使用redis从存储集获取所有唯一的分数?

pbwdgjma  于 2021-06-08  发布在  Redis
关注(0)|答案(1)|浏览(288)

使用以下代码添加地理数据

jedis.geoadd("storegeodata", 51.5074, 0.1278, "London");
//while member string is mostly json 
jedis.geoadd("storegeodata", 51.5074, 0.1278, "{place: London}");
jedis.geoadd("storegeodata", 51.5074, 0.1278, "{place: London, lat: 51.5074, lon: 0.1278}");

当geohash由于成员字符串不同而被复制时,如何检索基于geohash的唯一值以避免冗余

public Map<String, Object> checkRedisGeo(double lat, double lon, Jedis j) {
    Map<String, Object> result = new HashMap<>();
    try
    {
        GeoRadiusParam param = GeoRadiusParam.geoRadiusParam();
        param.sortAscending();
        param.withDist();
        param.count(1);
        System.out.println("lat :"+lat+" , lon :"+lon);
        List<GeoRadiusResponse> response =  j.georadius("commander",
                lon,lat, 150, GeoUnit.M, param);    
        //System.out.println(response.size()+" size");
      if(response.size() > 0)
      {
          for (GeoRadiusResponse geoRadiusResponse : response) {
                System.out.println("lat :"+lat+" , lon :"+lon+",  stringmember :"+geoRadiusResponse.getMemberByString());
                //System.out.println(geoRadiusResponse.getDistance());
                Object[] data= {geoRadiusResponse.getMemberByString()};
                System.out.println(data);
                result.put("result", data);

               }
      }else {
         // sendEvents(streamEvent, null, streamEventChunk);
          System.out.println("E");
      }

    } catch (Exception e) {
        LOGGER.error("checkRedisGeo err : "+e);
    }

  return result;

}

哪个检索结果但如何根据geohash/score值过滤掉,如何得到所有不同的分数?
下面是冗余数据示例

cwtwac6a

cwtwac6a1#

通过比较分数可以得到重复的和唯一的分数,因为分数是唯一的,并且为同一个分数存储了多个成员,下面是同一个分数的代码

Double previous_score = 0.0;
    int DuplincatesCount = 0;
    int UniqueCount = 0;
    Set<String> values = jedis.zrange("rediskey", 0, -1);// get all the members
    for (String member : values) {
        Double current_score = jedis.zscore("rediskey", member);//each member looped
        if (Double.compare(current_score, previous_score) == 0) { //comparing current score with previous
                                                                    // score
            DuplincatesCount++;
        } else {
            UniqueCount++;
        }
        previous_score = current_score;// score mapping
    }
    System.out.println("Duplincates entry " + DuplincatesCount);
    System.out.println("Unique entry " + UniqueCount);

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