我有两个HashMap对象,定义如下:

HashMap<String, Integer> map1 = new HashMap<String, Integer>();
HashMap<String, Integer> map2 = new HashMap<String, Integer>();

我还有第三个HashMap对象:

HashMap<String, Integer> map3;

如何将map1和map2合并为map3?


当前回答

您可以使用putAll函数用于Map,如下面的代码所示

HashMap<String, Integer> map1 = new HashMap<String, Integer>();
map1.put("a", 1);
map1.put("b", 2);
map1.put("c", 3);
HashMap<String, Integer> map2 = new HashMap<String, Integer>();
map1.put("aa", 11);
map1.put("bb", 12);
HashMap<String, Integer> map3 = new HashMap<String, Integer>();
map3.putAll(map1);
map3.putAll(map2);
map3.keySet().stream().forEach(System.out::println);
map3.values().stream().forEach(System.out::println);

其他回答

使用Java 8 Stream API的一行程序:

map3 = Stream.of(map1, map2).flatMap(m -> m.entrySet().stream())
       .collect(Collectors.toMap(Entry::getKey, Entry::getValue))

该方法的好处之一是能够传递一个merge函数,该函数将处理具有相同键的值,例如:

map3 = Stream.of(map1, map2).flatMap(m -> m.entrySet().stream())
       .collect(Collectors.toMap(Entry::getKey, Entry::getValue, Math::max))

你可以对其他类型使用Collection.addAll(),例如List, Set等。对于Map,您可以使用putAll。

假设输入如下:

    import java.util.stream.Stream;
    import java.util.stream.Collectors;
    import java.util.Map;
   
    ...

    var m1 = Map.of("k1", 1, "k2", 2);
    var m2 = Map.of("k3", 3, "k4", 4);

当你确定两个输入映射之间没有任何键冲突时,一个避免任何突变并产生不可变结果的简单表达式可以是:

    var merged = Stream.concat(
        m1.entrySet().stream(),
        m2.entrySet().stream()
    ).collect(Collectors.toUnmodifiableMap(Map.Entry::getKey, Map.Entry::getValue));

在可能发生键冲突的情况下,我们可以提供一个lambda来指定如何去重复它们。例如,如果我们想保留最大的值,以防两个输入中都有一个条目,我们可以:

    .collect(Collectors.toUnmodifiableMap(
        Map.Entry::getKey,
        Map.Entry::getValue,
        Math::max))  // any function  (Integer, Integer) -> Integer  is ok here
    HashMap<Integer,String> hs1 = new HashMap<>();
    hs1.put(1,"ram");
    hs1.put(2,"sita");
    hs1.put(3,"laxman");
    hs1.put(4,"hanuman");
    hs1.put(5,"geeta");

    HashMap<Integer,String> hs2 = new HashMap<>();
    hs2.put(5,"rat");
    hs2.put(6,"lion");
    hs2.put(7,"tiger");
    hs2.put(8,"fish");
    hs2.put(9,"hen");

    HashMap<Integer,String> hs3 = new HashMap<>();//Map is which we add

    hs3.putAll(hs1);
    hs3.putAll(hs2);

    System.out.println(" hs1 : " + hs1);
    System.out.println(" hs2 : " + hs2);
    System.out.println(" hs3 : " + hs3);

重复的项目将不会被添加(即重复的键),因为当我们将打印hs3时,我们将只获得键5的一个值,这将是最后一个添加的值,它将是rat。 **[设置不允许重复键,但值可以重复]

你可以使用HashMap<String, List<Integer>>来合并两个HashMap,避免丢失与相同键配对的元素。

HashMap<String, Integer> map1 = new HashMap<>();
HashMap<String, Integer> map2 = new HashMap<>();
map1.put("key1", 1);
map1.put("key2", 2);
map1.put("key3", 3);
map2.put("key1", 4);
map2.put("key2", 5);
map2.put("key3", 6);
HashMap<String, List<Integer>> map3 = new HashMap<>();
map1.forEach((str, num) -> map3.put(str, new ArrayList<>(Arrays.asList(num))));
//checking for each key if its already in the map, and if so, you just add the integer to the list paired with this key
for (Map.Entry<String, Integer> entry : map2.entrySet()) {
    Integer value = entry.getValue();
    String key = entry.getKey();
    if (map3.containsKey(key)) {
        map3.get(key).add(value);
    } else {
        map3.put(key, new ArrayList<>(Arrays.asList(value)));
    }
}
map3.forEach((str, list) -> System.out.println("{" + str + ": " + list + "}"));

输出:

{key1: [1, 4]}
{key2: [2, 5]}
{key3: [3, 6]}