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
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]
其他回答
方法1:将映射放在List中，然后连接
public class Test15 {
public static void main(String[] args) {
Map<String, List<String>> map1 = new HashMap<>();
map1.put("London", Arrays.asList("A", "B", "C"));
map1.put("Wales", Arrays.asList("P1", "P2", "P3"));
Map<String, List<String>> map2 = new HashMap<>();
map2.put("Calcutta", Arrays.asList("Protijayi", "Gina", "Gini"));
map2.put("London", Arrays.asList( "P4", "P5", "P6"));
map2.put("Wales", Arrays.asList( "P111", "P5555", "P677666"));
System.out.println(map1);System.out.println(map2);
// put the maps in an ArrayList
List<Map<String, List<String>>> maplist = new ArrayList<Map<String,List<String>>>();
maplist.add(map1);
maplist.add(map2);
/*
<T,K,U> Collector<T,?,Map<K,U>> toMap(
Function<? super T,? extends K> keyMapper,
Function<? super T,? extends U> valueMapper,
BinaryOperator<U> mergeFunction)
*/
Map<String, List<String>> collect = maplist.stream()
.flatMap(ch -> ch.entrySet().stream())
.collect(
Collectors.toMap(
//keyMapper,
Entry::getKey,
//valueMapper
Entry::getValue,
// mergeFunction
(list_a,list_b) -> Stream.concat(list_a.stream(), list_b.stream()).collect(Collectors.toList())
));
System.out.println("Final Result(Map after join) => " + collect);
/*
{Wales=[P1, P2, P3], London=[A, B, C]}
{Calcutta=[Protijayi, Gina, Gini], Wales=[P111, P5555, P677666], London=[P4, P5, P6]}
Final Result(Map after join) => {Calcutta=[Protijayi, Gina, Gini], Wales=[P1, P2, P3, P111, P5555, P677666], London=[A, B, C, P4, P5, P6]}
*/
}//main
}
方法二:法线映射合并
public class Test15 {
public static void main(String[] args) {
Map<String, List<String>> map1 = new HashMap<>();
map1.put("London", Arrays.asList("A", "B", "C"));
map1.put("Wales", Arrays.asList("P1", "P2", "P3"));
Map<String, List<String>> map2 = new HashMap<>();
map2.put("Calcutta", Arrays.asList("Protijayi", "Gina", "Gini"));
map2.put("London", Arrays.asList( "P4", "P5", "P6"));
map2.put("Wales", Arrays.asList( "P111", "P5555", "P677666"));
System.out.println(map1);System.out.println(map2);
/*
<T,K,U> Collector<T,?,Map<K,U>> toMap(
Function<? super T,? extends K> keyMapper,
Function<? super T,? extends U> valueMapper,
BinaryOperator<U> mergeFunction)
*/
Map<String, List<String>> collect = Stream.of(map1,map2)
.flatMap(ch -> ch.entrySet().stream())
.collect(
Collectors.toMap(
//keyMapper,
Entry::getKey,
//valueMapper
Entry::getValue,
// mergeFunction
(list_a,list_b) -> Stream.concat(list_a.stream(), list_b.stream()).collect(Collectors.toList())
));
System.out.println("Final Result(Map after join) => " + collect);
/*
{Wales=[P1, P2, P3], London=[A, B, C]}
{Calcutta=[Protijayi, Gina, Gini], Wales=[P111, P5555, P677666], London=[P4, P5, P6]}
Final Result(Map after join) => {Calcutta=[Protijayi, Gina, Gini], Wales=[P1, P2, P3, P111, P5555, P677666], London=[A, B, C, P4, P5, P6]}
*/
}//main
}
在Python中，HashMap被称为字典，我们可以很容易地合并它们。
x = {'Roopa': 1, 'Tabu': 2}
y = {'Roopi': 3, 'Soudipta': 4}
z = {**x,**y}
print(z)
{'Roopa': 1, 'Tabu': 2, 'Roopi': 3, 'Soudipta': 4}
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);
如果最终的映射不需要可变性，可以使用Guava的ImmutableMap及其Builder和putAll方法，与Java的map接口方法相反，ImmutableMap可以被链接。
使用示例:
Map<String, Integer> mergeMyTwoMaps(Map<String, Integer> map1, Map<String, Integer> map2) {
return ImmutableMap.<String, Integer>builder()
.putAll(map1)
.putAll(map2)
.build();
}
当然，这个方法可以更通用，使用可变参数和循环从参数中putAll map等，但我想展示一个概念。
此外，ImmutableMap和它的构建器有一些限制(或者可能是功能?):
他们是空敌对(抛出NullPointerException -如果map中的任何键或值为空) 生成器不接受重复的键(如果添加了重复的键则抛出IllegalArgumentException)。
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。
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]
