How to convert a Java 8 Stream to an Array?

Question

What is the easiest/shortest way to convert a Java 8 Stream into an array?

Semaphore

I'd suggest you to revert the rollback as the question was more complete and showed you had tried something. – Kutch

@Kutch Thanks! but i thought the attempted code does not really add more information to the question, and nobody has screamed "show us your attempt" yet =) – Semaphore

@skiwi: Although I usually shout at the do-my-homework-instead-of-me questions, this particular question seems to be clearer to me without any additional mess. Let's keep it tidy. – Lavonlavona

You can find a lot of answers and guidance in the official docs of the package: docs.oracle.com/javase/8/docs/api/java/util/stream/… – Aribold

Answer 1

The easiest method is to use the toArray(IntFunction<A[]> generator) method with an array constructor reference. This is suggested in the API documentation for the method.

String[] stringArray = stringStream.toArray(String[]::new);

What it does is find a method that takes in an integer (the size) as argument, and returns a String[], which is exactly what (one of the overloads of) new String[] does.

You could also write your own IntFunction:

Stream<String> stringStream = ...;
String[] stringArray = stringStream.toArray(size -> new String[size]);

The purpose of the IntFunction<A[]> generator is to convert an integer, the size of the array, to a new array.

Example code:

Stream<String> stringStream = Stream.of("a", "b", "c");
String[] stringArray = stringStream.toArray(size -> new String[size]);
Arrays.stream(stringArray).forEach(System.out::println);

Prints:

a
b
c

Kutch

and here is an explanation why and how the Array constructor reference actually work: #29448061 – Tildy

"Zenexer is right, the solution should be: stream.toArray(String[]::new);" ... Well ok, but one should understand that the method reference is logically and functionally equivalent to toArray(sz -> new String[sz]) so I'm not sure that one can really say what the solution should or must be. – Bullpup

The array constructor reference is more concise. The documentation for the toArray() method cites its concision and includes its use in the only example. When it was tacked on the end of the answer, I missed it on first reading. Moved it to the top, above the details, so that no one else will miss it. – Marzi

@Bullpup sz -> new String[sz] makes a new function where as the constructor reference does not. It depends how much you value Garbage Collection Churn I guess. – Ose

@Ose It does not. It (statically!) makes a new, private method, which cannot cause churn, and both versions need to create a new object. A reference creates an object that points directly at the target method; a lambda creates an object that points at the generated private one. A reference to a constructor should still perform better for lack of indirection and easier VM optimization, but churning has nothing to do with it. – Sebi

@Sebi you are correct, my apologise. It was infact my attempt to debug that was causing the churn that was causing the churn the first time I tried to do this, so I have had it stuck in my head that this is how it was. (Hate it when that happens). – Ose

"String[] stringArray = stringStream.toArray(size -> new String[size]);", where is the size come from? – Teal

Lambda should simplify things but java lambda like stream.toArray(size -> new String[size]); just looks akward and this is so unintuitive every developer will forget this syntax the next time he needs an array. Why do they make things so complicated where c# is making life pretty easy with lambda. My stack is java but since years java getting more and more behind of c#. – Saponify

Instead of stream.toArray(size -> new String[size]), you can shorten it using a method reference: stream.toArray(String[]::new), which accomplishes the same thing. Syntactic sugar, if you have a sweet tooth. – Airship

Answer 2

If you want to get an array of ints, with values from 1 to 10, from a Stream<Integer>, there is IntStream at your disposal.

Here we create a Stream with a Stream.of method and convert a Stream<Integer> to an IntStream using a mapToInt. Then we can call IntStream's toArray method.

Stream<Integer> stream = Stream.of(1,2,3,4,5,6,7,8,9,10);
//or use this to create our stream 
//Stream<Integer> stream = IntStream.rangeClosed(1, 10).boxed();
int[] array =  stream.mapToInt(x -> x).toArray();

Here is the same thing, without the Stream<Integer>, using only the IntStream:

int[]array2 =  IntStream.rangeClosed(1, 10).toArray();

Scamper

Answer 3

You can convert a java 8 stream to an array using this simple code block:

 String[] myNewArray3 = myNewStream.toArray(String[]::new);

But let's explain things more, first, let's Create a list of string filled with three values:

String[] stringList = {"Bachiri","Taoufiq","Abderrahman"};

Create a stream from the given Array :

Stream<String> stringStream = Arrays.stream(stringList);

we can now perform some operations on this stream Ex:

Stream<String> myNewStream = stringStream.map(s -> s.toUpperCase());

and finally convert it to a java 8 Array using these methods:

1-Classic method (Functional interface)

IntFunction<String[]> intFunction = new IntFunction<String[]>() {
    @Override
    public String[] apply(int value) {
        return new String[value];
    }
};


String[] myNewArray = myNewStream.toArray(intFunction);

2 -Lambda expression

 String[] myNewArray2 = myNewStream.toArray(value -> new String[value]);

3- Method reference

String[] myNewArray3 = myNewStream.toArray(String[]::new);

Method reference Explanation:

It's another way of writing a lambda expression that it's strictly equivalent to the other.

Grevera

Answer 4

Convert text to String array where separating each value by comma, and trim every field, for example:

String[] stringArray = Arrays.stream(line.split(","))          
                             .map(String::trim)
                             .toArray(String[]::new);

Shipman

Answer 5

import java.util.List;
import java.util.stream.Stream;

class Main {

    public static void main(String[] args) {
        // Create a stream of strings from list of strings
        Stream<String> myStreamOfStrings = List.of("lala", "foo", "bar").stream();

        // Convert stream to array by using toArray method
        String[] myArrayOfStrings = myStreamOfStrings.toArray(String[]::new);

        // Print results
        for (String string : myArrayOfStrings) {
            System.out.println(string);
        }
    }
}

Try it out online: https://repl.it/@SmaMa/Stream-to-array

Tallow

What is the difference between your answer and the accepted answer? – Pedersen

@LongNguyen It's a full example including an online replay scenario, not only a snippet. – Tallow

Answer 6

You can create a custom collector that convert a stream to array.

public static <T> Collector<T, ?, T[]> toArray( IntFunction<T[]> converter )
{
    return Collectors.collectingAndThen( 
                  Collectors.toList(), 
                  list ->list.toArray( converter.apply( list.size() ) ) );
}

and a quick use

List<String> input = Arrays.asList( ..... );

String[] result = input.stream().
         .collect( CustomCollectors.**toArray**( String[]::new ) );

Blanchblancha

Why would you use this instead of Stream.toArray(IntFunction)? – Nessy

I needed a collector to pass to the 2-arg Collectors.groupingBy so that I could map some attribute to arrays of objects per attribute value. This answer gives me exactly that. Also @DidierL. – Centroclinal

Since Java 11 the finisher in collectingAndThen can be written as list -> list.toArray(converter) due to addition of Collection.toArray​(IntFunction) – Sartor

Answer 7

Using the toArray(IntFunction<A[]> generator) method is indeed a very elegant and safe way to convert (or more correctly, collect) a Stream into an array of the same type of the Stream.

However, if the returned array's type is not important, simply using the toArray() method is both easier and shorter. For example:

    Stream<Object> args = Stream.of(BigDecimal.ONE, "Two", 3);
    System.out.printf("%s, %s, %s!", args.toArray());

Bizet

Answer 8

Stream<Integer> stream = Stream.of(1, 2, 3, 4, 5, 6);

int[] arr=   stream.mapToInt(x->x.intValue()).toArray();

Brigitta

Answer 9

// create a stream of string elements

Stream<String> stringStream = Stream.of("A", "B", "C", "D", "E");


// convert the stream to arrays of strings

String[] stringArray = stringStream.toArray(String[]::new);

Kermis

What was wrong with all the other answers saying the same? – Crupper

Answer 10

     Stream<Integer> stream = Stream.of(1, 2, 3, 4, 5, 6);

     Integer[] integers = stream.toArray(it->new Integer[it]);

Highspeed