如何构建一个新列表，其中包含现有列表中的所有条目以及修改了一个字段的每个条目的副本？

Question

I have a list of some Object called list. Using list.stream(), I need to create a new list of the same Object where the new list contains all of the original entries AND the new list contains a copy of each entry with one field modified. I know how to do this using list.stream() twice but I'd like to do it under 1 stream.

This is how I've accomplished the task using list.stream() twice

         newList = list.stream().collect(Collectors.toList());
         newList.addAll(list.stream().map(l -> {SomeObject a = new SomeObject(l);
                              a.setField1("New Value");
                              return a;
                              }).collect(Collectors.toList())
                        );

Answer 1

Use flatMap (assuming you don't mind the original and derived values being interleaved):

newList = list.stream()
    .flatMap(l -> {
      SomeObject a = new SomeObject(l);
      a.setField1("New Value");
      return Stream.of(l, a);
    })
    .collect(toList());

If it's just that you don't want to use stream() twice, you could avoid the first one using addAll ; and the unnecessary collect for the second:

newList = new ArrayList<>(list.size() * 2);
newList.addAll(list);
list.stream()
    .map(l -> {
        SomeObject a = new SomeObject(l);
        a.setField1("New Value");
        return a;
    })
    .forEach(newList::add);

Answer 2

If you want objct then processed object and so on, in your new list then, you can try something like this:

List<String> list = List.of("a", "b", "c" );
        
List<String> answer = list.stream()
             .map(s -> List.of(s,s+"1"))
             .flatMap(List::stream).collect(Collectors.toList());
        
System.out.println(answer);

Output:

[a, a1, b, b1, c, c1]