Java:如何从List <T>转换为Map <f1(T),List(f2(T))>而无需迭代

Question

我有一个对象列表,我需要转换为一个映射,其中键是每个元素的函数,值是每个元素的另一个函数的列表.实际上,这是通过它们的功能对元素进行分组.

例如,假设一个简单的元素类:

class Element {
    int f1() { ... }
    String f2() { ... }
}

以及这些的清单:

[
    { f1=100, f2="Alice" },
    { f1=200, f2="Bob" },
    { f1=100, f2="Charles" },
    { f1=300, f2="Dave" }
]

那我想要一张地图如下:

{
    {key=100, value=[ "Alice", "Charles" ]},
    {key=200, value=[ "Bob" ]},
    {key=300, value=[ "Dave" ]}
}

任何人都可以建议一种简洁的方式在Java中执行此操作而不进行迭代吗？LambdaJ的group方法与Guava的组合Maps.transform几乎可以实现,但group不生成地图.

Answer 1

Guava有Maps.uniqueIndex(Iterable values,Function keyFunction)和Multimaps.index(Iterable values,Function keyFunction),但它们不会转换值.还有一些 请求添加你想要什么实用的方法,但现在,你必须使用Multimaps.index()和Multimaps.transformValues()将自己滚吧:

static class Person {
    private final Integer age;
    private final String name;

    public Person(Integer age, String name) {
        this.age = age;
        this.name = name;
    }

    public Integer getAge() {
        return age;
    }

    public String getName() {
        return name;
    }
}

private enum GetAgeFunction implements Function<Person, Integer> {
    INSTANCE;

    @Override
    public Integer apply(Person person) {
        return person.getAge();
    }
}

private enum GetNameFunction implements Function<Person, String> {
    INSTANCE;

    @Override
    public String apply(Person person) {
        return person.getName();
    }
}

public void example() {
    List<Person> persons = ImmutableList.of(
            new Person(100, "Alice"),
            new Person(200, "Bob"),
            new Person(100, "Charles"),
            new Person(300, "Dave")
    );

    ListMultimap<Integer, String> ageToNames = getAgeToNamesMultimap(persons);

    System.out.println(ageToNames);

    // prints {100=[Alice, Charles], 200=[Bob], 300=[Dave]}
}

private ListMultimap<Integer, String> getAgeToNamesMultimap(List<Person> persons) {
    ImmutableListMultimap<Integer, Person> ageToPersons = Multimaps.index(persons, GetAgeFunction.INSTANCE);
    ListMultimap<Integer, String> ageToNames = Multimaps.transformValues(ageToPersons, GetNameFunction.INSTANCE);

    // Multimaps.transformValues() returns a *lazily* transformed view of "ageToPersons"
    // If we want to iterate multiple times over it, it's better to create a copy
    return ImmutableListMultimap.copyOf(ageToNames);
}

可重用的实用方法可以是:

public static <E, K, V> ImmutableListMultimap<K, V> keyToValuesMultimap(Iterable<E> elements, Function<E, K> keyFunction, Function<E, V> valueFunction) {
    ImmutableListMultimap<K, E> keysToElements = Multimaps.index(elements, keyFunction);
    ListMultimap<K, V> keysToValuesLazy = Multimaps.transformValues(keysToElements, valueFunction);
    return ImmutableListMultimap.copyOf(keysToValuesLazy);
}

我想我们可以通过使用Function<? extends E, K>或者其他东西来改进签名中的泛型,但我没有时间进一步钻研......

Answer 2

现在使用Java8,您可以这样做:

static class Element {
    final int f1;
    final String f2;

    Element(int f1, String f2) {
        this.f1 = f1;
        this.f2 = f2;
    }

    int f1() { return f1;}
    String f2() { return f2; }
}

public static void main(String[] args) {
    List<Element> elements = new ArrayList<>();
    elements.add(new Element(100, "Alice"));
    elements.add(new Element(200, "Bob"));
    elements.add(new Element(100, "Charles"));
    elements.add(new Element(300, "Dave"));

    elements.stream()
            .collect(Collectors.groupingBy(
                    Element::f1,
                    Collectors.mapping(Element::f2, Collectors.toList())
                    ))
            .forEach((f1, f2) -> System.out.println("{"+f1.toString() + ", value="+f2+"}"));
}