Using Java Stream (stream) to map the values to filter, sort operation

Foreword

For the Java8 stream (stream) This elegant, convenient, fast, efficient operation is already Xinshou stick to the bar, but only List, Set. Do I have to face not Map Well? Required to operate the data transfer from the Map List at work, so there will be this essay.

Create a new Userclass, pay attention to the use of the following lombok; if you are idea, please download the plug-in and rely on, here I use the mavenquote here a little detail.

/**
 * @program: strategy-demo
 * @description: stream操作
 * @author: YuAoTian
 * @create: 2020-01-06 23:23
 **/
@Data
@AllArgsConstructor
class User{
    private Integer id;
    private String name;
    private String city;
    private Integer age;
}

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Continue to look at the code and see Mainwhich method

 public static void main(String[] args) {
        List<User> userList = Arrays.asList(
                new User((int) (Math.random()*10),"小明","北京",18),
                new User((int) (Math.random()*10),"小红", "南京", 17),
                new User((int) (Math.random()*10),"小丫", "湖北", 14),
                new User((int) (Math.random()*10),"小黑", "深圳", 22),
                new User((int) (Math.random()*10),"小玉", "广州", 19),
                new User((int) (Math.random()*10),"小小", "江西", 21)
                );
      Map<Integer, List<User>> userMap = userList.stream().collect(Collectors.groupingBy(User::getId));//已ID作为Key
    }

Map Sorting

Positive displacement

Map<Integer, List<User>> map = userMap.entrySet().stream().sorted(Comparator.comparing(o -> o.getValue().get(0).getAge())).map(entry -> {
            Map<Integer, List<User>> result = new LinkedHashMap<>();
            result.put(entry.getKey(), entry.getValue());
            return result;
        }).reduce((map1, map2) -> {
            map2.forEach(map1::put);
            return map1;
        }).get();
     System.out.print(map);

Export

{5=[User(id=5, name=小丫, city=湖北, age=14), User(id=5, name=小玉, city=广州, age=19), User(id=5, name=小小, city=江西, age=21)], 0=[User(id=0, name=小红, city=南京, age=17)], 8=[User(id=8, name=小明, city=北京, age=18)], 9=[User(id=9, name=小黑, city=深圳, age=22)]}

Inverted

 Map<Integer, List<User>> map = userMap.entrySet().stream().sorted(Comparator.comparing(o -> {
            //倒排中 reversed() 方法 是object 对象，需要在里面强制回 Entry 就行。
            Map.Entry<Integer, List<User>> temp = (Map.Entry<Integer, List<User>>) o;
            return temp.getValue().get(0).getAge();
        }).reversed()).map(entry -> {
            Map<Integer, List<User>> result = new LinkedHashMap<>();
            result.put(entry.getKey(), entry.getValue());
            return result;
        }).reduce((map1, map2) -> {
            map2.forEach(map1::put);
            return map1;
        }).get();
    System.out.print(map);

Export

{3=[User(id=3, name=小黑, city=深圳, age=22)], 5=[User(id=5, name=小玉, city=广州, age=19)], 0=[User(id=0, name=小明, city=北京, age=18), User(id=0, name=小丫, city=湖北, age=14)], 2=[User(id=2, name=小红, city=南京, age=17), User(id=2, name=小小, city=江西, age=21)]}

These are for age from small to large sums of sort, and the last return is a Optional, if not understand can next look at some examples.

Optional<Map<Integer, List<User>>> map = userMap.entrySet().stream().sorted(Comparator.comparing(o -> {
            //倒排中 reversed() 方法 是object 对象，需要在里面强制回 Entry 就行。
            Map.Entry<Integer, List<User>> temp = (Map.Entry<Integer, List<User>>) o;
            return temp.getValue().get(0).getAge();
        }).reversed()).map(entry -> {
            Map<Integer, List<User>> result = new LinkedHashMap<>();
            result.put(entry.getKey(), entry.getValue());
            return result;
        }).reduce((map1, map2) -> {
            map2.forEach(map1::put);
            return map1;
        });
        //为null返回false，否则true，即可调用get()返回~
        if(map.isPresent()){
            Map<Integer, List<User>> listMap = map.get();//调用get()返回一个Map
        }
        System.out.print(map);

optimization

Map<Integer, List<User>> map = userMap.entrySet().stream().sorted(Comparator.comparing(o -> {
            //倒排中 reversed() 方法 是object 对象，需要在里面强制回 Entry 就行。
            Map.Entry<Integer, List<User>> temp = (Map.Entry<Integer, List<User>>) o;
            return temp.getValue().get(0).getAge();
        }).reversed()).collect(LinkedHashMap::new,(m,e) -> m.put(e.getKey(),e.getValue()),LinkedHashMap::putAll);

The world is not instantaneous feeling a lot of peace and quiet - you can deal with what some of the indentation, can he reduced to one line!

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Map filter

map.entrySet().stream().filter(entry -> entry.getValue().get(0).getAge() > 16).collect(LinkedHashMap::new, (m, e) -> m.put(e.getKey(), e.getValue()), LinkedHashMap::putAll);
 System.out.println(map);

Export

{1=[User(id=1, name=小丫, city=湖北, age=14)]}

Filter out the age of <16, the filter is very simple I do not provide other examples of; a filter and then directly collect on the line ~

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In this paper it is over, if there is a better way to inform Hope ~