简体   繁体   English

如何使用Java 8流遍历多个列表?

[英]How to traverse multiple list using Java 8 stream?

I have three lists, 我有三个清单,

List<Double> list1= new ArrayList(List.of(1, 1.5));
List<Double> list2= new ArrayList(List.of(30, 25));
List<Double> list3= new ArrayList(List.of(30, 25));

I want to traverse through each simultaneously, and print 我想同时遍历每个,并打印

1    30  30
1.5  25  25

How can I do this using java-8 stream API? 我怎么能用java-8流API做到这一点?

You could do (assuming all lists have the same size): 你可以做(​​假设所有列表都有相同的大小):

IntStream.range(0, list1.size())
         .forEach(x -> Syso(list1.get(x) + " " + list2.get(x) + " " + list3.get(x)))

But there is no real benefit here, just use a plain for loop. 但是这里没有真正的好处,只需使用普通的for循环。

And btw don't wrap these with new ArrayList(List.of(30, 25)); 并且btw不会用new ArrayList(List.of(30, 25));包装它们new ArrayList(List.of(30, 25)); , this would be enough: ,这就足够了:

 List<Double> list1 = List.of(1D, 1.5D)

List.of(..) is a static factory returning a List itself, there is no need to wrap it into a ArrayList::new constructor. List.of(..)是一个返回List本身的静态工厂,不需要将它包装到ArrayList::new构造函数中。

Here is the code you probably need, however, don't underestimate the power for a for-loop : 以下是您可能需要的代码,但是,不要低估for-loop

// Here you need to get the max of all the List sizes to traverse all Lists
int maxSize = Math.max(list1.size(), Math.max(list2.size(), list3.size()));      

IntStream.range(0, maxSize)                                    // Iterate 0 .. minSize-1
         .mapToObj(i -> Stream.of(list1, list2, list3)         // Map each index
                              .filter(list -> i < list.size()) // Secure the existing index
                              .map(list -> list.get(i))        // Get the item
                              .collect(Collectors.toList()))   // Collect to List
                              .forEach(System.out::println);   // And iterate (print)

Output: 输出:

[1, 30, 30]
[1.5, 25, 25]

In case of variable list size, the traversal iteration is safe because I touch only the existing indices. 在变量列表大小的情况下,遍历迭代是安全的,因为我只触摸现有索引。

No need for streams - just transpose the lists: 不需要流 - 只需转置列表:

private static class Transpose<T> extends AbstractList<List<T>> implements List<List<T>> {
    private final List<T> [] lists;
    public Transpose(List<T>... lists) {
        this.lists = lists;
    }

    private class Column extends AbstractList<T> implements List<T> {
        private final int column;

        public Column(int column) {
            this.column = column;
        }

        @Override
        public T get(int row) {
            return lists[row].get(column);
        }

        @Override
        public int size() {
            return lists.length;
        }
    }
    @Override
    public List<T> get(int column) {
        return new Column(column);
    }

    @Override
    public int size() {
        return lists[0].size();
    }
}

private void test() {
    List<Double> list1= Arrays.asList(1.0, 1.5);
    List<Double> list2= Arrays.asList(30.0, 25.0);
    List<Double> list3= Arrays.asList(30.0, 25.0);

    List<List<Double>> matrix = new Transpose(list1, list2, list3);
    System.out.println(matrix);
}

Prints: 打印:

[[1.0, 30.0, 30.0], [1.5, 25.0, 25.0]] [[1.0,30.0,30.0],[1.5,25.0,25.0]]

Obviously this would be more difficult if the lists aren't all the same length. 显然,如果列表长度不同,这将更加困难。

This works when the lists' size are same or different: 这在列表的大小相同或不同时有效:

List<Double> list1 = List.of(1D, 1.5D);
List<Double> list2 = List.of(30D, 25D);
List<Double> list3 = List.of(30D, 25D);
Stream<List<Double>> listStream = Stream.of(list1, list2, list3);

int maxSize = listStream.mapToInt(List::size).max().orElse(0);

IntStream.range(0, maxSize)
        .forEach(index -> {
            listStream
                    .filter(list -> list.size() > index)
                    .forEach(list -> System.out.print(list.get(index) + " "));
            System.out.println();
        });

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM