使用Stream比较两个集合-anyMatch [英] Comparing two collections using Stream - anyMatch

问题描述

我想比较list1中是否存在list2中的任何对象.

我可以遍历两个列表并使用.contains()比较所有元素，但我想知道是否没有更有效的方法.我发现了此，我正在尝试实施建议的方法:

List<Item> list1;
List<Item> list2;

boolean anyMatch = list1.stream().anyMatch(x -> x.equals(list2.stream()));
System.out.println(anyMatch);

执行此操作时，即使我希望使用true，我也总是得到false.怎么会来?

解决方案

根据您的评论，您有两个列表，分别为list1和list2.您想确定list1中是否至少包含list2中的一个元素.

使用Stream API，您可以获得list2的Stream.然后，调用 anyMatch(predicate) 返回此流的元素之一是否与给定谓词匹配，在这种情况下，该谓词将测试该元素是否包含在list1中.

boolean anyMatch = list2.stream().anyMatch(list1::contains);

这使用方法参考作为谓词. /p>

通过将list1转换为Set，可以保证恒定的查找时间，从而获得更好的性能:

boolean anyMatch = list2.stream().anyMatch(new HashSet<>(list1)::contains);

I want to compare if any of the objects in a list2 is present in a list1.

I could iterate over both lists and compare all elements using .contains() but I am wondering if there is not a more efficient way. I found this and I am trying to implement the suggested method:

List<Item> list1;
List<Item> list2;

boolean anyMatch = list1.stream().anyMatch(x -> x.equals(list2.stream()));
System.out.println(anyMatch);

When I do this I constantly get false, even when I'd expect a true. How come?

解决方案

From your comments, you have two lists, list1 and list2. You want to find out if at least one of the element in list2 is contained in list1.

With the Stream API, you can acquire a Stream of list2. Then, a call anyMatch(predicate) returns whether one of the element of this stream matches the given predicate, which, in this case, tests whether the element is contained in list1.

boolean anyMatch = list2.stream().anyMatch(list1::contains);

This uses a method reference as the predicate.

You would have better performance by converting the list1 into a Set, which guarantees constant-time look-up:

boolean anyMatch = list2.stream().anyMatch(new HashSet<>(list1)::contains);

这篇关于使用Stream比较两个集合-anyMatch的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！