如何快速打好Java基础?

2020/02/12

Predicate 是 Java 8 新增的一个函数式接口(通过 @FunctionalInterface 注解定义),因此可以将一个 Lambda 表达式赋值于它。

来看这样一个例子:

Predicate<Integer> noGreaterThan5 = x -> x > 5;

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);

List<Integer> collect = list.stream()
        .filter(noGreaterThan5)
        .collect(Collectors.toList());

System.out.println(collect); // [6, 7, 8, 9, 10]

还可以使用 and() 方法对条件进行拼接:

Predicate<Integer> noGreaterThan5 = x -> x > 5;
Predicate<Integer> noLessThan8 = x -> x < 8;

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);

List<Integer> collect = list.stream()
        .filter(noGreaterThan5.and(noLessThan8))
        .collect(Collectors.toList());

System.out.println(collect); // [6, 7]

除了 and(),还有 or

Predicate<String> lengthIs3 = x -> x.length() == 3;
Predicate<String> startWithA = x -> x.startsWith("A");

List<String> list = Arrays.asList("A", "AA", "AAA", "B", "BB", "BBB");

List<String> collect = list.stream()
        .filter(lengthIs3.or(startWithA))
        .collect(Collectors.toList());

System.out.println(collect); // [A, AA, AAA, BBB]

还有 negate()(相反):

Predicate<String> startWithA = x -> x.startsWith("A");

List<String> list = Arrays.asList("A", "AA", "AAA", "B", "BB", "BBB");

List<String> collect = list.stream()
        .filter(startWithA.negate())
        .collect(Collectors.toList());

System.out.println(collect); //[B, BB, BBB]

再来看一下 test()

Predicate<String> startWithA = x -> x.startsWith("王");

// 以王或者沉开头
boolean result = startWithA.or(x -> x.startsWith("沉")).test("沉默王二");
System.out.println(result);     // true

(转载本站文章请注明作者和出处 沉默王二

Show Disqus Comments

Post Directory

扫码或搜索:沉默王二
发送 290992
即可立即永久解锁本站全部文章