MarcG MarcG - 9 days ago 7
Java Question

How to convert List<V> into Map<K, List<V>>, with Java 8 streams and custom List and Map suppliers?

It's easy to convert

List<V>
into
Map<K, List<V>>
. For example:

public Map<Integer, List<String>> getMap(List<String> strings) {
return
strings.stream()
.collect(Collectors.groupingBy(String::length));
}


But I want to do it with my own
List
and
Map
suppliers.

I have come up with this:

public Map<Integer, List<String>> getMap(List<String> strings) {
return strings.stream()
.collect(Collectors.toMap(
String::length,
item -> {List<String> list = new ArrayList<>(); list.add(item); return list;},
(list1, list2) -> {list1.addAll(list2); return list1;},
HashMap::new));
}


Question: Is there an easier, less verbose, or more efficient way of doing it? For example, something like this (which doesn't work):

return strings.stream()
.collect(Collectors.toMap(
String::length,
ArrayList::new,
HashMap::new));


And what if I only need to define the
List
supplier, but not the
Map
supplier?

Answer

You could have the following:

public Map<Integer, List<String>> getMap(List<String> strings) {
    return strings.stream().collect(
      Collectors.groupingBy(String::length, HashMap::new, Collectors.toCollection(ArrayList::new))
    );
}

The collector groupingBy(classifier, mapFactory, downstream) can be used to specify which type of map is wanted, by passing it a supplier of the wanted map for the mapFactory. Then, the downstream collector, which is used to collect elements grouped to the same key, is toCollection(collectionFactory), which enables to collect into a collection obtained from the given supplier.

This makes sure that the map returned is a HashMap and that the lists, in each value, are ArrayList. Note that if you want to return specific implementations of map and collection, then you most likely want the method to return those specific types as well, so you can use their properties.

If you only want to specify a collection supplier, and keep groupingBy default map, you can just omit the supplier in the code above and use the two arguments overload:

public Map<Integer, List<String>> getMap(List<String> strings) {
    return strings.stream().collect(
      Collectors.groupingBy(String::length, Collectors.toCollection(ArrayList::new))
    );
}

As a side-note, you could have a generic method for that:

public <K, V, C extends Collection<V>, M extends Map<K, C>> M getMap(List<V> list,
        Function<? super V, ? extends K> classifier, Supplier<M> mapSupplier, Supplier<C> collectionSupplier) {
    return list.stream().collect(
        Collectors.groupingBy(classifier, mapSupplier, Collectors.toCollection(collectionSupplier))
    );
}

The advantage with this declaration is that you can now use it to have specific HashMap of ArrayLists as result, or LinkedHashMap of LinkedListss, if the caller wishes it:

HashMap<Integer, ArrayList<String>> m = getMap(Arrays.asList("foo", "bar", "toto"),
        String::length, HashMap::new, ArrayList::new);
LinkedHashMap<Integer, LinkedList<String>> m2 = getMap(Arrays.asList("foo", "bar", "toto"),
        String::length, LinkedHashMap::new, LinkedList::new);

but, at that point, it may be simpler to directly use the groupingBy in the code...

Comments