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在嵌套数据结构上使用Java8流创建地图

[英]Creating a Map using Java8 streams on a nested Data Structure

I am searching for an elegant equivalent of this piece of code using Java 8's streams: 我正在使用Java 8的流来搜索这段代码的优雅用法:

Collection<X> xs = ...;
Map<B, A> map = new SomeMap<>();

for (X x : xs) {
    A a = x.getA();
    Collection<B> bs = x.getBs();

    for (B b : bs)
        map.put(b, a);
}

That one is a bit too tricky for me as I can't think of a combination using flatMap and Collectors.toMap which would implement the desired functionality. 这对我来说有点棘手,因为我想不出使用flatMap和Collectors.toMap的组合来实现所需的功能。

Compilable example: 编译示例:

import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.Map;

public class Application {

    public static class A {}

    public static class B {}

    public static class X {
        private A a;
        private Collection<B> bs;

        public X(A a, Collection<B> bs) {
            this.a = a;
            this.bs = bs;
        }

        public A getA() {
            return a;
        }

        public Collection<B> getBs() {
            return bs;
        }
    }

    public static void main(String[] args) {
        X x1 = new X(new A(), Arrays.asList(new B(), new B()));
        X x2 = new X(new A(), Arrays.asList(new B()));

        Collection<X> xs = Arrays.asList(x1, x2);

        Map<B, A> map = new HashMap<>();

        for (X x : xs) {
            A a = x.getA();
            Collection<B> bs = x.getBs();

            for (B b : bs)
                map.put(b, a);
        }
    }

}

As you thought you can do it with a mixture of flatMap and toMap: 如您所愿,可以将flatMap和toMap混合使用:

Map<B, A> map = xs.stream()
                  .flatMap(x -> x.getBs().stream()
                                 .map(b -> new SimpleEntry<> (b, x.getA())))
                  .collect(toMap(Entry::getKey, Entry::getValue));

Note that this code differs from your original if there are duplicate Bs: your code will keep overwriting the corresponding value whereas this code will throw an exception. 请注意,如果B重复,则此代码与原始代码不同:您的代码将继续覆盖相应的值,而此代码将引发异常。

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