Reputation: 5917

Retrieving a List from a java.util.stream.Stream in Java 8

I was playing around with Java 8 lambdas to easily filter collections. But I did not find a concise way to retrieve the result as a new list within the same statement. Here is my most concise approach so far:

List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
List<Long> targetLongList = new ArrayList<>();
sourceLongList.stream().filter(l -> l > 100).forEach(targetLongList::add);

Examples on the net did not answer my question because they stop without generating a new result list. There must be a more concise way. I would have expected, that the Stream class has methods as toList(), toSet(), …

Is there a way that the variables targetLongList can be directly be assigned by the third line?

Upvotes: 509

Answers (15)

MohamedSanaulla

Reputation: 6252

One approach is to use Collectors.toList to collect the stream into a list:

targetLongList = 
    sourceLongList.stream().
    filter(l -> l > 100).
    collect(Collectors.toList());

If a specific List implementation is desired, Collectors.toCollection can be used instead:

targetLongList = 
    sourceLongList.stream().
    filter(l -> l > 100).
    collect(Collectors.toCollection(ArrayList::new));

Upvotes: 198

ZhekaKozlov

Reputation: 39634

`Stream#toList`

There is a new method Stream.toList() in Java 16+, which accumulates the stream elements into an unmodifiable list:

List<Long> targetLongList = sourceLongList
         .stream()
         .filter(l -> l > 100)
         .toList();

Upvotes: 17

Maurice Naftalin

Reputation: 10533

What you are doing may be the simplest way, provided your stream stays sequential—otherwise you will have to put a call to sequential() before forEach.

The reason the call to sequential() is necessary is that the code as it stands (forEach(targetLongList::add)) would be racy if the stream was parallel. Even then, it will not achieve the effect intended, as forEach is explicitly nondeterministic—even in a sequential stream the order of element processing is not guaranteed. You would have to use forEachOrdered to ensure correct ordering. The intention of the Stream API designers is that you will use collector in this situation, as below:

targetLongList = sourceLongList.stream()
    .filter(l -> l > 100)
    .collect(Collectors.toList());

Upvotes: 673

user_3380739

Reputation: 1284

Here is code by abacus-common

LongStream.of(1, 10, 50, 80, 100, 120, 133, 333).filter(e -> e > 100).toList();

Disclosure： I'm the developer of abacus-common.

Upvotes: 0

Saikat

Reputation: 16930

To collect in a mutable list:

targetList = sourceList.stream()
                       .filter(i -> i > 100) //apply filter
                       .collect(Collectors.toList());

To collect in a immutable list:

targetList = sourceList.stream()
                       .filter(i -> i > 100) //apply filter
                       .collect(Collectors.toUnmodifiableList());

Explanation of collect from the JavaDoc:

Performs a mutable reduction operation on the elements of this stream using a Collector. A Collector encapsulates the functions used as arguments to collect(Supplier, BiConsumer, BiConsumer), allowing for reuse of collection strategies and composition of collect operations such as multiple-level grouping or partitioning. If the stream is parallel, and the Collector is concurrent, and either the stream is unordered or the collector is unordered, then a concurrent reduction will be performed (see Collector for details on concurrent reduction.)

This is a terminal operation.

When executed in parallel, multiple intermediate results may be instantiated, populated, and merged so as to maintain isolation of mutable data structures. Therefore, even when executed in parallel with non-thread-safe data structures (such as ArrayList), no additional synchronization is needed for a parallel reduction.

Upvotes: 2

Pramod

Reputation: 417

collect(Collectors.toList());

This is the call which you can use to convert any Stream to List.

more concretely:

    List<String> myList = stream.collect(Collectors.toList());

from:

https://www.geeksforgeeks.org/collectors-tolist-method-in-java-with-examples/

Upvotes: 9

Pratik Pawar

Reputation: 91

You can rewrite code as below :

List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
List<Long> targetLongList = sourceLongList.stream().filter(l -> l > 100).collect(Collectors.toList());

Upvotes: 0

Vaneet Kataria

Reputation: 605

There is an another variant of collect method provided by LongStream class and similarly by IntStream and DoubleStream classes too .

<R> R collect(Supplier<R> supplier,
              ObjLongConsumer<R> accumulator,
              BiConsumer<R,R> combiner)

Performs a mutable reduction operation on the elements of this stream. A mutable reduction is one in which the reduced value is a mutable result container, such as an ArrayList, and elements are incorporated by updating the state of the result rather than by replacing the result. This produces a result equivalent to:

R result = supplier.get();
  for (long element : this stream)
       accumulator.accept(result, element);
  return result;

Like reduce(long, LongBinaryOperator), collect operations can be parallelized without requiring additional synchronization. This is a terminal operation.

And answer to your question with this collect method is as below :

    LongStream.of(1L, 2L, 3L, 3L).filter(i -> i > 2)
    .collect(ArrayList::new, (list, value) -> list.add(value)
    , (list1, list2) -> list1.addAll(list2));

Below is the method reference variant which is quite smart but some what tricky to understand :

     LongStream.of(1L, 2L, 3L, 3L).filter(i -> i > 2)
    .collect(ArrayList::new, List::add , List::addAll);

Below will be the HashSet variant :

     LongStream.of(1L, 2L, 3L, 3).filter(i -> i > 2)
     .collect(HashSet::new, HashSet::add, HashSet::addAll);

Similarly LinkedList variant is like this :

     LongStream.of(1L, 2L, 3L, 3L)
     .filter(i -> i > 2)
     .collect(LinkedList::new, LinkedList::add, LinkedList::addAll);

Upvotes: 1

sharath

Reputation: 225

String joined = 
                Stream.of(isRead?"read":"", isFlagged?"flagged":"", isActionRequired?"action":"", isHide?"hide":"")
                      .filter(s -> s != null && !s.isEmpty())
                      .collect(Collectors.joining(","));

Upvotes: 0

Lii

Reputation: 12131

I like to use a util method that returns a collector for ArrayList when that is what I want.

I think the solution using Collectors.toCollection(ArrayList::new) is a little too noisy for such a common operation.

Example:

ArrayList<Long> result = sourceLongList.stream()
    .filter(l -> l > 100)
    .collect(toArrayList());

public static <T> Collector<T, ?, ArrayList<T>> toArrayList() {
    return Collectors.toCollection(ArrayList::new);
}

With this answer I also want to demonstrate how simple it is to create and use custom collectors, which is very useful generally.

Upvotes: 13

msayag

Reputation: 8675

A little more efficient way (avoid the creating the source List and the auto-unboxing by the filter):

List<Long> targetLongList = LongStream.of(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L)
    .filter(l -> l > 100)
    .boxed()
    .collect(Collectors.toList());

Upvotes: 5

Debapriya Biswas

Reputation: 1349

If you don't use parallel() this will work

List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);

List<Long> targetLongList =  new ArrayList<Long>();

sourceLongList.stream().peek(i->targetLongList.add(i)).collect(Collectors.toList());

Upvotes: -3

Nikhil Nanivadekar

Reputation: 1152

If you have an array of primitives, you can use the primitive collections available in Eclipse Collections.

LongList sourceLongList = LongLists.mutable.of(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
LongList targetLongList = sourceLongList.select(l -> l > 100);

If you can't change the sourceLongList from List:

List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
List<Long> targetLongList = 
    ListAdapter.adapt(sourceLongList).select(l -> l > 100, new ArrayList<>());

If you want to use LongStream:

long[] sourceLongs = new long[]{1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L};
LongList targetList = 
    LongStream.of(sourceLongs)
    .filter(l -> l > 100)
    .collect(LongArrayList::new, LongArrayList::add, LongArrayList::addAll);

Note: I am a contributor to Eclipse Collections.

Upvotes: 5

John McClean

Reputation: 5313

If you don't mind using 3rd party libraries, AOL's cyclops-react lib (disclosure I am a contributor) has extensions for all JDK Collection types, including List. The ListX interface extends java.util.List and adds a large number of useful operators, including filter.

You can simply write-

ListX<Long> sourceLongList = ListX.of(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
ListX<Long> targetLongList = sourceLongList.filter(l -> l > 100);

ListX also can be created from an existing List (via ListX.fromIterable)

Upvotes: 1

Kashyap

Reputation: 17544

In case someone (like me) out there is looking for ways deal with Objects instead of primitive types then use mapToObj()

String ss = "An alternative way is to insert the following VM option before "
        + "the -vmargs option in the Eclipse shortcut properties(edit the "
        + "field Target inside the Shortcut tab):";

List<Character> ll = ss
                        .chars()
                        .mapToObj(c -> new Character((char) c))
                        .collect(Collectors.toList());

System.out.println("List type: " + ll.getClass());
System.out.println("Elem type: " + ll.get(0).getClass());
ll.stream().limit(50).forEach(System.out::print);

prints:

List type: class java.util.ArrayList
Elem type: class java.lang.Character
An alternative way is to insert the following VM o

Upvotes: 0

Retrieving a List from a java.util.stream.Stream in Java 8

Answers (15)

Stream#toList

Related Questions

`Stream#toList`