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;
}
}