, V> map;
    TreeMultiset<V> multiset;
    public V put(K key, V value) {
        removeValue(map.put(key, value));
        multiset.add(value);
    }
    public boolean remove(K key) {
        removeValue(map.remove(key));
    }
    /** removes value that was removed/replaced in map */
    private removeValue(V value) {
        if (value != null) {
            multiset.remove(value);
        }
    }
    public Set<K> keySet() {
        return Collections.unmodifiableSet(map.keySet());
    }
    public Collection<V> values() {
        return Collections.unmodifiableCollection(multiset);
    }
    // many more methods to be implemented, e.g. count, isEmpty etc.
    // but these are fairly simple
}

This way, you have a sorted Multiset returned from values(). However, if you need it to be a list (e.g. you need the array-like get(index) method), you'd need something more complex.

For brevity, I only return unmodifiable collections. What @Lino mentioned is correct, and modifying the keySet or values collection as it is would make it inconsistent. I don't know any consistent way to make the values mutable, but the keySet could support remove if it uses the remove method from the MyCustomCollection class above.

Answer 5

I believe you can write it as such:

coll.stream().collect(Collectors.toList())

Answer 6

Collections.sort( new ArrayList( coll ) );

Answer 7

Something like this should work, calling the ArrayList constructor that takes a Collection:

List theList = new ArrayList(coll);

Answer 8

Here is a sub-optimal solution as a one-liner:

Collections.list(Collections.enumeration(coll));

Answer 9

@Kunigami: I think you may be mistaken about Guava's newArrayList method. It does not check whether the Iterable is a List type and simply return the given List as-is. It always creates a new list:

@GwtCompatible(serializable = true)
public static <E> ArrayList<E> newArrayList(Iterable<? extends E> elements) {
  checkNotNull(elements); // for GWT
  // Let ArrayList's sizing logic work, if possible
  return (elements instanceof Collection)
      ? new ArrayList<E>(Collections2.cast(elements))
      : newArrayList(elements.iterator());
}

Answer 10

List list = new ArrayList(coll);
Collections.sort(list);

As Erel Segal Halevi says below, if coll is already a list, you can skip step one. But that would depend on the internals of TreeBidiMap.

List list;
if (coll instanceof List)
  list = (List)coll;
else
  list = new ArrayList(coll);

Answer 11

I think Paul Tomblin's answer may be wasteful in case coll is already a list, because it will create a new list and copy all elements. If coll contains many elemeents, this may take a long time.

My suggestion is:

List list;
if (coll instanceof List)
  list = (List)coll;
else
  list = new ArrayList(coll);
Collections.sort(list);