Question

我可以从Map<Integer, Integer>到List语句构建for，如此

List<Integer> integers = Arrays.asList(1, 2, 53, 66, 55, 99, 6989, 99, 33);

Map<Integer, Integer> map = new HashMap<>();
for (Integer integer : integers) {
    map.put(integer, integer);
}

System.out.println(map);

但是当我尝试使用流

时

map = integers.stream().distinct().map(p -> p).collect(Collectors.toMap(integers::get, integers::get));

代码抛出此异常

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 53
    at java.util.Arrays$ArrayList.get(Unknown Source)
    at java.util.stream.Collectors.lambda$toMap$58(Unknown Source)
    at java.util.stream.ReduceOps$3ReducingSink.accept(Unknown Source)
    at java.util.stream.ReferencePipeline$3$1.accept(Unknown Source)
    at java.util.stream.DistinctOps$1$2.accept(Unknown Source)
    at java.util.Spliterators$ArraySpliterator.forEachRemaining(Unknown Source)
    at java.util.stream.AbstractPipeline.copyInto(Unknown Source)
    at java.util.stream.AbstractPipeline.wrapAndCopyInto(Unknown Source)
    at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(Unknown Source)
    at java.util.stream.AbstractPipeline.evaluate(Unknown Source)
    at java.util.stream.ReferencePipeline.collect(Unknown Source)
    at com.me.lamda.LamdaOps.main(LamdaOps.java:30)

为什么呢？

Answer 1

使用Integer::intValue代替integers::get，因为它希望参数类似于ClassName::methodNameToBeCalledToGetTheDesiredValue

此外，map（p-> p）在这里是多余的，因为您没有将值映射到其他内容。

试试这个

    List<Integer> integers= Arrays.asList(1,2,53,66,55,99,6989,99,33);
    Map<Integer, Integer> map = integers.stream().distinct().collect(Collectors.toMap(Integer::intValue, Integer::intValue));
    System.out.println(map);

输出：{33 = 33,1 = 1,66 = 66,2 = 2,99 = 99,53 = 53,55 = 55,6989 = 6989}

将Integer类型的列表转换为java8中的映射？

1 个答案: