java将list中某一属性相同的数据分组

客官°小女子只卖身不卖艺 2023-10-08 23:01 55阅读 0赞

场景：

list列表中数据如下图所示，有很多属性相同的数据，现在需要将这些数据分组

![9455bc48f4da4f848a0cdd8272f7474c.png][]

解决方案：

Map<String, List<RiskAssessForTrendVo>> assessResultMap = 
    assessResultList.stream()
    .collect(Collectors.groupingBy(RiskAssessForTrendVo::getCreateTime));

解释：

//（第一个*表示实体类，第二个*表示实体类中属性，如果不是实体类，可直接写）
    Map< x,x > Map = list.stream().collect(groupingBy(*::get*));

分组后数据如下

![e7aa2260c80b42a6be1b5007d720e22b.png][]

之后可以对分组后的map进行遍历

for (Map.Entry<String, List<RiskAssessForTrendVo>> entry : assessResultMap.entrySet()) {
                //key是分组的日期 value是List<RiskAssessForTrendVo>
                String createTime = entry.getKey();
                double totalScoreAll = 0;
                List<RiskAssessForTrendVo> riskAssessRecordList = assessResultMap.get(createTime);
                if (CollectionUtil.isNotEmpty(riskAssessRecordList)) {
                    for (RiskAssessForTrendVo riskAssessRecord : riskAssessRecordList) {
                        double totalScore = riskAssessRecord.getTotalScore();
                        totalScoreAll += totalScore;
                    }
                    SynthesizeSafeTrendVo safeTrendVo = new SynthesizeSafeTrendVo();
                    safeTrendVo.setScore(Math.rint(totalScoreAll / riskAssessRecordList.size()));
                    safeTrendVo.setTime(createTime);
                    trendList.add(safeTrendVo);
                }
            }

[9455bc48f4da4f848a0cdd8272f7474c.png]: https://img-blog.csdnimg.cn/9455bc48f4da4f848a0cdd8272f7474c.png
[e7aa2260c80b42a6be1b5007d720e22b.png]: https://img-blog.csdnimg.cn/e7aa2260c80b42a6be1b5007d720e22b.png