使用流，您可以使用 Collectors.groupingBy 以及 Collectors.mapping ：

With streams, you could use Collectors.groupingBy along with Collectors.mapping:

Map<String, Set<String>> map = list.stream()
    .collect(Collectors.groupingBy(
        Student::getName,
        Collectors.mapping(student -> getNum(student.getAddr()),
            Collectors.toSet())));

我选择创建集合而不是列表地图，因为它似乎你不希望列表中有重复项。

I've chosen to create a map of sets instead of a map of lists, as it seems that you don't want duplicates in the lists.

如果你确实需要列表而不是集合，那么首先更有效率收集到集合，然后将集合转换为列表：

If you do need lists instead of sets, it's more efficient to first collect to sets and then convert the sets to lists:

Map<String, List<String>> map = list.stream()
    .collect(Collectors.groupingBy(
        Student::getName,
        Collectors.mapping(s -> getNum(s.getAddr()),
            Collectors.collectingAndThen(Collectors.toSet(), ArrayList::new))));

这使用 Collectors.collectingAndThen ，首先收集然后转换结果。

This uses Collectors.collectingAndThen, which first collects and then transforms the result.

另一种更紧凑的方式，没有流：

Another more compact way, without streams:

Map<String, Set<String>> map = new HashMap<>(); // or LinkedHashMap
list.forEach(s -> 
    map.computeIfAbsent(s.getName(), k -> new HashSet<>()) // or LinkedHashSet
        .add(getNum(s.getAddr())));

此变体使用 Iterable.forEach 迭代列表和 Map.computeIfAbsent 按学生名称对转换后的地址进行分组。

This variant uses Iterable.forEach to iterate the list and Map.computeIfAbsent to group transformed addresses by student name.