Java8 stream操作toMap的key重复问题原

待我称王封你为后i 2021-11-09 03:04 351阅读 0赞

## Java8 stream操作toMap的key重复问题  ##

准备以下User对象集合 ，构造方法User（Long Id, String username）

List<User> userList = new ArrayList<>();
            userList.add(new User(1L, "aaa"));
            userList.add(new User(2L, "bbb"));
            userList.add(new User(3L, "ccc"));
            userList.add(new User(2L, "ddd"));
            userList.add(new User(3L, "eee"));

当进行普通toMap操作时

Map<Long, String> map = userList.stream()
                .collect(Collectors.toMap(User::getId, User::getUsername);

就会报错，说明处理到已存在的key，其对应value为bbb

java.lang.IllegalStateException: Duplicate key bbb
    	at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)
    	at java.util.HashMap.merge(HashMap.java:1253)
    	at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)
    	at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
    	at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1374)
    	at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481)
    	at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471)
    	at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
    	at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
    	at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)

解决方式就在Java8提供的Collectors.toMap() 方法中，其第三个参数就是当出现 duplicate key的时候的处理方案

方案一： 出现重复时，取前面value的值，或者取后面放入的value值，则覆盖先前的value值

Map<Long, String> map = userList.stream()
            .collect(Collectors.toMap(User::getId, User::getUsername, (v1, v2) -> v1));

Map<Long, String> map = userList.stream()
            .collect(Collectors.toMap(User::getId, User::getUsername, (v1, v2) -> v2));

方案二： Map的value可以储存一个list，把重复key的值放入list，再存到value中

userList.stream().collect(Collectors.toMap(User::getId,
                    e -> Arrays.asList(e.getUsername()),
                    (List<String> oldList, List<String> newList) -> {
                        oldList.addAll(newList);
                        return oldList;
                    }));

....