Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
251 views
in Technique[技术] by (71.8m points)

java - Grouping elements of a list into sublists (maybe by using guava)

I want to group elements of a list. I'm currently doing it this way:

public static <E> List<List<E>> group(final List<E> list, final GroupFunction<E> groupFunction) {

    List<List<E>> result = Lists.newArrayList();

    for (final E element : list) {

        boolean groupFound = false;
        for (final List<E> group : result) {
            if (groupFunction.sameGroup(element, group.get(0))) {
                group.add(element);
                groupFound = true;
                break;
            }
        }
        if (! groupFound) {

            List<E> newGroup = Lists.newArrayList();
            newGroup.add(element);
            result.add(newGroup);
        }
    }

    return result;
}

public interface GroupFunction<E> {
    public boolean sameGroup(final E element1, final E element2);
}

Is there a better way to do this, preferably by using guava?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Sure it is possible, and even easier with Guava :) Use Multimaps.index(Iterable, Function):

ImmutableListMultimap<E, E> indexed = Multimaps.index(list, groupFunction);

If you give concrete use case it would be easier to show it in action.

Example from docs:

List<String> badGuys =
   Arrays.asList("Inky", "Blinky", "Pinky", "Pinky", "Clyde");
Function<String, Integer> stringLengthFunction = ...;
Multimap<Integer, String> index =
   Multimaps.index(badGuys, stringLengthFunction);
System.out.println(index);

prints

{4=[Inky], 6=[Blinky], 5=[Pinky, Pinky, Clyde]}

In your case if GroupFunction is defined as:

GroupFunction<String> groupFunction = new GroupFunction<String>() {
  @Override public String sameGroup(final String s1, final String s2) {
    return s1.length().equals(s2.length());
  }
}

then it would translate to:

Function<String, Integer> stringLengthFunction = new Function<String, Integer>() {
  @Override public Integer apply(final String s) {
    return s.length();
  }
}

which is possible stringLengthFunction implementation used in Guava's example.


Finally, in Java 8, whole snippet could be even simpler, as lambas and method references are concise enough to be inlined:

ImmutableListMultimap<E, E> indexed = Multimaps.index(list, String::length);

For pure Java 8 (no Guava) example using Collector.groupingBy see Jeffrey Bosboom's answer, although there are few differences in that approach:

  • it doesn't return ImmutableListMultimap but rather Map with Collection values,
  • There are no guarantees on the type, mutability, serializability, or thread-safety of the Map returned (source),

  • it's a bit more verbose than Guava + method reference.

EDIT: If you don't care about indexed keys you can fetch grouped values:

List<List<E>> grouped = Lists.transform(indexed.keySet().asList(), new Function<E, List<E>>() {
        @Override public List<E> apply(E key) {
            return indexed.get(key);
        }
});

// or the same view, but with Java 8 lambdas:
List<List<E>> grouped = Lists.transform(indexed.keySet().asList(), indexed::get);

what gives you Lists<List<E>> view which contents can be easily copied to ArrayList or just used as is, as you wanted in first place. Also note that indexed.get(key) is ImmutableList.

// bonus: similar as above, but not a view, instead collecting to list using streams:
List<List<E>> grouped = indexed.keySet().stream()
    .map(indexed::get)
    .collect(Collectors.toList());

EDIT 2: As Petr Gladkikh mentions in comment below, if Collection<List<E>> is enough, above example could be simpler:

Collection<List<E>> grouped = indexed.asMap().values();

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...