Question

我有一个对象Ob列表，定义为

class Ob {
   private String type;
   private List<String> attire;
   // standard getter and setters

   public Ob (String type){
        this.type=type;
    }

    public Ob addAttrire(String att){
        if(attire == null){
            attire = new ArrayList<>();
        }
        attire.add(att);
        return this;
    }
}

我收到的对象为

[{
"type" : "upper"
attires : [{"t1","t2"}]
},
{
"type" : "upper"
attires : ["t3","t4"]
},
{
"type" : "lower"
attires : ["l1","l2"]
}]

我必须合并为

[{
"type" : "upper"
attires : ["t1","t2","t3","t4"]
},{
"type" : "lower"
attires : ["l1","l2"]
}]

我如何使用流来做到这一点。减少帮助吗？可以使用的流是

List<Ob> coll = new ArrayList<>();


 coll.add(new Ob("a").addAttrire("1").addAttrire("2").addAttrire("3"));
        coll.add(new Ob("a").addAttrire("1").addAttrire("2").addAttrire("3"));
        coll.add(new Ob("a").addAttrire("1").addAttrire("2").addAttrire("3"));
        coll.add(new Ob("b").addAttrire("1").addAttrire("2").addAttrire("3"));
        coll.add(new Ob("b").addAttrire("1").addAttrire("2").addAttrire("3"));
        coll.add(new Ob("b").addAttrire("1").addAttrire("2").addAttrire("3"));



Collection<Ob> values = coll.stream()
                .collect(toMap(Ob::getType, Function.identity(), (o1, o2) -> {
                    o1.getAttire().addAll(o2.getAttire());
                    return o1;
                })).values();

用Ruben解决方案更新了问题。不需要删除重复项，但可以使用Ob中的set来完成重复项。目前的解决方案运作完美。

Answer 1

此解决方案使用groupingBy收集器，然后使用单独的步骤创建新的Ob，这是合并具有相同Ob的所有type的结果

我认为鲁本斯解决方案比这个答案有优势，因为它更短更简单。我认为这个答案有一个优势，因为它没有修改原始的Ob，因此更具功能性。

public static void testTrickyStreamSet() {
    Stream<Ob> s = Stream.of(
        new Ob("a", "1", "2"),
        new Ob("b", "1", "4"),
        new Ob("a", "1", "3"),
        new Ob("b", "1", "5"));

     List<Ob> os = s.collect(groupingBy(o -> o.type))
         .entrySet().stream()
         .map(e -> new Ob(e.getKey(), 
             e.getValue().stream().flatMap(o -> o.attire.stream()).collect(toList())))
         .collect(toList());

     // Prints [<Ob type=a, attire=[1, 2, 3]>, <Ob type=b, attire=[1, 4, 5]>]
     System.out.println(os);
}

public static class Ob {
    public String type;
    public List<String> attire;

    public Ob(String type, String... attire) {
        this.type = type;
        this.attire = Arrays.asList(attire);
    }

    public Ob(String type, List<String> attire) {
        this.type = type;
        this.attire = new ArrayList<>(attire);
    }

    @Override
    public String toString() {
        return "<Ob type=" + type + ", attire=" + attire + ">";
    }
}

Answer 2

您可以使用合并列表的合并函数收集toMap

    Collection<Ob> values = coll.stream()
     .collect(toMap(Ob::getType, Function.identity(), (o1, o2) -> {
        o1.getAttire().addAll(o2.getAttire());
        return o1;
    })).values();

通过Java-8中的公共字段减少对象集合

2 个答案: