Lambda如何实现条件去重distinct List,如何实现条件分组groupBy List
条件去重
我们知道, Java8 lambda自带的去重为 distinct 方法, 但是只能过滤整体对象, 不能实现对象里的某个值进行判定去重, 比如:
List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 5, 5, 5, 6, 7);
List<Integer> distinctNumbers = numbers.stream()
.distinct()
.collect(Collectors.toList());
System.out.println(distinctNumbers);//1, 2, 3, 4, 5, 6, 7
List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 5, 5, 5, 6, 7);
List<Integer> distinctNumbers = numbers.stream()
.distinct()
.collect(Collectors.toList());
System.out.println(distinctNumbers);//1, 2, 3, 4, 5, 6, 7
但是, 如果我们有一个 List
我们想要的效果是这样的:
List<User> distinctUsers = users.stream()
.distinct(User::getName)
.collect(Collectors.toList());
但是很遗憾, distinct()方法并不能设置条件. 解决方案如下:
首先定义一个过滤器:
public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
Map<Object, Boolean> seen = new ConcurrentHashMap<>();
return object -> seen.putIfAbsent(keyExtractor.apply(object), Boolean.TRUE) == null;
}
然后就可以进行条件去重啦:
List<User> users = new LinkedList<>();
users.add(new User("Jim"));
users.add(new User("Jim"));
users.add(new User("Tom"));
users.add(new User("Leo"));
List<User> distinctUsers = users.stream()
.filter(distinctByKey(User::getName))
.collect(Collectors.toList());
System.out.println(distinctUsers);//[Jim, Tom, Leo]
条件分组
还是一样的例子, 我们有一个 List
直接上代码:
List<User> users = new LinkedList<>();
users.add(new User("Jim", 12));
users.add(new User("John", 18));
users.add(new User("Tom", 21));
users.add(new User("Leo", 30));
users.add(new User("Kate", 44));
users.add(new User("Lio", 50));
Map<String, List<User>> tripleUsers = users.stream()
.collect(Collectors.groupingBy((Function<User, String>) user -> {
String key;
if (user.getAge() <= 20) {
key = "less20";
} else if (user.getAge() <= 40) {
key = "less40";
} else {
key = "more40";
}
return key;
}, Collectors.toList()));
System.out.println(tripleUsers);
//{more40=[Kate, Lio], less40=[Tom, Leo], less20=[Jim, John]}