用Java8流API合并两个Map<String，Integer>

Question

我有两个（或更多）Map<String, Integer>对象。我希望将它们与Java8流API合并，使公共键的值应该是这些值中的最大值

@Test
public void test14() throws Exception {
    Map<String, Integer> m1 = ImmutableMap.of("a", 2, "b", 3);
    Map<String, Integer> m2 = ImmutableMap.of("a", 3, "c", 4);
    List<Map<String, Integer>> list = newArrayList(m1, m2);

    Map<String, Integer> mx = list.stream()... // TODO

    Map<String, Integer> expected = ImmutableMap.of("a", 3, "b", 3, "c", 4);
    assertEquals(expected, mx);
}

如何使此测试方法变为绿色

我已经玩了collect和Collectors一阵子都没有成功

（ImmutableMap和newArrayList来自谷歌番石榴。）

Answer 1

@Test
public void test14() throws Exception {
    Map<String, Integer> m1 = ImmutableMap.of("a", 2, "b", 3);
    Map<String, Integer> m2 = ImmutableMap.of("a", 3, "c", 4);

    Map<String, Integer> mx = Stream.of(m1, m2)
        .map(Map::entrySet)          // converts each map into an entry set
        .flatMap(Collection::stream) // converts each set into an entry stream, then
                                     // "concatenates" it in place of the original set
        .collect(
            Collectors.toMap(        // collects into a map
                Map.Entry::getKey,   // where each entry is based
                Map.Entry::getValue, // on the entries in the stream
                Integer::max         // such that if a value already exist for
                                     // a given key, the max of the old
                                     // and new value is taken
            )
        )
    ;

    /* Use the following if you want to create the map with parallel streams
    Map<String, Integer> mx = Stream.of(m1, m2)
        .parallel()
        .map(Map::entrySet)          // converts each map into an entry set
        .flatMap(Collection::stream) // converts each set into an entry stream, then
                                     // "concatenates" it in place of the original set
        .collect(
            Collectors.toConcurrentMap(        // collects into a map
                Map.Entry::getKey,   // where each entry is based
                Map.Entry::getValue, // on the entries in the stream
                Integer::max         // such that if a value already exist for
                                     // a given key, the max of the old
                                     // and new value is taken
            )
        )
    ;
    */

    Map<String, Integer> expected = ImmutableMap.of("a", 3, "b", 3, "c", 4);
    assertEquals(expected, mx);
}

Answer 2

我在proton pack library中添加了我的贡献，其中包含流API的实用方法。以下是你如何实现自己的目标：

Map<String, Integer> mx = MapStream.ofMaps(m1, m2).mergeKeys(Integer::max).collect();

基本上mergeKeys将在一个新映射中收集键值对（提供一个可选的合并函数，否则最终将得到一个Map<String, List<Integer>>），并在entrySet()上调用stream()以获得一个新的MapStream。然后使用collect()获得结果图

Answer 3

我已经为可能感兴趣的任何人创建了@srborlongan所做工作的视觉表现

Answer 4

mx = list.stream().collect(HashMap::new,
        (a, b) -> b.forEach((k, v) -> a.merge(k, v, Integer::max)),
        Map::putAll);

这涵盖了任何大小列表的一般情况，应该适用于任何类型，只需根据需要交换Integer::max和/或HashMap::new

如果您不在乎合并中会出现哪个值，那么有一个更简洁的解决方案：

mx = list.stream().collect(HashMap::new, Map::putAll, Map::putAll);

作为一般方法：

public static <K, V> Map<K, V> mergeMaps(Stream<? extends Map<K, V>> stream) {
    return stream.collect(HashMap::new, Map::putAll, Map::putAll);
}

public static <K, V, M extends Map<K, V>> M mergeMaps(Stream<? extends Map<K, V>> stream,
        BinaryOperator<V> mergeFunction, Supplier<M> mapSupplier) {
    return stream.collect(mapSupplier,
            (a, b) -> b.forEach((k, v) -> a.merge(k, v, mergeFunction)),
            Map::putAll);
}

Answer 5

Map<String, Integer> mx = new HashMap<>(m1);
m2.forEach((k, v) -> mx.merge(k, v, Integer::max));

Answer 6

使用StreamEx可以执行以下操作：

StreamEx.of(m1, m2)
    .flatMapToEntry(x -> x)
    .grouping(IntCollector.max())