Create Java 8 Stream from ArrayNode

Question

Is it possible to create stream from com.fasterxml.jackson.databind.node.ArrayNode?
I tried:

ArrayNode files = (ArrayNode) json.get("files");
Stream<JsonNode> stream = Stream.of(files);

But it will actually give stream of one element, the initial ArrayNode object.
Correct result should be Stream<JsonNode>, can I achieve it?

Will the entire array here be loaded in memory? I believe the answer is "yes". So what is the point to have a stream here? — Andrii Karaivanskyi, Sep 20 '16 at 07:43
@AndreyKarayvansky it's not about performance but about ability to use `Java 8 Stream API` methods to process `ArrayNode` collection in a "functional way" (using methods like `map`, `filter`, `collect`...). — icl7126, Sep 20 '16 at 11:12
Lists.newArrayList(arrayNode.elements()).stream() //by Guava — Anderson, Aug 11 '21 at 08:11

score 46 · Accepted Answer · edited Mar 24 '22 at 14:23

46

ArrayNode implements Iterable. Iterable has a spliterator() method. You can create a sequential Stream from a Spliterator using

ArrayNode arrayNode = (ArrayNode) json.get("xyz");
StreamSupport.stream(arrayNode.spliterator(), false)

edited Mar 24 '22 at 14:23

Robert

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answered Sep 20 '15 at 20:12

JB Nizet

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I tried your solution and it works but I don't understand why Stream.of(files) didn't while spliterator did. – progonkpa Apr 18 '19 at 20:46
2

Because Stream.of() expects an array as argument, and an ArrayNode is not an array. – JB Nizet Apr 18 '19 at 21:44

score 18 · Answer 2 · answered Sep 21 '15 at 02:22

18

An ArrayNode class provides random access: you can get size() and an element by index (using get(index)). This is all you need to create a good stream:

Stream<JsonNode> nodes = IntStream.range(0, files.size()).mapToObj(files::get);

Note that this solution is better than using default spliterator (as suggested by other answerers) as it can split well and reports the size properly. Even if you don't care about parallel processing, some operations like toArray() will work more effectively as knowing the size in advance will help to allocate an array of proper size.

answered Sep 21 '15 at 02:22

Tagir Valeev

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I agree with all you said and this is in many ways better solution. But I would expect to see such solution directly in `ArrayNode` class or as some `Utils` method. Streams should make code easier to read and understand. Using this solution for simple filter-map-collect operation would slightly increase complexity of code. Anyway, very nice thinking, you got my +1. – icl7126 Sep 21 '15 at 17:36
@klerik, It's not so easy to provide stream API methods while maintaining pre-Java-8 compatibility. – Tagir Valeev Sep 22 '15 at 00:59
I like this as it means I don't have to cast my `JsonNode` to an `ArrayNode` either. – Bernie May 23 '18 at 22:37

John McClean · Answer 3 · 2015-09-20T20:24:55.953

ArrayNode#elements returns an Iterator over it's elements you can use that to create a Stream (by leveraging StreamSupport). StreamSupport requires a Spliterator and to create a Spliterator from an Iterator you can use the Spliterators class.

  ArrayNode files = (ArrayNode) json.get("files");
  Stream<JsonNode>  elementStream = StreamSupport.stream(Spliterators
                  .spliteratorUnknownSize(files.elements(),
                        Spliterator.ORDERED),false);

cyclops-streams has a StreamUtils class has a static method that makes this a bit cleaner (I am the author).

 ArrayNode files = (ArrayNode) json.get("files");
 Stream<JsonNode>  elementStream = StreamUtils.stream(files.elements());

Taking into account @JB Nizet's answer that ArrayNode is an iterable with StreamUtils you can pass in the ArrayNode and get the Stream back directly.

Stream<JsonNode>  elementStream = StreamUtils.stream((ArrayNode) json.get("files"));

Create Java 8 Stream from ArrayNode

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