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java8合并两个Map

JavaSE 来源:tangyaya8 1次浏览

合并两个Map

需求:如果两个Map有冲突则Value取第二个Map中的Value

例子准备

实体类

import lombok.*;
@Getter
@Setter
@ToString
@AllArgsConstructor
@NoArgsConstructor
public class Employee { 
    private Long id;
    private String name;
}

新建两个Map<String,Employee>

private static Map<String, Employee> map1 = new HashMap<>();
    private static Map<String, Employee> map2 = new HashMap<>();
    static { 
        Employee employee1 = new Employee(1L, "Henry");
        map1.put(employee1.getName(), employee1);
        Employee employee2 = new Employee(22L, "Annie");
        map1.put(employee2.getName(), employee2);
        Employee employee3 = new Employee(8L, "John");
        map1.put(employee3.getName(), employee3);
        Employee employee4 = new Employee(2L, "George");
        map2.put(employee4.getName(), employee4);
        Employee employee5 = new Employee(3L, "Henry");
        map2.put(employee5.getName(), employee5);
    }

在java8之前我们合并

@Test
    public void fun0() { 
        Map<String, Employee> map3 = new HashMap<>(map1);
        for (Map.Entry<String, Employee> entry : map2.entrySet()) { 
            if (map3.containsKey(entry.getKey())) { 
                 map3.put(entry.getKey(), new Employee(map3.get(entry.getKey()).getId(), entry.getValue().getName()));
            }else{ 
                map3.put(entry.getKey(), entry.getValue());
            }
        }
    }

java8之后

1.用map提供的merge()方法来合并

@Test
    public void fun1() { 
        Map<String, Employee> map3 = new HashMap<>(map1);
        map2.forEach((key, value) -> map3.merge(key, value, (v1, v2) -> new Employee(v1.getId(),v2.getName())));
        map3.forEach((k,v)-> System.out.println(k+"=:"+v));
    }

merge方法有三个参数,第一个是要合并的Key,第二个是合并的Value,第三个是一个接收两个参数的函数,它的作用就如我们上面的对于重复的key处理的逻辑一样,你可以自定义,v1是map2中的值,v2是map3中重复的值。

2.用Stream.concat来合并

Stream.contcat接收两个值,顺序的将两个集合中的数据加入流中,然后再用收集器变成Map

 @Test
    public void fun2() { 
        Stream<Map.Entry<String, Employee>> concat = Stream.concat(map1.entrySet().stream(), map2.entrySet().stream());
        Map<String, Employee> collect = concat.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (value1, value2) -> new Employee(value1.getId(), value1.getName())));
        collect.forEach((k,v)-> System.out.println(k+"=:"+v));
    }

3.用Stream.of()创建流后合并

和contact不同的是stream.of可以初始化多个元素,然后用扁平化的处理成需要的流,然后用收集器来转为Map

@Test
    public void fun3() { 
        Map<String, Employee> collect = Stream.of(map1, map2)
                .flatMap(x -> x.entrySet().stream())
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (v1, v2) -> new Employee(v1.getId(), v2.getName())));
        collect.forEach((k,v)-> System.out.println(k+"=:"+v));
    }

4.直接用Collector来收集

@Test
    public void fun4() { 
        Map<String, Employee> map3 = map2.entrySet()
                .stream()
                .collect(Collectors.toMap(
                        Map.Entry::getKey,
                        Map.Entry::getValue,
                        (v1, v2) -> new Employee(v1.getId(), v2.getName()),() -> new HashMap<>(map1)));
        map3.forEach((k,v)-> System.out.println(k+"=:"+v));

    }

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