通过项本身的属性索引集合的最简洁方法,该项本身就是一个集合

Question

我有一个List<Foo>并且想要一个番石榴Multimap<String, Foo>,我们Foo按照它们的Collection<String> getTags()功能的每个标签对它们进行分组.

我正在使用java 8,所以lambdas和方法引用很好/鼓励.

例如,如果我有:

foo1, tags=a,b,c
foo2, tags=c,d
foo3, tags=a,c,e

我会得到一个Multimap<String, Foo>:

a -> foo1, foo3
b -> foo1
c -> foo1, foo2, foo3
d -> foo2
e -> foo3

Answer 1

您可以使用自定义收集器:

Multimap<String, Foo> map = list.stream().collect(
    ImmutableMultimap::builder,
    (builder, value) -> value.getTags().forEach(tag -> builder.put(tag, value)),
    (builder1, builder2) -> builder1.putAll(builder2.build())
).build();

这不会引起额外的副作用(参见此处),并发且更具惯用性.

您还可以将这些特殊的lambda提取到一个成熟的收集器中,如下所示:

public static <T, K> Collector<T, ?, Multimap<K, T>> toMultimapByKey(Function<? super T, ? extends Iterable<? extends K>> keysMapper) {
    return new MultimapCollector<>(keysMapper);
}

private static class MultimapCollector<T, K> implements Collector<T, ImmutableMultimap.Builder<K, T>, Multimap<K, T>> {
    private final Function<? super T, ? extends Iterable<? extends K>> keysMapper;

    private MultimapCollector(Function<? super T, ? extends Iterable<? extends K>> keysMapper) {
        this.keysMapper = keysMapper;
    }

    @Override
    public Supplier<ImmutableMultimap.Builder<K, T>> supplier() {
        return ImmutableMultimap::builder;
    }

    @Override
    public BiConsumer<ImmutableMultimap.Builder<K, T>, T> accumulator() {
        return (builder, value) -> keysMapper.apply(value).forEach(k -> builder.put(k, value));
    }

    @Override
    public BinaryOperator<ImmutableMultimap.Builder<K, T>> combiner() {
        return (b1, b2) -> b1.putAll(b2.build());
    }

    @Override
    public Function<ImmutableMultimap.Builder<K, T>, Multimap<K, T>> finisher() {
        return ImmutableMultimap.Builder<K, T>::build;
    }

    @Override
    public Set<Characteristics> characteristics() {
        return Collections.emptySet();
    }
}

然后集合看起来像这样:

Multimap<String, Foo> map = list.stream().collect(toMultimapByKey(Foo::getTags));

如果订单对您不重要,您也可以EnumSet.of(Characteristics.UNORDERED)从characteristics()方法返回.这可以使内部收集机械更有效地发挥作用,特别是在并行减少的情况下.

Answer 2

ImmutableMultimap.Builder<String, Foo> builder = ImmutableMultimap.builder();
list.forEach(foo -> foo.getTags().forEach(tag -> builder.put(tag, foo));
return builder.build();