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  • [转]jdk8中map新增的merge方法

    1 前言

    转摘于:jdk8中map新增的merge方法介绍

    2 内容

    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。

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  • 原文地址:https://www.cnblogs.com/fanbi/p/15634499.html
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