List转换Map的一些操作

一、List转Map

1. List 转 Map<Integer,List>

Map<Integer, List<User>> subLineInfoMap = userInfos .stream().collect(Collectors.groupingBy(User::getSex));

2. List 转 Map<Integer,User>

Map<Long, User>userMap = userInfos.stream().collect(Collectors.toMap(User::getStuNum,Function.identity()));

3. List 转 Map<Integer,String>

Map<Integer, String> collect = userInfos.stream().collect(Collectors.toMap(User::geUserNum, User::getUserName));

4. list 转 map 保持顺序

LinkedHashMap<String, User> userMap = users.stream().collect(LinkedHashMap::new, (map, item) -> map.put(item.getAccountId(), item), Map::putAll);

5. 将list转成map 并排序

将list 排序，并按照排序后的结果进行有序分组

LinkedHashMap<String, List<AlarmData>> alarmMap = alarmDataList.stream().sorted(Comparator.comparing(t->t.getId().getData_time())).collect(Collectors.groupingBy(t->t.getId().getVirtualPointId(), LinkedHashMap::new, Collectors.toList()));

将map排序，并且每个key对应的list里面也是排序好的

6. 我们在利用Lambda 将list转成Map时就会出现 Duplicate key xxxx 的异常,意思就是对要转为map的key有重复了,除了进行for循环去重之外,我们还有其它方式能够优雅的处理它.

key重复时直接用后面的值(使用最新的或最老的值)

Map<String, Long> collect = list.stream().collect(Collectors.toMap(User::getExternalUserId, User::getUserId, (val1, val2) -> val2));

将两个值拼接起来

Map<String, Long> collect = list.stream().collect(Collectors.toMap(User::getExternalUserId, User::getUserId, (val1, val2) -> val1+val2));

将重复key的value变成一个集合，注意null值处理

Map<String, List<String>> tagMap = list.stream().collect(Collectors.toMap(User::getExternalUserId, s -> { List<String> tags = new ArrayList<>(); tags.add(s.getFollowTags()); return tags; }, (List<String> val1, List<String> val2) -> { val1.addAll(val2); return val1; } ));

list中有null元素的话会报错

对于处理的对象的属性为null的话,也可以进行,就是把null一起带进来

Person p1 = new Person(1, "张三", 16);
Person p2 = new Person(1, "李四", 16);
Person p3 = new Person(2, "王五", 16);
Person p4 = new Person(3, "张六", 16);
Person p5 = new Person(1, "张三", 16);
Person p7 = new Person(2, null, 16);
Person p8 = new Person(4, null, 16);
Person p9 = null;
ps.removeAll(Collections.singleton(null));
Map<Integer, List<String>> personmap = ps.stream().collect(Collectors.toMap(Person::getId, s -> { List<String> names = new ArrayList<>(); names.add(s.getName()); return names; }, (List<String> v1, List<String> v2) -> { v1.addAll(v2); return v1; } ));
System.out.println(personmap.toString());

这段代码输出

{1=[张三, 李四, 张三], 2=[王五, null], 3=[张六], 4=[null]}

结果不带null的处理

Map<Integer, List<String>> personmap = ps.stream().collect(Collectors.toMap(Person::getId, s -> { List<String> names = new ArrayList<>(); if(s.getName()!=null){ names.add(s.getName()); } return names; }, ( v1, v2) -> { v1.addAll(v2); HashSet h = new HashSet(v1); v1.clear(); v1.addAll(h); return v1; } ));
System.out.println(personmap.toString());

这段代码输出

{1=[李四, 张三], 2=[王五], 3=[张六], 4=[]}