merge()用法初试
假设有这么一段业务逻辑,有一个学生成绩对象的列表,对象包含学生姓名、科目、科目分数三个属性,要求求得每个学生的总成绩。加入列表如下:
1、初始化数据
private List<StudentScore> buildATestList() {List<StudentScore> studentScoreList = new ArrayList<>();StudentScore studentScore1 = new StudentScore() {{setStuName("张三");setSubject("语文");setScore(70);}};StudentScore studentScore2 = new StudentScore() {{setStuName("张三");setSubject("数学");setScore(80);}};StudentScore studentScore3 = new StudentScore() {{setStuName("张三");setSubject("英语");setScore(65);}};StudentScore studentScore4 = new StudentScore() {{setStuName("李四");setSubject("语文");setScore(68);}};StudentScore studentScore5 = new StudentScore() {{setStuName("李四");setSubject("数学");setScore(70);}};StudentScore studentScore6 = new StudentScore() {{setStuName("李四");setSubject("英语");setScore(90);}};StudentScore studentScore7 = new StudentScore() {{setStuName("王五");setSubject("语文");setScore(80);}};StudentScore studentScore8 = new StudentScore() {{setStuName("王五");setSubject("数学");setScore(85);}};StudentScore studentScore9 = new StudentScore() {{setStuName("王五");setSubject("英语");setScore(70);}};studentScoreList.add(studentScore1);studentScoreList.add(studentScore2);studentScoreList.add(studentScore3);studentScoreList.add(studentScore4);studentScoreList.add(studentScore5);studentScoreList.add(studentScore6);studentScoreList.add(studentScore7);studentScoreList.add(studentScore8);studentScoreList.add(studentScore9);return studentScoreList;}
2、常规的分组计算
ObjectMapper objectMapper = new ObjectMapper();List<StudentScore> studentScoreList = buildATestList();Map<String, Integer> studentScoreMap = new HashMap<>();studentScoreList.forEach(studentScore -> {if (studentScoreMap.containsKey(studentScore.getStuName())) {studentScoreMap.put(studentScore.getStuName(),studentScoreMap.get(studentScore.getStuName()) + studentScore.getScore());} else {studentScoreMap.put(studentScore.getStuName(), studentScore.getScore());}});System.out.println(objectMapper.writeValueAsString(studentScoreMap));// 结果如下:// {"李四":228,"张三":215,"王五":235}
3、merge()的分组计算
Map<String, Integer> studentScoreMap2 = new HashMap<>();studentScoreList.forEach(studentScore -> studentScoreMap2.merge(studentScore.getStuName(),studentScore.getScore(),Integer::sum));System.out.println(objectMapper.writeValueAsString(studentScoreMap2));// 结果如下:// {"李四":228,"张三":215,"王五":235}
merge()简介
merge() 可以这么理解:它将新的值赋值到 key (如果不存在)或更新给定的key 值对应的 value,其源码如下:
default V merge(K key, V value, BiFunction<? super V, ? super V, ? extends V> remappingFunction) {Objects.requireNonNull(remappingFunction);Objects.requireNonNull(value);V oldValue = this.get(key);V newValue = oldValue == null ? value : remappingFunction.apply(oldValue, value);if (newValue == null) {this.remove(key);} else {this.put(key, newValue);}return newValue;}
可以看到原理也是很简单的,该方法接收三个参数,一个 key 值,一个 value,一个 remappingFunction ,如果给定的key不存在,它就变成了 put(key, value) 。
但是,如果 key 已经存在一些值,remappingFunction 可以选择合并的方式,然后将合并得到的 newValue 赋值给原先的 key。
使用场景
这个使用场景相对来说还是比较多的,比如分组求和这类的操作,虽然 stream 中有相关 groupingBy() 方法,但如果想在循环中做一些其他操作的时候,merge() 还是一个挺不错的选择的。
其他
除了 merge() 方法之外,Java 8 中 map 相关的其他方法,比如 putIfAbsent、compute() 、computeIfAbsent() 、computeIfPresent,这些方法看名字应该就知道是什么意思了,这里贴一下 compute()(Map.class) 的源码,其返回值是计算后得到的新值:
default V compute(K key, BiFunction<? super K, ? super V, ? extends V> remappingFunction) {Objects.requireNonNull(remappingFunction);V oldValue = this.get(key);V newValue = remappingFunction.apply(key, oldValue);if (newValue == null) {if (oldValue == null && !this.containsKey(key)) {return null;} else {this.remove(key);return null;}} else {this.put(key, newValue);return newValue;}}
