Java 8 stream reverse order

Question

General question: What's the proper way to reverse a stream? Assuming that we don't know what type of elements that stream consists of, what's the generic way to reverse any stream?

Specific question:

IntStream provides range method to generate Integers in specific range IntStream.range(-range, 0), now that I want to reverse it switching range from 0 to negative won't work, also I can't use Integer::compare

List<Integer> list = Arrays.asList(1,2,3,4);
list.stream().sorted(Integer::compare).forEach(System.out::println);

with IntStream I'll get this compiler error

Error:(191, 0) ajc: The method sorted() in the type IntStream is not applicable for the arguments (Integer::compare)

what am I missing here?

Answer 1

For the specific question of generating a reverse IntStream, try something like this:

static IntStream revRange(int from, int to) {
    return IntStream.range(from, to)
                    .map(i -> to - i + from - 1);
}

This avoids boxing and sorting.

For the general question of how to reverse a stream of any type, I don't know of there's a "proper" way. There are a couple ways I can think of. Both end up storing the stream elements. I don't know of a way to reverse a stream without storing the elements.

This first way stores the elements into an array and reads them out to a stream in reverse order. Note that since we don't know the runtime type of the stream elements, we can't type the array properly, requiring an unchecked cast.

@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
    Object[] temp = input.toArray();
    return (Stream<T>) IntStream.range(0, temp.length)
                                .mapToObj(i -> temp[temp.length - i - 1]);
}

Another technique uses collectors to accumulate the items into a reversed list. This does lots of insertions at the front of ArrayList objects, so there's lots of copying going on.

Stream<T> input = ... ;
List<T> output =
    input.collect(ArrayList::new,
                  (list, e) -> list.add(0, e),
                  (list1, list2) -> list1.addAll(0, list2));

It's probably possible to write a much more efficient reversing collector using some kind of customized data structure.

UPDATE 2016-01-29

Since this question has gotten a bit of attention recently, I figure I should update my answer to solve the problem with inserting at the front of ArrayList. This will be horribly inefficient with a large number of elements, requiring O(N^2) copying.

It's preferable to use an ArrayDeque instead, which efficiently supports insertion at the front. A small wrinkle is that we can't use the three-arg form of Stream.collect(); it requires the contents of the second arg be merged into the first arg, and there's no "add-all-at-front" bulk operation on Deque. Instead, we use addAll() to append the contents of the first arg to the end of the second, and then we return the second. This requires using the Collector.of() factory method.

The complete code is this:

Deque<String> output =
    input.collect(Collector.of(
        ArrayDeque::new,
        (deq, t) -> deq.addFirst(t),
        (d1, d2) -> { d2.addAll(d1); return d2; }));

The result is a Deque instead of a List, but that shouldn't be much of an issue, as it can easily be iterated or streamed in the now-reversed order.

Answer 2

Elegant solution

List<Integer> list = Arrays.asList(1,2,3,4);
list.stream()
    .sorted(Collections.reverseOrder()) // Method on Stream<Integer>
    .forEach(System.out::println);

Answer 3

General Question:

Stream does not store any elements.

So iterating elements in the reverse order is not possible without storing the elements in some intermediate collection.

Stream.of("1", "2", "20", "3")
      .collect(Collectors.toCollection(ArrayDeque::new)) // or LinkedList
      .descendingIterator()
      .forEachRemaining(System.out::println);

Update: Changed LinkedList to ArrayDeque (better) see here for details

Prints:

3

20

2

1

By the way, using sort method is not correct as it sorts, NOT reverses (assuming stream may have unordered elements)

Specific Question:

I found this simple, easier and intuitive(Copied @Holger comment)

IntStream.iterate(to - 1, i -> i - 1).limit(to - from)

Answer 4

Many of the solutions here sort or reverse the IntStream, but that unnecessarily requires intermediate storage. Stuart Marks's solution is the way to go:

static IntStream revRange(int from, int to) {
    return IntStream.range(from, to).map(i -> to - i + from - 1);
}

It correctly handles overflow as well, passing this test:

@Test
public void testRevRange() {
    assertArrayEquals(revRange(0, 5).toArray(), new int[]{4, 3, 2, 1, 0});
    assertArrayEquals(revRange(-5, 0).toArray(), new int[]{-1, -2, -3, -4, -5});
    assertArrayEquals(revRange(1, 4).toArray(), new int[]{3, 2, 1});
    assertArrayEquals(revRange(0, 0).toArray(), new int[0]);
    assertArrayEquals(revRange(0, -1).toArray(), new int[0]);
    assertArrayEquals(revRange(MIN_VALUE, MIN_VALUE).toArray(), new int[0]);
    assertArrayEquals(revRange(MAX_VALUE, MAX_VALUE).toArray(), new int[0]);
    assertArrayEquals(revRange(MIN_VALUE, MIN_VALUE + 1).toArray(), new int[]{MIN_VALUE});
    assertArrayEquals(revRange(MAX_VALUE - 1, MAX_VALUE).toArray(), new int[]{MAX_VALUE - 1});
}

Answer 5

How NOT to do it:

• Don't use .sorted(Comparator.reverseOrder()) or .sorted(Collections.reverseOrder()), because it will just sort elements in the descending order.
  Using it for given Integer input:
  [1, 4, 2, 5, 3]
  the output would be as follows:
  [5, 4, 3, 2, 1]
  For String input:
  ["A", "D", "B", "E", "C"]
  the output would be as follows:
  [E, D, C, B, A]
• Don't use .sorted((a, b) -> -1) (explanation at the end)

The easiest way to do it properly:

List<Integer> list = Arrays.asList(1, 4, 2, 5, 3);
Collections.reverse(list);
System.out.println(list);

Output:
[3, 5, 2, 4, 1]

The same for String:

List<String> stringList = Arrays.asList("A", "D", "B", "E", "C");
Collections.reverse(stringList);
System.out.println(stringList);

Output:
[C, E, B, D, A]

Don't use .sorted((a, b) -> -1)!
It breaks comparator contract and might work only for some cases ie. only on single thread but not in parallel.
yankee explanation:

(a, b) -> -1 breaks the contract for Comparator. Whether this works depends on the implementation of the sort algorithm. The next release of the JVM might break this. Actually I can already break this reproduciblly on my machine using IntStream.range(0, 10000).parallel().boxed().sorted((a, b) -> -1).forEachOrdered(System.out::println);

//Don't use this!!!
List<Integer> list = Arrays.asList(1, 4, 2, 5, 3);
List<Integer> reversedList = list.stream()
        .sorted((a, b) -> -1)
        .collect(Collectors.toList());
System.out.println(reversedList);

Output in positive case:
[3, 5, 2, 4, 1]

Possible output in parallel stream or with other JVM implementation:
[4, 1, 2, 3, 5]

The same for String:

//Don't use this!!!
List<String> stringList = Arrays.asList("A", "D", "B", "E", "C");
List<String> reversedStringList = stringList.stream()
        .sorted((a, b) -> -1)
        .collect(Collectors.toList());
System.out.println(reversedStringList);

Output in positive case:
[C, E, B, D, A]

Possible output in parallel stream or with other JVM implementation:
[A, E, B, D, C]

Answer 6

without external lib...

import java.util.List;
import java.util.Collections;
import java.util.stream.Collector;

public class MyCollectors {

    public static <T> Collector<T, ?, List<T>> toListReversed() {
        return Collectors.collectingAndThen(Collectors.toList(), l -> {
            Collections.reverse(l);
            return l;
        });
    }

}

Answer 7

You could define your own collector that collects the elements in reverse order:

public static <T> Collector<T, List<T>, List<T>> inReverse() {
    return Collector.of(
        ArrayList::new,
        (l, t) -> l.add(t),
        (l, r) -> {l.addAll(r); return l;},
        Lists::<T>reverse);
}

And use it like:

stream.collect(inReverse()).forEach(t -> ...)

I use an ArrayList in forward order to efficiently insert collect the items (at the end of the list), and Guava Lists.reverse to efficiently give a reversed view of the list without making another copy of it.

Here are some test cases for the custom collector:

import static org.hamcrest.MatcherAssert.assertThat;
import static org.hamcrest.Matchers.*;

import java.util.ArrayList;
import java.util.List;
import java.util.function.BiConsumer;
import java.util.function.BinaryOperator;
import java.util.function.Function;
import java.util.function.Supplier;
import java.util.stream.Collector;

import org.hamcrest.Matchers;
import org.junit.Test;

import com.google.common.collect.Lists;

public class TestReverseCollector {
    private final Object t1 = new Object();
    private final Object t2 = new Object();
    private final Object t3 = new Object();
    private final Object t4 = new Object();

    private final Collector<Object, List<Object>, List<Object>> inReverse = inReverse();
    private final Supplier<List<Object>> supplier = inReverse.supplier();
    private final BiConsumer<List<Object>, Object> accumulator = inReverse.accumulator();
    private final Function<List<Object>, List<Object>> finisher = inReverse.finisher();
    private final BinaryOperator<List<Object>> combiner = inReverse.combiner();

    @Test public void associative() {
        final List<Object> a1 = supplier.get();
        accumulator.accept(a1, t1);
        accumulator.accept(a1, t2);
        final List<Object> r1 = finisher.apply(a1);

        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        final List<Object> a3 = supplier.get();
        accumulator.accept(a3, t2);
        final List<Object> r2 = finisher.apply(combiner.apply(a2, a3));

        assertThat(r1, Matchers.equalTo(r2));
    }

    @Test public void identity() {
        final List<Object> a1 = supplier.get();
        accumulator.accept(a1, t1);
        accumulator.accept(a1, t2);
        final List<Object> r1 = finisher.apply(a1);

        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        accumulator.accept(a2, t2);
        final List<Object> r2 = finisher.apply(combiner.apply(a2, supplier.get()));

        assertThat(r1, equalTo(r2));
    }

    @Test public void reversing() throws Exception {
        final List<Object> a2 = supplier.get();
        accumulator.accept(a2, t1);
        accumulator.accept(a2, t2);

        final List<Object> a3 = supplier.get();
        accumulator.accept(a3, t3);
        accumulator.accept(a3, t4);

        final List<Object> r2 = finisher.apply(combiner.apply(a2, a3));

        assertThat(r2, contains(t4, t3, t2, t1));
    }

    public static <T> Collector<T, List<T>, List<T>> inReverse() {
        return Collector.of(
            ArrayList::new,
            (l, t) -> l.add(t),
            (l, r) -> {l.addAll(r); return l;},
            Lists::<T>reverse);
    }
}

Answer 8

If implemented Comparable<T> (ex. Integer, String, Date), you can do it using Comparator.reverseOrder().

List<Integer> list = Arrays.asList(1, 2, 3, 4);
list.stream()
     .sorted(Comparator.reverseOrder())
     .forEach(System.out::println);

Answer 9

cyclops-react StreamUtils has a reverse Stream method (javadoc).

  StreamUtils.reverse(Stream.of("1", "2", "20", "3"))
             .forEach(System.out::println);

It works by collecting to an ArrayList and then making use of the ListIterator class which can iterate in either direction, to iterate backwards over the list.

If you already have a List, it will be more efficient

  StreamUtils.reversedStream(Arrays.asList("1", "2", "20", "3"))
             .forEach(System.out::println);

Answer 10

Here's the solution I've come up with:

private static final Comparator<Integer> BY_ASCENDING_ORDER = Integer::compare;
private static final Comparator<Integer> BY_DESCENDING_ORDER = BY_ASCENDING_ORDER.reversed();

then using those comparators:

IntStream.range(-range, 0).boxed().sorted(BY_DESCENDING_ORDER).forEach(// etc...

Answer 11

I would suggest using jOOλ, it's a great library that adds lots of useful functionality to Java 8 streams and lambdas.

You can then do the following:

List<Integer> list = Arrays.asList(1,2,3,4);    
Seq.seq(list).reverse().forEach(System.out::println)

Simple as that. It's a pretty lightweight library, and well worth adding to any Java 8 project.

Answer 12

How about this utility method?

public static <T> Stream<T> getReverseStream(List<T> list) {
    final ListIterator<T> listIt = list.listIterator(list.size());
    final Iterator<T> reverseIterator = new Iterator<T>() {
        @Override
        public boolean hasNext() {
            return listIt.hasPrevious();
        }

        @Override
        public T next() {
            return listIt.previous();
        }
    };
    return StreamSupport.stream(Spliterators.spliteratorUnknownSize(
            reverseIterator,
            Spliterator.ORDERED | Spliterator.IMMUTABLE), false);
}

Seems to work with all cases without duplication.

Answer 13

With regard to the specific question of generating a reverse IntStream:

starting from Java 9 you can use the three-argument version of the IntStream.iterate(...):

IntStream.iterate(10, x -> x >= 0, x -> x - 1).forEach(System.out::println);

// Out: 10 9 8 7 6 5 4 3 2 1 0

where:

IntStream.iterate​(int seed, IntPredicate hasNext, IntUnaryOperator next);

• seed - the initial element;
• hasNext - a predicate to apply to elements to determine when the stream must terminate;
• next - a function to be applied to the previous element to produce a new element.

Answer 14

Simplest way (simple collect - supports parallel streams):

public static <T> Stream<T> reverse(Stream<T> stream) {
    return stream
            .collect(Collector.of(
                    () -> new ArrayDeque<T>(),
                    ArrayDeque::addFirst,
                    (q1, q2) -> { q2.addAll(q1); return q2; })
            )
            .stream();
}

Advanced way (supports parallel streams in an ongoing way):

public static <T> Stream<T> reverse(Stream<T> stream) {
    Objects.requireNonNull(stream, "stream");

    class ReverseSpliterator implements Spliterator<T> {
        private Spliterator<T> spliterator;
        private final Deque<T> deque = new ArrayDeque<>();

        private ReverseSpliterator(Spliterator<T> spliterator) {
            this.spliterator = spliterator;
        }

        @Override
        @SuppressWarnings({"StatementWithEmptyBody"})
        public boolean tryAdvance(Consumer<? super T> action) {
            while(spliterator.tryAdvance(deque::addFirst));
            if(!deque.isEmpty()) {
                action.accept(deque.remove());
                return true;
            }
            return false;
        }

        @Override
        public Spliterator<T> trySplit() {
            // After traveling started the spliterator don't contain elements!
            Spliterator<T> prev = spliterator.trySplit();
            if(prev == null) {
                return null;
            }

            Spliterator<T> me = spliterator;
            spliterator = prev;
            return new ReverseSpliterator(me);
        }

        @Override
        public long estimateSize() {
            return spliterator.estimateSize();
        }

        @Override
        public int characteristics() {
            return spliterator.characteristics();
        }

        @Override
        public Comparator<? super T> getComparator() {
            Comparator<? super T> comparator = spliterator.getComparator();
            return (comparator != null) ? comparator.reversed() : null;
        }

        @Override
        public void forEachRemaining(Consumer<? super T> action) {
            // Ensure that tryAdvance is called at least once
            if(!deque.isEmpty() || tryAdvance(action)) {
                deque.forEach(action);
            }
        }
    }

    return StreamSupport.stream(new ReverseSpliterator(stream.spliterator()), stream.isParallel());
}

Note you can quickly extends to other type of streams (IntStream, ...).

Testing:

// Use parallel if you wish only
revert(Stream.of("One", "Two", "Three", "Four", "Five", "Six").parallel())
    .forEachOrdered(System.out::println);

Results:

Six
Five
Four
Three
Two
One

Additional notes: The simplest way it isn't so useful when used with other stream operations (the collect join breaks the parallelism). The advance way doesn't have that issue, and it keeps also the initial characteristics of the stream, for example SORTED, and so, it's the way to go to use with other stream operations after the reverse.

Answer 15

Not purely Java8 but if you use guava's Lists.reverse() method in conjunction, you can easily achieve this:

List<Integer> list = Arrays.asList(1,2,3,4);
Lists.reverse(list).stream().forEach(System.out::println);

Answer 16

ArrayDeque are faster in the stack than a Stack or LinkedList. "push()" inserts elements at the front of the Deque

 protected <T> Stream<T> reverse(Stream<T> stream) {
    ArrayDeque<T> stack = new ArrayDeque<>();
    stream.forEach(stack::push);
    return stack.stream();
}

Answer 17

List newStream = list.stream().sorted(Collections.reverseOrder()).collect(Collectors.toList());
        newStream.forEach(System.out::println);

Answer 18

One could write a collector that collects elements in reversed order:

public static <T> Collector<T, ?, Stream<T>> reversed() {
    return Collectors.collectingAndThen(Collectors.toList(), list -> {
        Collections.reverse(list);
        return list.stream();
    });
}

And use it like this:

Stream.of(1, 2, 3, 4, 5).collect(reversed()).forEach(System.out::println);

Original answer (contains a bug - it does not work correctly for parallel streams):

A general purpose stream reverse method could look like:

public static <T> Stream<T> reverse(Stream<T> stream) {
    LinkedList<T> stack = new LinkedList<>();
    stream.forEach(stack::push);
    return stack.stream();
}

Answer 19

For reference I was looking at the same problem, I wanted to join the string value of stream elements in the reverse order.

itemList = { last, middle, first } => first,middle,last

I started to use an intermediate collection with collectingAndThen from comonad or the ArrayDeque collector of Stuart Marks, although I wasn't happy with intermediate collection, and streaming again

itemList.stream()
        .map(TheObject::toString)
        .collect(Collectors.collectingAndThen(Collectors.toList(),
                                              strings -> {
                                                      Collections.reverse(strings);
                                                      return strings;
                                              }))
        .stream()
        .collect(Collector.joining());

So I iterated over Stuart Marks answer that was using the Collector.of factory, that has the interesting finisher lambda.

itemList.stream()
        .collect(Collector.of(StringBuilder::new,
                             (sb, o) -> sb.insert(0, o),
                             (r1, r2) -> { r1.insert(0, r2); return r1; },
                             StringBuilder::toString));

Since in this case the stream is not parallel, the combiner is not relevant that much, I'm using insert anyway for the sake of code consistency but it does not matter as it would depend of which stringbuilder is built first.

I looked at the StringJoiner, however it does not have an insert method.

Answer 20

Reversing string or any Array

(Stream.of("abcdefghijklm 1234567".split("")).collect(Collectors.collectingAndThen(Collectors.toList(),list -> {Collections.reverse(list);return list;}))).stream().forEach(System.out::println);

split can be modified based on the delimiter or space

Answer 21

This method works with any Stream and is Java 8 compliant:

Stream<Integer> myStream = Stream.of(1, 2, 3, 4, 5);
myStream.reduce(Stream.empty(),
        (Stream<Integer> a, Integer b) -> Stream.concat(Stream.of(b), a),
        (a, b) -> Stream.concat(b, a))
        .forEach(System.out::println);

Answer 22

How about reversing the Collection backing the stream prior?

import java.util.Collections;
import java.util.List;

public void reverseTest(List<Integer> sampleCollection) {
    Collections.reverse(sampleCollection); // remember this reverses the elements in the list, so if you want the original input collection to remain untouched clone it first.

    sampleCollection.stream().forEach(item -> {
      // you op here
    });
}

Answer 23

Answering specific question of reversing with IntStream, below worked for me:

IntStream.range(0, 10)
  .map(x -> x * -1)
  .sorted()
  .map(Math::abs)
  .forEach(System.out::println);

Answer 24

In all this I don't see the answer I would go to first.

This isn't exactly a direct answer to the question, but it's a potential solution to the problem.

Just build the list backwards in the first place. If you can, use a LinkedList instead of an ArrayList and when you add items use "Push" instead of add. The list will be built in the reverse order and will then stream correctly without any manipulation.

This won't fit cases where you are dealing with primitive arrays or lists that are already used in various ways but does work well in a surprising number of cases.

Answer 25

the simplest solution is using List::listIterator and Stream::generate

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
ListIterator<Integer> listIterator = list.listIterator(list.size());

Stream.generate(listIterator::previous)
      .limit(list.size())
      .forEach(System.out::println);

Answer 26

Based on @stuart-marks's answer, but without casting, function returning stream of list elements starting from end:

public static <T> Stream<T> reversedStream(List<T> tList) {
    final int size = tList.size();
    return IntStream.range(0, size)
            .mapToObj(i -> tList.get(size - 1 - i));
}

// usage
reversedStream(list).forEach(System.out::println);

Answer 27

This is how I do it.

I don't like the idea of creating a new collection and reverse iterating it.

The IntStream#map idea is pretty neat, but I prefer the IntStream#iterate method, for I think the idea of a countdown to Zero better expressed with the iterate method and easier to understand in terms of walking the array from back to front.

import static java.lang.Math.max;

private static final double EXACT_MATCH = 0d;

public static IntStream reverseStream(final int[] array) {
    return countdownFrom(array.length - 1).map(index -> array[index]);
}

public static DoubleStream reverseStream(final double[] array) {
    return countdownFrom(array.length - 1).mapToDouble(index -> array[index]);
}

public static <T> Stream<T> reverseStream(final T[] array) {
    return countdownFrom(array.length - 1).mapToObj(index -> array[index]);
}

public static IntStream countdownFrom(final int top) {
    return IntStream.iterate(top, t -> t - 1).limit(max(0, (long) top + 1));
}

Here are some tests to prove it works:

import static java.lang.Integer.MAX_VALUE;
import static org.junit.Assert.*;

@Test
public void testReverseStream_emptyArrayCreatesEmptyStream() {
    Assert.assertEquals(0, reverseStream(new double[0]).count());
}

@Test
public void testReverseStream_singleElementCreatesSingleElementStream() {
    Assert.assertEquals(1, reverseStream(new double[1]).count());
    final double[] singleElementArray = new double[] { 123.4 };
    assertArrayEquals(singleElementArray, reverseStream(singleElementArray).toArray(), EXACT_MATCH);
}

@Test
public void testReverseStream_multipleElementsAreStreamedInReversedOrder() {
    final double[] arr = new double[] { 1d, 2d, 3d };
    final double[] revArr = new double[] { 3d, 2d, 1d };
    Assert.assertEquals(arr.length, reverseStream(arr).count());
    Assert.assertArrayEquals(revArr, reverseStream(arr).toArray(), EXACT_MATCH);
}

@Test
public void testCountdownFrom_returnsAllElementsFromTopToZeroInReverseOrder() {
    assertArrayEquals(new int[] { 4, 3, 2, 1, 0 }, countdownFrom(4).toArray());
}

@Test
public void testCountdownFrom_countingDownStartingWithZeroOutputsTheNumberZero() {
    assertArrayEquals(new int[] { 0 }, countdownFrom(0).toArray());
}

@Test
public void testCountdownFrom_doesNotChokeOnIntegerMaxValue() {
    assertEquals(true, countdownFrom(MAX_VALUE).anyMatch(x -> x == MAX_VALUE));
}

@Test
public void testCountdownFrom_givesZeroLengthCountForNegativeValues() {
    assertArrayEquals(new int[0], countdownFrom(-1).toArray());
    assertArrayEquals(new int[0], countdownFrom(-4).toArray());
}

Answer 28

What's the proper generic way to reverse a stream?

If the stream does not specify an encounter order, don't. (!s.spliterator().hasCharacteristics(java.util.Spliterator.ORDERED))

Answer 29

The only way to reverse any general stream is to read through the stream, capturing all values, and then stream through the captured values in reverse order. This means that at minimum, a full copy of the stream data must be stored in order to stream through it in reverse order.

JEP 431: Sequenced Collections in the upcoming version 21 of Java adds (among other things) a reversed() method List, which provides a reversed view of the list. Using this feature makes reversing a general Stream a one-liner:

Stream<T> reversed = myStream.stream().toList().reversed().toStream();

Accomplishing the same thing with Java version 20 without using a customized collection type is more verbose, typically requiring an extra step of reversing the resulting list. The following Java 20 implementation streams to a List, reverses the list, and then return a stream using .toStream().

List<T> list = myStream.stream().collect(Collectors.toCollection(ArrayList::new));
Collections.reverse(list);
Stream<T> reversed = list.stream();

Answer 30

The most generic and the easiest way to reverse a list will be :

public static <T> void reverseHelper(List<T> li){

 li.stream()
.sorted((x,y)-> -1)
.collect(Collectors.toList())
.forEach(System.out::println);

    }

Answer 31

Java 8 way to do this:

    List<Integer> list = Arrays.asList(1,2,3,4);
    Comparator<Integer> comparator = Integer::compare;
    list.stream().sorted(comparator.reversed()).forEach(System.out::println);