java8函数式编程（一）

曾经终败给现在 2022-05-09 03:06 182阅读 0赞

## 需求 ##

将List集合中选出每种颜色相同的元素中size最大的元素，将这些元素组成一个新集合。

## List<Frog> ##

public Frog(String name, Integer age, String color, Double size, String desc) { this.name = name; this.age = age; this.color = color; this.size = size; this.desc = desc; } List<Frog> list = new ArrayList<>(); list.add(new Frog("aaa", 1, "green", 5.0, "111")); list.add(new Frog("bbb", 2, "red", 5.1, "222")); list.add(new Frog("ccc", 1, "blue", 5.2, "333")); list.add(new Frog("ddd", 3, "blue", 5.2, "444")); list.add(new Frog("eee", 3, "blue", 5.0, "555")); list.add(new Frog("fff", 1, "green", 5.3, "666"));

需要将name为bbb，fff，ccc或ddd（其中一个）组成一个新集合

## Java8 ##

List<Frog> result = new ArrayList<>();
            List<Frog> list = before();
            list.stream()
                    .collect(Collectors.groupingBy(Frog::getColor))
                    .forEach((k, v) -> {
                        result.add(v.stream().max(Comparator.comparingDouble(Frog::getSize)).get());
                    });
    
            System.out.println("insert: " + list);
            System.out.println("result: " + result);

结果

insert: [Frog{name='aaa', age=1, color='green', size=5.0, desc='111'}, Frog{name='bbb', age=2, color='red', size=5.1, desc='222'}, Frog{name='ccc', age=1, color='blue', size=5.2, desc='333'}, Frog{name='ddd', age=3, color='blue', size=5.2, desc='444'}, Frog{name='eee', age=3, color='blue', size=5.0, desc='555'}, Frog{name='fff', age=1, color='green', size=5.3, desc='666'}]
    result: [Frog{name='bbb', age=2, color='red', size=5.1, desc='222'}, Frog{name='fff', age=1, color='green', size=5.3, desc='666'}, Frog{name='ccc', age=1, color='blue', size=5.2, desc='333'}]

**我的思路是先将集合按color分组组成一个（color，list）的map，然后遍历map，将map的value（分组后的list）中size最大的元素找出，放到已经创建好的result集合中。**  
**最近刚开始接触java8的新特性，请路过的各位朋友指点，感激不尽！**