暴操java.util.stream.Stream

java.util.stream.Stream

1. filter 过滤

  private List<Integer> integers = Lists.list(30, 40, 10, 20);
    Set<Integer> collect = integers.stream()
        .filter(i -> i > 20).collect(Collectors.toSet());
    assertEquals(Sets.newTreeSet(30, 40), collect);

2. map 对每个元素执行操作,不会改变原本数量

  private List<Integer> integers = Lists.list(30, 40, 10, 20);

  List<String> collect = integers.stream().map(i -> i + "呵呵").collect(Collectors.toList());

    assertEquals(Lists.list("30呵呵", "40呵呵", "10呵呵", "20呵呵"), collect);

3. mapToInt 转成IntSream

    ArrayList<Integer> list = Lists.list(2, 5, 1, 6);
    IntStream intStream = list.stream().mapToInt(i -> i);
    List<Integer> collect = intStream.boxed()
        .collect(Collectors.toList()); //  boxed() 把int转成Integer??
    System.out.println(collect);
//    mapToDouble mapToLong 同理

4. flatMap 类似把多个集合压平成一个?

    ArrayList<ArrayList<Integer>> list = Lists.list(Lists.list(1, 2, 3), Lists.list(1, 2, 3));

    Stream<Integer> integerStream = list.stream().flatMap(Collection::stream);
    List<Integer> collect = integerStream.collect(Collectors.toList());
    System.out.println(collect);//[1, 2, 3, 1, 2, 3]

//    list.stream().flatMapToInt() TODO

5. distinct 去重

List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);
    List<Integer> collect = integers.stream().distinct().collect(Collectors.toList());
    System.out.println(collect);//[30, 40, 10, 20, 50]

6. sorted 排序

collect = integers.stream().sorted().collect(Collectors.toList());
    System.out.println(collect);//  [10, 10, 20, 20, 20, 30, 40, 50]

    collect = integers.stream().sorted((x, y) -> {//这里认为40 不吉利 自定义排到最后
      if (y == 40) {
        return -1;  //  注意 排序比较时 结果 1排在前,-1排在后
      }else if (x==40){
        return 1;
      }
      return x.compareTo(y);
    }).collect(Collectors.toList());
    System.out.println(collect);// [10, 10, 20, 20, 20, 30, 50, 40]

7. peek 不影响主流的情况下 对当前流元素进行提取

 List<String> list = Lists.list("刘德华", "x蔡徐坤", "x吴亦凡", "朱大胖", "外星人");

    List<String> earth = new ArrayList<>();
    List<String> noFat = new ArrayList<>();
    List<String> noBadMan =
        list.stream().filter(i -> !"外星人".equals(i)).peek(earth::add)
            .filter(i -> !i.equals("朱大胖")).peek(noFat::add).filter(i -> !i.startsWith("x"))
            .collect(Collectors.toList());
    System.out.println("earth : " + earth); // earth : [刘德华, x蔡徐坤, x吴亦凡, 朱大胖]

    System.out.println("noFat : " + noFat); // noFat : [刘德华, x蔡徐坤, x吴亦凡]

    System.out.println("noBadMan : " + noBadMan); // noBadMan : [刘德华]

8. limit 限定流元素数量

 List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);
    List<Integer> collect = integers.stream().limit(3).collect(Collectors.toList());
    System.out.println(collect); // [30, 40, 10]

9. skip 跳过多少个元素

 List<Integer> q = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);
    List<Integer> collect5 = q.stream().skip(2).collect(Collectors.toList());
    System.out.println(collect5); // [10, 20, 20, 10, 50, 20]

10. foreach foreachOrdered 遍历

并行流时 foreach不会遵守原本顺序 foreachOrdered遵从原本顺序
不并行无区别

 Stream<Integer> limit = Stream.iterate(0,x->x+1).limit(100);
    List list = new ArrayList();

//    limit.parallel().forEachOrdered(list::add);// [0, 1, 2, 3, 4...
    limit.parallel().forEach(list::add);// [62, 63, 64, 65,...
    System.out.println(list);

11. reduce 合并?累加

a,b,c -> (a+b),c -> (a+b+c) -> result

    List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);

  Integer reduce = integers.stream().reduce(0, (x, y) -> x + y);
    Integer integer = integers.stream().reduce((x, y) -> x + y).orElse(-1);
    Integer reduce1 = integers.stream().reduce(0, (x, y) -> x + y,Integer::sum);
    System.out.println(reduce); // 200
    System.out.println(integer); // 200
    System.out.println(reduce1); // 200

12. collect 收集

    List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);

 ArrayList<Object> collect1 = integers.stream()
        .collect(ArrayList::new, ArrayList::add, ArrayList::add);
    // 在combiner使用时，你的Stream是平行的，因为在这种情况下，
    // 多个线程收集的元素Stream到最终输出的子列表ArrayList，
    // 并且这些子列表必须被组合以产生最终的ArrayList。

    System.out.println(collect1);// [30, 40, 10, 20, 20, 10, 50, 20]

13. min 最小 count总数

    List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);

 Integer integer1 = integers.stream().min(Integer::compareTo).orElse(-1);
    System.out.println(integer1); //10
    long count = integers.stream().count();
    System.out.println(count);//8

14. findFirst 第一个 findAny 随便一个

    List<Integer> integers = Lists.list(30, 40, 10, 20, 20, 10, 50, 20);

 Integer integer2 = integers.stream().findFirst().orElse(-1);
    System.out.println(integer2);
    Integer integer3 = integers.stream().findAny().orElse(-1);
    System.out.println(integer3);//30

15. Stream.iterate Stream.generate 生成一个无限流

后者可以自定义实现方法
必须有limit限制,否则无限

 List<Integer> collect3 = Stream.iterate(0, (x) -> ++x).limit(100).collect(Collectors.toList());
    System.out.println(collect3);

    List<Integer> collect4 = Stream.generate(MyStreamTest::get).limit(200)
        .collect(Collectors.toList());
    System.out.println(collect4);
  

  private static Integer get() {
    return RandomUtils.nextInt(1, 1000);
  }

16. Stream.concat 组合俩流

如果两个都排好序 那组合的也排好序

Stream<Integer> limit = Stream.generate(MyStreamTest::get).limit(10).sorted();
    Stream<Integer> limit1 = Stream.generate(MyStreamTest::get).limit(10).sorted();
    List<Integer> collect = Stream.concat(limit, limit1).collect(Collectors.toList());
    System.out.println(collect);