How can I turn a List of Lists into a List in Java 8?

Question

If I have a List<List<Object>>, how can I turn that into a List<Object> that contains all the objects in the same iteration order by using the features of Java 8?

Answer 1

Below code should work:-

    List<Object> result = new ArrayList<>();
    listOfLists.forEach(result::addAll);

Answer 2

Since java-16, you can use Stream#mapMulti

List<Object> result = listOfLists.stream()
                                 .mapMulti((List<Object> list, Consumer<Object> consumer) -> {
                                     list.forEach(consumer::accept);
                                 })
                                 .collect(Collectors.toList());

If you need an immutable List you can even use toList() as terminal operation

List<Object> result = listOfLists.stream()
                                 .mapMulti((List<Object> list, Consumer<Object> consumer) -> {
                                     list.forEach(consumer::accept);
                                 })
                                 .toList();

Answer 3

List<List> list = map.values().stream().collect(Collectors.toList());

    List<Employee> employees2 = new ArrayList<>();
    
     list.stream().forEach(
             
             n-> employees2.addAll(n));

Answer 4

Method to convert a List<List> to List :

listOfLists.stream().flatMap(List::stream).collect(Collectors.toList());

See this example:

public class Example {

    public static void main(String[] args) {
        List<List<String>> listOfLists = Collections.singletonList(Arrays.asList("a", "b", "v"));
        List<String> list = listOfLists.stream().flatMap(List::stream).collect(Collectors.toList());

        System.out.println("listOfLists => " + listOfLists);
        System.out.println("list => " + list);
    }

}       

It prints:

listOfLists => [[a, b, c]]
list => [a, b, c]

In Python this can be done using List Comprehension.

list_of_lists = [['Roopa','Roopi','Tabu', 'Soudipta'],[180.0, 1231, 2112, 3112], [130], [158.2], [220.2]]

flatten = [val for sublist in list_of_lists for val in sublist]

print(flatten)

['Roopa', 'Roopi', 'Tabu', 'Soudipta', 180.0, 1231, 2112, 3112, 130, 158.2, 220.2]

Answer 5

An expansion on Eran's answer that was the top answer, if you have a bunch of layers of lists, you can keep flatmapping them.

This also comes with a handy way of filtering as you go down the layers if needed as well.

So for example:

List<List<List<List<List<List<Object>>>>>> multiLayeredList = ...

List<Object> objectList = multiLayeredList
    .stream()
    .flatmap(someList1 -> someList1
        .stream()
        .filter(...Optional...))
    .flatmap(someList2 -> someList2
        .stream()
        .filter(...Optional...))
    .flatmap(someList3 -> someList3
        .stream()
        .filter(...Optional...))
    ...
    .collect(Collectors.toList())

This is would be similar in SQL to having SELECT statements within SELECT statements.

Answer 6

We can use flatmap for this, please refer below code :

 List<Integer> i1= Arrays.asList(1, 2, 3, 4);
 List<Integer> i2= Arrays.asList(5, 6, 7, 8);

 List<List<Integer>> ii= Arrays.asList(i1, i2);
 System.out.println("List<List<Integer>>"+ii);
 List<Integer> flat=ii.stream().flatMap(l-> l.stream()).collect(Collectors.toList());
 System.out.println("Flattened to List<Integer>"+flat);

Answer 7

I just want to explain one more scenario like List<Documents>, this list contains a few more lists of other documents like List<Excel>, List<Word>, List<PowerPoint>. So the structure is

class A {
  List<Documents> documentList;
}

class Documents {
  List<Excel> excels;
  List<Word> words;
  List<PowerPoint> ppt;
}

Now if you want to iterate Excel only from documents then do something like below..

So the code would be

 List<Documents> documentList = new A().getDocumentList();

 //check documentList as not null

 Optional<Excel> excelOptional = documentList.stream()
                         .map(doc -> doc.getExcel())
                         .flatMap(List::stream).findFirst();
 if(excelOptional.isPresent()){
   Excel exl = optionalExcel.get();
   // now get the value what you want.
 }

I hope this can solve someone's issue while coding...

Answer 8

flatmap is better but there are other ways to achieve the same

List<List<Object>> listOfList = ... // fill

List<Object> collect = 
      listOfList.stream()
                .collect(ArrayList::new, List::addAll, List::addAll);

Answer 9

Just as @Saravana mentioned:

flatmap is better but there are other ways to achieve the same

 listStream.reduce(new ArrayList<>(), (l1, l2) -> {
        l1.addAll(l2);
        return l1;
 });

To sum up, there are several ways to achieve the same as follows:

private <T> List<T> mergeOne(Stream<List<T>> listStream) {
    return listStream.flatMap(List::stream).collect(toList());
}

private <T> List<T> mergeTwo(Stream<List<T>> listStream) {
    List<T> result = new ArrayList<>();
    listStream.forEach(result::addAll);
    return result;
}

private <T> List<T> mergeThree(Stream<List<T>> listStream) {
    return listStream.reduce(new ArrayList<>(), (l1, l2) -> {
        l1.addAll(l2);
        return l1;
    });
}

private <T> List<T> mergeFour(Stream<List<T>> listStream) {
    return listStream.reduce((l1, l2) -> {
        List<T> l = new ArrayList<>(l1);
        l.addAll(l2);
        return l;
    }).orElse(new ArrayList<>());
}

private <T> List<T> mergeFive(Stream<List<T>> listStream) {
    return listStream.collect(ArrayList::new, List::addAll, List::addAll);
}

Answer 10

The flatMap method on Stream can certainly flatten those lists for you, but it must create Stream objects for element, then a Stream for the result.

You don't need all those Stream objects. Here is the simple, concise code to perform the task.

// listOfLists is a List<List<Object>>.
List<Object> result = new ArrayList<>();
listOfLists.forEach(result::addAll);

Because a List is Iterable, this code calls the forEach method (Java 8 feature), which is inherited from Iterable.

Performs the given action for each element of the Iterable until all elements have been processed or the action throws an exception. Actions are performed in the order of iteration, if that order is specified.

And a List's Iterator returns items in sequential order.

For the Consumer, this code passes in a method reference (Java 8 feature) to the pre-Java 8 method List.addAll to add the inner list elements sequentially.

Appends all of the elements in the specified collection to the end of this list, in the order that they are returned by the specified collection's iterator (optional operation).

Answer 11

You can use flatMap to flatten the internal lists (after converting them to Streams) into a single Stream, and then collect the result into a list:

List<List<Object>> list = ...
List<Object> flat = 
    list.stream()
        .flatMap(List::stream)
        .collect(Collectors.toList());

Answer 12

You can use the flatCollect() pattern from Eclipse Collections.

MutableList<List<Object>> list = Lists.mutable.empty();
MutableList<Object> flat = list.flatCollect(each -> each);

If you can't change list from List:

List<List<Object>> list = new ArrayList<>();
List<Object> flat = ListAdapter.adapt(list).flatCollect(each -> each);

Note: I am a contributor to Eclipse Collections.