Sorting ArrayList with Lambda in Java 8

Question

Could somebody show me a quick example how to sort an ArrayList alphabetically in Java 8 using the new lambda syntax.

Answer 1

If you have an array with elements that have natural ordering (i.e String, int, double); then it can be achieved by:

List<String> myList = new ArrayList<>();
myList.add("A");
myList.add("D");
myList.add("C");
myList.add("B");
myList.sort(Comparator.comparing(s -> s));
myList.forEach(System.out::println);

If on the another hand you have an array of objects and you want to sort base on some sort of object field, then you can use:

class User {
    double score;
    // Constructor // Getters // Setters
}

List<User> users = new ArrayList<>();
users.add(new User(19d));
users.add(new User(67d));
users.add(new User(50d));
users.add(new User(91d));

List<User> sortedUsers = users
        .stream()
        .sorted(Comparator.comparing(User::getScore))
        .collect(Collectors.toList());

sortedUsers.forEach(System.out::println);

If the sorting is more complex, then you would have to write your own comparator and pass that in.

Answer 2

A really generic solution would be to introduce some StreamUtil like

public class StreamUtil {

    private StreamUtil() {
    }       

    @SuppressWarnings({ "rawtypes", "unchecked" })
    public static <TYPE> Comparator<TYPE> sort(Function<TYPE, ? extends Comparable> getterFunction, boolean descending) {
        if (descending) {
            return (o1, o2) -> getterFunction.apply(o2).compareTo(getterFunction.apply(o1));
        }
        return (o1, o2) -> getterFunction.apply(o1).compareTo(getterFunction.apply(o2));
    }

}

The call would look something like

list.stream().sorted(sort(YourClass::getSortProperty, true));

Answer 3

Most concise:

Collections.sort(stringList, String::compareToIgnoreCase);

Answer 4

Lambdas shouldn't be the goal. In your case, you can sort it the same way as in Java 1.2:

Collections.sort(list); // case sensitive
Collections.sort(list, String.CASE_INSENSITIVE_ORDER); // case insensitive

If you want to do it in Java 8 way:

list.sort(Comparator.naturalOrder()); // case sensitive
list.sort(String.CASE_INSENSITIVE_ORDER); // case insensitive

You can also use list.sort(null) but I don't recommend this because it's not type-safe.

Answer 5

In functional programming, you're not using the old objects to operate on them, but creating the new one in such a fashion:

list.stream().sorted().map(blah-blah).filter(...)...