如何将集合转换为按嵌套集合属性的元素分组的Guava Multimap?

问题描述：

我有一个List<Foo>，想要番石榴

I have a List<Foo> and want a Guava Multimap<String, Foo> where we've grouped the Foos by each tag of their Collection<String> getTags() function.

我正在使用Java 8，因此可以很好地鼓励使用lambda和方法引用.

I am using Java 8, so lambdas and method references are fine/encouraged.

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

答

您可以为此使用自定义收集器:

You can use custom collector for this:

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

This does not cause extra side effects (see here on this), is concurrent and more idiomatic.

您还可以将这些临时lambda提取到成熟的收集器中，如下所示:

You can also extract these ad-hoc lambdas into a full-fledged collector, something like this:

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:

Then the collection would look like this:

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

如果顺序对您而言不重要，则还可以从characteristics()方法返回EnumSet.of(Characteristics.UNORDERED).这可以使内部收集机制更有效地发挥作用，尤其是在并行缩减的情况下.

You can also return EnumSet.of(Characteristics.UNORDERED) from characteristics() method if the order is not important for you. This can make internal collection machinery act more efficiently, especially in case of parallel reduction.

如何将集合转换为按嵌套集合属性的元素分组的Guava Multimap?

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