两个list求和_java8 stream自定义分组求和并排序

喜欢ヅ旅行 2023-01-07 10:04 176阅读 0赞

public static void main(String[] args) {        List list = new ArrayList<>();        GroupDetailDTO dto1 = new GroupDetailDTO();        dto1.setHeadsetId(1);        dto1.setTime("2020-01-03");        dto1.setActConcreteTime("a");        dto1.setPlayed(1);        list.add(dto1);        GroupDetailDTO dto2 = new GroupDetailDTO();        dto2.setHeadsetId(1);        dto2.setTime("2020-01-01");        dto2.setActConcreteTime("b");        dto2.setPlayed(1);        list.add(dto2);        GroupDetailDTO dto3 = new GroupDetailDTO();        dto3.setHeadsetId(1);        dto3.setTime("2020-01-02");        dto3.setActConcreteTime("c");        dto3.setPlayed(1);        list.add(dto3);        GroupDetailDTO dto4 = new GroupDetailDTO();        dto4.setHeadsetId(2);        dto4.setTime("2020-01-01");        dto4.setActConcreteTime("d");        dto4.setPlayed(4);        list.add(dto4);        Map collect = list.stream().collect(Collectors.groupingBy(o -> new GroupDetailDTO(o.getHeadsetId(), o.getTime()), TreeMap::new, Collectors.summingInt(o -> o.getPlayed())));        collect.forEach((k, v) -> {            System.out.println(k + "@@@" + v);        });    }

@Datapublic class GroupDetailDTO implements Comparable {    private Integer headsetId;    private String actConcreteTime;    private String time;    private Integer played;    public GroupDetailDTO() {    }    public GroupDetailDTO(Integer headsetId, String time) {        this.headsetId = headsetId;        this.time = time;    }    @Override    public int compareTo(GroupDetailDTO o) {        String a = this.headsetId + "|" + this.getTime();        String b = o.getHeadsetId() + "|" + o.getTime();        return a.compareTo(b);//        return this.getTime().compareTo(o.gtTime());    }}

结果：

![2ebaff6c6512fee375d379c39883caab.png][]

上述代码实现对对象GroupDetailDTO按设备和时间分组求played的和，并且按时间排序了。当初想着既然是按时间排序那么compareTo方法直接使用  
return this.getTime().compareTo(o.getTime());  
但是结果不正确

![513754fb0177ae3cea5e56abe534b6d1.png][]

分析：分组求和时如果使用了排序，那么分组时对相同属性的合并将直接根据compareTo返回的结果合并对象，所以如果用上述方法将产生相同的时间直接合并，而不是new GroupDetailDTO(o.getHeadsetId(), o.getTime())里面的两个属性都相同才合并。

[2ebaff6c6512fee375d379c39883caab.png]: /images/20221119/739f95da9bc0457e9433bb36c77190eb.png
[513754fb0177ae3cea5e56abe534b6d1.png]: /images/20221119/14f3382eef5745fcb79ef3560b80bc00.png