是否有Java数据结构在对象集合中返回一组特定属性？

Question

I have a collection of objects which are all the same object type. Within that type there is an attribute that I want to access - in this case it's to add it to a comma separated list:

String sep = ", ";
List<Item> items = db.findItems();

for (Item item : items) 
{
    result.append(item.getDescription() + sep);
} //for

return result.substring(0, results.length - sep.length());

It would be nice if I could just directly access that attribute from all the objects in the collections so that I could call Google Guava's Joiner function:

return Joiner.on(", ").join(/*Description attribute collection here*/);

The kind of structure I'm thinking of is like a 2D array where each column represents an attribute and each row represents an object and so I want to be able to call either a specific row which returns the object or a specific column (attribute) that returns a collection of the attributes from all objects.

Does Java or a good third party have a data structure like this or would I need to create my own?

Cheers,

Alexei Blue.

Answer 1

Why don't you use Google's Guava to the fullest then?

Joiner.on(",").join(Collections2.transform(items, new Function<Item, String>() {
    public String apply(Item input) {
        return item.getDescription();
    }
}));

Answer 2

You can write an inner Iterable subclass that reaches into each object in a collection and emits the correct attribute. See @Hiery's response, he got to the code faster...

Answer 3

Comming soon in JDK 8 with lambda expressions and method references

List<Jedi> jedis = asList(new Jedi("Obiwan",80), new Jedi("Luke", 35));
List<String> jediNames = jedis.map(Jedi::getName).into(new ArrayList<>()); //with metdod ref
Iterable<Integer> jediAges = jedis.map(j -> j.getAge()); //with lambdas

Depending on what you're doing you may even use the JDK 8 preview provided in the link.

Alternatively, the library Lambdaj offers a way to do that.

Answer 4

Silly, but I would opt for Item.toString(). Doesn't there exist a map so you can do join(map)?