java8函数式编程（二）

水深无声 2022-05-08 02:12 244阅读 0赞

List常用操作

## 准备数据 ##

private static List<Frog> init() { 
            List<Frog> list = new ArrayList<>();
            list.add(new Frog("a", 2 , "red", 5.1));
            list.add(new Frog("b", 3 , "red", 5.0));
            list.add(new Frog("c", 2 , "blue", 5.5));
            list.add(new Frog("d", 1 , "blue", 5.3));
            list.add(new Frog("e", 5 , "yellow", 5.1));
            list.add(new Frog("f", 3 , "yellow", 5.2));
            return list;
        }

## forEach ##

循环打印集合中每个元素

private static void foreach() { 
            //init().forEach(t -> System.out.println(t));
            init().forEach(System.out::println);
        }

## toList ##

private static void toList() { 
            List<String> list = Stream.of("a", "b", "c")
                    .collect(Collectors.toList());
            System.out.println(list);
        }

## filter ##

计算集合中age＞4的元素个数

private static void filter() { 
            long count = init().stream()
                    .filter(frog -> frog.getAge() < 4)
                    .count();
            System.out.println(count);
        }

筛选出颜色为blue的元素

private static void filter() { 
            // 1
            List<Frog> list = init().stream()
                    .filter(frog -> "blue".equals(frog.getColor()))
                    .collect(Collectors.toList());
            System.out.println(list);
            // 2
            List<Frog> frogs = init();
            frogs.removeIf(frog -> !"blue".equals(frog.getColor()));
            System.out.println(frogs);
        }

## map ##

将集合中元素name重新组合为一个集合

private static void map() { 
            List<String> nameList = init().stream()
                    .map(Frog::getName)
                    .collect(Collectors.toList());
            System.out.println(nameList);
        }

## max ##

集合中size最大的元素

private static void max() { 
            Frog frog = init().stream()
                    .max(Comparator.comparingDouble(Frog::getSize))
                    .get();
            System.out.println(frog);
        }

## reduce ##

对集合中元素的age求和

private static void reduce() { 
            int value = init().stream()
                    .mapToInt(Frog::getAge)
                    .reduce(0, (x, y) -> x + y);
            System.out.println(value);
        }

## flatMap ##

使用flatMap将多个stream合并为一个stream

private static void flatMap() { 
            List<Cool> cools = new ArrayList<>();
            cools.add(new Cool("cool1", init()));
            cools.add(new Cool("cool2", init()));
    
            Set<String> set = new HashSet<>();
            for (Cool cool: cools) { 
                for (Frog frog: cool.getList()) { 
                    if (frog.getAge() > 4) { 
                        set.add(frog.getName());
                    }
                }
            }
            System.out.println(set);
    
            
            Set<String> set1 = cools.stream()
                    .flatMap(cool -> cool.getList().stream())
                    .filter(frog -> frog.getAge() > 4)
                    .map(Frog::getName)
                    .collect(Collectors.toSet());
            System.out.println(set1);
        }

## Optional ##

isPresent 该方法表示一个Optional 对象里是否有值  
orElse 当Optional 对象为空时，该方法提供一个备选值  
orElseGet 方法适用当备选值在计算上繁琐时使用  
empty和ofNullable为工厂方法，可将一个空值转换成Optional 对象

private static void optional() { 
            Frog frog1 = init().stream()
                    .filter(frog -> frog.getAge() > 6)
                    .findFirst()
                    .orElseGet(Frog::new);
            System.out.println(frog1);
    
            String str = init().stream()
                    .filter(frog -> frog.getAge() > 6)
                    .map(Frog::getName)
                    .findFirst()
                    .orElse("is null");
            System.out.println(str);
        }

如果为空创建一个新对象；如果为空打印“is null”  
**备注**：尽量避免使用下面1的书写方式，可以考虑2

1：	Frog frog =	init().stream()
                    .filter(frog -> frog.getAge() > 6)
                    .findFirst()
                    .orElse(null);
        if (frog != null){
    		// 业务处理；
    	}
    2：	init().stream()
                    .filter(frog -> frog.getAge() > 6)
                    .findFirst()
                    .isPresent( e -> { // 业务处理; });

## Collectors ##

private static void collect() { 
            // 按颜色分组
            Map<String, List<Frog>> map = init().stream()
                    .collect(Collectors.groupingBy(Frog::getColor));
            System.out.println(map);
    
            // 将名称按指定格式拼接成字符串
            String str = init().stream()
                    .map(Frog::getName)
                    .collect(Collectors.joining("、","{","}"));
            System.out.println(str);
    
            // 每种颜色的数量
            Map<String, Long> map1 = init().stream()
                    .collect(Collectors.groupingBy(Frog::getColor, Collectors.counting()));
            System.out.println(map1);
    
            // 每种颜色中的名称
            Map<String, List<String>> map2 = init().stream()
                    .collect(Collectors.groupingBy(
                            Frog::getColor,
                            Collectors.mapping(Frog::getName, Collectors.toList())));
            System.out.println(map2);
    
            // 每种颜色中的size最大元素
            Map<String, Optional<Frog>> map3 = init().stream()
                    .collect(Collectors.groupingBy(
                            Frog::getColor,
                            Collectors.maxBy(Comparator.comparingDouble(Frog::getSize))));
            System.out.println(map3);
        }

执行结果

{ red=[Frog{ name='a', age=2, color='red', size=5.1}, Frog{ name='b', age=3, color='red', size=5.0}], blue=[Frog{ name='c', age=2, color='blue', size=5.5}, Frog{ name='d', age=1, color='blue', size=5.3}], yellow=[Frog{ name='e', age=5, color='yellow', size=5.1}, Frog{ name='f', age=3, color='yellow', size=5.2}]}
    { a、b、c、d、e、f}
    { red=2, blue=2, yellow=2}
    { red=[a, b], blue=[c, d], yellow=[e, f]}
    { red=Optional[Frog{ name='a', age=2, color='red', size=5.1}], blue=Optional[Frog{ name='c', age=2, color='blue', size=5.5}], yellow=Optional[Frog{ name='f', age=3, color='yellow', size=5.2}]}