Convert a List

Question

For example, for two classes like this:

Class RawItem {
    private String category;
    private int code;
    private String name;
}

Class Item {
    private int code;
    private String name;
}

And list of RawItem as following:

[ { "category":"a", "code":1, "name":"item1" }, { "category":"a", "code":1, "name":"item1" }, { "category":"a", "code":2, "name":"item2" }, { "category":"b", "code":1, "name":"item1" }, { "category":"b", "code":1, "name":"item1" } ]

Turn it to a Map<String, List<Item>> like this:

{
   "a":[
      {
         "code":1,
         "name":"item1"
      },
      {
         "code":2,
         "name":"item2"
      }
   ],
   "b":[
      {
         "code":1,
         "name":"item1"
      }
   ]
}

Any reply will be greatly appreciated. thanks.

If it is mostly code then you should consider describing your problem more, telling where did you find the problem (which would require posting your attempt) — Cleptus, Mar 19 '18 at 14:23
`List` to `Map>` would be nicest though `Item` would need equals+hashCode. — Joop Eggen, Mar 19 '18 at 14:49
It is mostly code because I quoted some non-code as code, that’s it. The problem is already described clearly — Wuaner, Mar 19 '18 at 16:19

Alexandre Dupriez · Accepted Answer · 2018-03-20T10:38:41.257

3

The following groupingBy should work. As part of your additional requirements, we know that the equals method of RawItem objects cannot be used, which is why we are using a filter instead of the Stream distinct method (*):

    Set<Object> seen = ConcurrentHashMap.newKeySet();

    Map<String, List<Item>> m = 
       l.stream()
        .filter(item -> seen.add(item.category + ":" + item.name + ":" + item.code))
        .collect(Collectors.groupingBy(item -> item.category,
            Collectors.mapping(item -> new Item(item.code, item.name), 
                               Collectors.toList())));

Note that in order to allow the method distinct to work and remove duplicates, you need to have a proper implementation of the hashCode and equals methods in the RawItem class.

As an example, the following test:

List<RawItem> list = Arrays.asList(new RawItem("a", 1, "dfg"),
                                   new RawItem("a", 1, "dfg"),
                                   new RawItem("a", 1, "fdgdfdfgdg"),
                                   new RawItem("b", 1, "dfg"));

Map<String, List<Item>> map = // the above

System.out.println(map);

gives

{
   a=[Item{code=1, name='dfg'}, Item{code=1, name='fdgdfdfgdg'}],
   b=[Item{code=1, name='dfg'}]
}

(*) Edit:
As shared by the OP, the following posts help to design a solution based on a custom equivalence relationship with the stream API, when distinct() cannot be used:

edited Mar 20 '18 at 10:38

answered Mar 19 '18 at 14:12

Alexandre Dupriez

3,026
20
25

This does not fully solve the problem since it does not get rid of items with identical content (duplicates) – Claudiu Guja Mar 19 '18 at 14:35
@ClaudiuGuja Indeed, fixed the post. – Alexandre Dupriez Mar 19 '18 at 14:42
Thanks for prompt response, I will try it tmrw because I already off my work after I asked this question, I’m using my phone to commenting now...but since distinct() need a proper implementation of equals, I doubt if it works since I can’t use category, code, name to uniquely identify a RawItem, it have other key fields that not listed here. So any other solution as a supplementary? – Wuaner Mar 19 '18 at 15:44
@Wuaner: Sure, I updated the post to take into account this additional requirement – Alexandre Dupriez Mar 19 '18 at 16:47
@AlexandreDupriez So brilliant~ can’t test it now since no computer around, but I know it will work! Accepted and thanks again. – Wuaner Mar 19 '18 at 20:05
It works~ saved my day, and I extract the Set _seen_ as a Predicate as [1](https://stackoverflow.com/a/27872852/1635855) [2](https://stackoverflow.com/a/27872086/1635855) [3](https://stackoverflow.com/a/42819145/1635855) illustrated. – Wuaner Mar 20 '18 at 02:49
@Wuaner Thank you for the feedback, I added these references to the post. I can also update it to replace the set *seen* with a predicate as you mention? – Alexandre Dupriez Mar 20 '18 at 10:40
@AlexandreDupriez I already done my work with predicate, but it would be great if we post it here for some others, I think?. Many thanks for this solution, btw. – Wuaner Mar 21 '18 at 03:11

1 Answers1