如何将集合转换为按嵌套集合属性的元素分组的Guava Multimap？

您可以为此使用定制收集器：

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（请参见此处），并且是并发的并且更加惯用。

您还可以将这些临时lambda提取到成熟的收集器中，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();    }}

然后该集合将如下所示：

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

如果顺序对您不重要，则也可以

EnumSet.of(Characteristics.UNORDERED)

characteristics()