我希望能够将List转换为HashMap,其中键是elementName,值是随机的列表(在本例中是元素名)。所以简而言之,我想要(A->List(A), B->List(B), C-> List(C))。我尝试使用toMap()并将它传递给keyMapper和ValueMapper,但是我得到了一个编译错误。如果有人能帮我,我会很感激的。
谢谢!
public static void main(String[] args) {
// TODO Auto-generated method stub
List<String> list = Arrays.asList("A","B","C","D");
Map<String, List<String>> map = list.stream().map((element)->{
Map<String, List<String>> map = new HashMap<>();
map.put(element, Arrays.asList(element));
return map;
}).collect(??);
}
Function<Map<String, String>, String> key = (map) -> {
return map.keySet().stream().findFirst().get();
};
Function<Map<String, String>, String> value = (map) -> {
return map.values().stream().findFirst().get();
};=== ,这对我有用,
谢谢大家的帮助!@izstas“他们应该对元素进行操作”帮助了很多人:)事实上,这正是我想要找的东西
public static void test2 (){
Function<Entry<String, List<String>>, String> key = (entry) -> {
return entry.getKey();
};
Function<Entry<String, List<String>>, List<String>> value = (entry) -> {
return new ArrayList<String>(entry.getValue());
};
BinaryOperator<List<String>> merge = (old, latest)->{
old.addAll(latest);
return old;
};
Map<String, List<String>> map1 = new HashMap<>();
map1.put("A", Arrays.asList("A1", "A2"));
map1.put("B", Arrays.asList("B1"));
map1.put("D", Arrays.asList("D1"));
Map<String, List<String>> map2 = new HashMap<>();
map2.put("C", Arrays.asList("C1","C2"));
map2.put("D", Arrays.asList("D2"));
Stream<Map<String, List<String>>> stream =Stream.of(map1, map2);
System.out.println(stream.flatMap((map)->{
return map.entrySet().stream();
}).collect(Collectors.toMap(key, value, merge)));
}发布于 2014-07-23 17:57:55
您在代码中定义的函数key和value是不正确的,因为它们应该操作列表中的元素,而您的元素不是Map的。
以下代码适用于我:
List<String> list = Arrays.asList("A", "B", "C", "D");
Map<String, List<String>> map = list.stream()
.collect(Collectors.toMap(Function.identity(), Arrays::asList));Collectors.toMap的第一个参数定义了如何从list元素中生成键(保持原样),第二个参数定义了如何创建值(使用单个元素创建ArrayList )。
发布于 2015-07-16 14:09:03
您可以使用groupingBy方法来管理聚合,例如:
public static void main(String[] args) {
List<String> list = Arrays.asList("A", "B", "C", "D", "A");
Map<String, List<String>> map = list.stream().collect(Collectors.groupingBy(Function.identity()));
}如果您希望具有更大的灵活性(例如,映射值并返回一个集合而不是列表),您可以始终使用groupingBy方法,并使用javadoc中指定的更多参数:
Map<City, Set<String>> namesByCity = people.stream().collect(Collectors.groupingBy(Person::getCity, mapping(Person::getLastName, toSet())));发布于 2014-07-24 13:53:08
谢谢大家的帮助!@izstas“他们应该对元素进行操作”帮助了很多人:)事实上,这正是我想要找的东西
public static void test2 (){
Function<Entry<String, List<String>>, String> key = (entry) -> {
return entry.getKey();
};
Function<Entry<String, List<String>>, List<String>> value = (entry) -> {
return new ArrayList<String>(entry.getValue());
};
BinaryOperator<List<String>> merge = (old, latest)->{
old.addAll(latest);
return old;
};
Map<String, List<String>> map1 = new HashMap<>();
map1.put("A", Arrays.asList("A1", "A2"));
map1.put("B", Arrays.asList("B1"));
map1.put("D", Arrays.asList("D1"));
Map<String, List<String>> map2 = new HashMap<>();
map2.put("C", Arrays.asList("C1","C2"));
map2.put("D", Arrays.asList("D2"));
Stream<Map<String, List<String>>> stream =Stream.of(map1, map2);
System.out.println(stream.flatMap((map)->{
return map.entrySet().stream();
}).collect(Collectors.toMap(key, value, merge)));
}https://stackoverflow.com/questions/24917053
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