public class AnswerApp { public static void main(String[] args) throws Exception { List<String> names = Lists.newArrayList("Answer", "AnswerAIL", "AI"); Map<String, Integer> map = names.stream().collect(Collectors.toMap(v -> v, v -> 1)); System.out.println(map); } }
程序运行输出
{Answer=1, AnswerAIL=1, AI=1}
public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { users.add(new User("answer" + i, new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); Map<String, Integer> map = users.stream().collect(Collectors.toMap(User::getName, User::getAge)); System.out.println(map); }
程序运行输出
[{"age":78,"name":"answer0"},{"age":89,"name":"answer1"},{"age":72,"name":"answer2"}] {answer2=72, answer1=89, answer0=78}
实现方式1
public class AnswerApp { public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { // 改为此代码, 转map时会报错 Duplicate key User // users.add(new User("answer", new Random().nextInt(100))); users.add(new User("answer" + i, new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity())); System.out.println(JSON.toJSONString(map)); } }
该方式如果 map 的 key(如上述例子的 User::getName 的值) 重复, 会抛错java.lang.IllegalStateException: Duplicate key User
程序运行输出
[{"age":22,"name":"answer0"},{"age":79,"name":"answer1"},{"age":81,"name":"answer2"}] {"answer2":{"age":81,"name":"answer2"},"answer1":{"age":79,"name":"answer1"},"answer0":{"age":22,"name":"answer0"}}
实现方式2
public class AnswerApp { public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { users.add(new User("answer", new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); // 如果 key 重复, 则根据 冲突方法 ·(key1, key2) -> key2· 判断. 解释: key1 key2 冲突时 取 key2 Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity(), (key1, key2) -> key2)); System.out.println(JSON.toJSONString(map)); } }
程序运行输出
[{"age":24,"name":"answer"},{"age":89,"name":"answer"},{"age":68,"name":"answer"}] {"answer":{"age":68,"name":"answer"}}
如果改为 (key1, key2) -> key1 则输出 {"answer":{"age":24,"name":"answer"}}
User 实体
@Data @NoArgsConstructor @AllArgsConstructor public class User { private Long id; private String name; private Integer age; public User(String name) { this.name = name; } public User(String name, Integer age) { this.name = name; this.age = age; } }
补充:java8中使用Lambda表达式将list中实体类的两个字段转Map
代码:
List<Entity> list = new ArrayList<>(); Map<Integer, String> map = list.stream().collect(Collectors.toMap(Entity::getId, Entity::getType));
** * List -> Map * 需要注意的是: * toMap 如果集合对象有重复的key,会报错Duplicate key .... * apple1,apple12的id都为1。 * 可以用 (k1,k2)->k1 来设置,如果有重复的key,则保留key1,舍弃key2 */ Map<Integer, Apple> appleMap = appleList.stream().collect(Collectors.toMap(Apple::getId, a -> a,(k1,k2)->k1)); 安照某一字段去重 list = list.stream().filter(distinctByKey(p -> ((ModCreditColumn) p).getFieldCode())).collect(Collectors.toList()); List<Double> unitNetValue = listIncreaseDto.stream().map(IncreaseDto :: getUnitNetValue).collect(Collectors.toList()); //求和 对象List BigDecimal allFullMarketPrice = entityList.stream().filter(value -> value.getFullMarketPrice()!= null).map(SceneAnalysisRespVo::getFullMarketPrice).reduce(BigDecimal.ZERO, BigDecimal::add); List<BigDecimal> naturalDayList; BigDecimal total = naturalDayList.stream().reduce(BigDecimal.ZERO, BigDecimal::add); 分组函数 Map<String, List<SceneAnalysisRespVo>> groupMap = total.getGroupList().stream().collect(Collectors.groupingBy(SceneAnalysisRespVo::getVmName)); //DV01之和 BigDecimal allDV01 = values.stream().filter(sceneAnalysisRespVo -> sceneAnalysisRespVo.getDv() != null).map(SceneAnalysisRespVo::getDv).reduce(BigDecimal.ZERO, BigDecimal::add);
以上为个人经验,希望能给大家一个参考,也希望大家多多支持呐喊教程。如有错误或未考虑完全的地方,望不吝赐教。
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