jdk8中map新增的merge方法介绍

1.Map.merge方法介绍

  jdk8对于许多常用的类都扩展了一些面向函数,lambda表达式,方法引用的功能,使得java面向函数编程更为方便。其中Map.merge方法就是其中一个,merge方法有三个参数,key:map中的键,value:使用者传入的值,remappingFunction:BiFunction函数接口(该接口接收两个值,执行自定义功能并返回最终值)。当map中不存在指定的key时,便将传入的value设置为key的值,当key存在值时,执行一个方法该方法接收key的旧值和传入的value,执行自定义的方法返回最终结果设置为key的值。

//map.merge方法源码
default V merge(K key, V value, BiFunction<? super V, ? super V, ? extends V> remappingFunction) { Objects.requireNonNull(remappingFunction); Objects.requireNonNull(value); V oldValue = get(key); V newValue = (oldValue == null) ? value : remappingFunction.apply(oldValue, value); if(newValue == null) { remove(key); } else { put(key, newValue); } return newValue; }

 比如以下代码的含义:当name不存在时设置name的值为1,当name的值存在时,将值加1赋给name

public static void main(String[] args) {
        Map<String, Integer> map = new HashMap<>();
        map.put("name", 1);
        map.merge("name", 1, (oldValue, newValue) -> oldValue + newValue);
        map.merge("count", 1, (oldValue, newValue) -> oldValue + newValue);
        System.out.println(map);
    }
//返回结果
//{count=1, name=2}

2.map.merge()方法使用场景

  merge方法在统计时用的场景比较多,这里举一个统计学生总成绩的例子来说明。现在有一个学生各科成绩的集合,要统计每个学生的总成绩,以下给出使用merge方法与不使用的写法

 

public class StudentScoreSum {

    @Data
    static class StudentScore {
        private Integer sid;
        private String scoreName;
        private Integer score;

        public StudentScore(Integer sid, String scoreName, Integer score) {
            this.sid = sid;
            this.scoreName = scoreName;
            this.score = score;
        }

        public StudentScore() {
        }
    }

    public static void main(String[] args) {
        List<StudentScore> list = new ArrayList<>();
        list.add(new StudentScore(1, "chinese", 110));
        list.add(new StudentScore(1, "english", 120));
        list.add(new StudentScore(1, "math", 135));
        list.add(new StudentScore(2, "chinese", 99));
        list.add(new StudentScore(2, "english", 100));
        list.add(new StudentScore(2, "math", 133));
        list.add(new StudentScore(3, "chinese", 88));
        list.add(new StudentScore(3, "english", 140));
        list.add(new StudentScore(3, "math", 90));
        list.add(new StudentScore(4, "chinese", 108));
        list.add(new StudentScore(4, "english", 123));
        list.add(new StudentScore(4, "math", 114));
        list.add(new StudentScore(5, "chinese", 116));
        list.add(new StudentScore(5, "english", 133));
        list.add(new StudentScore(5, "math", 135));

        System.out.println(sum1(list));
        System.out.println(sum2(list));
    }
    //传统写法
    public static Map<Integer, Integer> sum1(List<StudentScore> list) {
        Map<Integer, Integer> map = new HashMap<>();
        for (StudentScore studentScore : list) {
            if (map.containsKey(studentScore.getSid())) {
                map.put(studentScore.getSid(),
                        map.get(studentScore.getSid()) + studentScore.getScore());
            } else {
                map.put(studentScore.getSid(), studentScore.getScore());
            }
        }
        return map;
    }

    //merger写法
    public static Map<Integer, Integer> sum2(List<StudentScore> list) {
        Map<Integer, Integer> map = new HashMap<>();
        list.stream().forEach(studentScore -> map.merge(studentScore.getSid()
                , studentScore.getScore(), Integer::sum));
        return map;
    }
}


//输出结果

{1=365, 2=332, 3=318, 4=345, 5=384}
{1=365, 2=332, 3=318, 4=345, 5=384}

 

3.总结

  merger方法使用起来确实在一定程度上减少了代码量,使得代码更加简洁。可见,java8新增的函数是编程确实能让我们少些点模板代码,更加关注与业务实现。

 

注意:本文仅代表个人理解和看法哟!和本人所在公司和团体无任何关系!

 

posted @ 2019-05-30 22:18  freeTimeWY  阅读(20851)  评论(0编辑  收藏  举报