Question

我有一个场景，其中N个项目在N个位置（纬度，长度）交付。我想找到一组彼此靠近（100米）的物品。比如说（Item1，Item9，... Item499），他们之间的距离小于100。 Item9可以距离Item499超过100米，但如果距离该组中的任何一个项目的距离小于100米，它将在该组中。

查找所有此类群组。我希望有人指出正确的算法。然后，我将继续努力。优先语言是C＃

Answer 1

以下是解决此问题的算法，我发现here

public class ClusterCreator
{
    public Dictionary<int, SimpleCluster> GetClusters(List<SimpleCluster> clusterList, int maxDistance)
    {
        //LOAD DATA
        //var clusterList = new List<SimpleCluster>(); // LoadSimpleClusterList();
        //var latitudeSensitivity = 3;
        //var longitutdeSensitivity = 3;

        //CLUSTER THE DATA
        return ClusterTheData(clusterList, maxDistance);
    }

    public Dictionary<int, SimpleCluster> ClusterTheData(List<SimpleCluster> clusterList, int maxDistance)
    {
        //CLUSTER DICTIONARY
        var clusterDictionary = new Dictionary<int, SimpleCluster>();

        //Add the first node to the cluster list
        if (clusterList.Count > 0)
        {
            clusterDictionary.Add(clusterList[0].ID, clusterList[0]);
        }

        //ALGORITHM
        for (int i = 1; i < clusterList.Count; i++)
        {
            SimpleCluster combinedCluster = null;
            SimpleCluster oldCluster = null;
            foreach (var clusterDict in clusterDictionary)
            {
                //Check if the current item belongs to any of the existing clusters
                if (CheckIfInCluster(clusterDict.Value, clusterList[i], maxDistance))
                {
                    //If it belongs to the cluster then combine them and copy the cluster to oldCluster variable;
                    combinedCluster = CombineClusters(clusterDict.Value, clusterList[i]);
                    oldCluster = new SimpleCluster(clusterDict.Value);
                }
            }

            //This check means that no suitable clusters were found to combine, so the current item in the list becomes a new cluster.
            if (combinedCluster == null)
            {
                //Adding new cluster to the cluster dictionary 
                clusterDictionary.Add(clusterList[i].ID, clusterList[i]);
            }
            else
            {
                //We have created a combined cluster. Now it is time to remove the old cluster from the dictionary and instead of it add a new cluster.
                clusterDictionary.Remove(oldCluster.ID);
                clusterDictionary.Add(combinedCluster.ID, combinedCluster);
            }
        }
        return clusterDictionary;
    }

    public static SimpleCluster CombineClusters(SimpleCluster home, SimpleCluster imposter)
    {
        //Deep copy of the home object
        var combinedCluster = new SimpleCluster(home);
        combinedCluster.LAT_LON_LIST.AddRange(imposter.LAT_LON_LIST);
        combinedCluster.NAMES.AddRange(imposter.NAMES);

        //Combine the data of both clusters
        //combinedCluster.LAT_LON_LIST.AddRange(imposter.LAT_LON_LIST);
        //combinedCluster.NAMES.AddRange(imposter.NAMES);

        //Recalibrate the new center
        combinedCluster.LAT_LON_CENTER = new LAT_LONG
        {
            LATITUDE = ((home.LAT_LON_CENTER.LATITUDE + imposter.LAT_LON_CENTER.LATITUDE) / 2.0),
            LONGITUDE = ((home.LAT_LON_CENTER.LONGITUDE + imposter.LAT_LON_CENTER.LONGITUDE) / 2.0)
        };

        return combinedCluster;
    }

    public bool CheckIfInCluster(SimpleCluster home, SimpleCluster imposter, int maxDistance)
    {
        foreach (var item in home.LAT_LON_LIST)
        {
            var sCoord = new GeoCoordinate(item.LATITUDE, item.LONGITUDE);
            var eCoord = new GeoCoordinate(imposter.LAT_LON_CENTER.LATITUDE, imposter.LAT_LON_CENTER.LONGITUDE);
            var distance = sCoord.GetDistanceTo(eCoord);
            if (distance <= maxDistance)
                return true;
        }
        return false;


        //if ((home.LAT_LON_CENTER.LATITUDE + latitudeSensitivity) > imposter.LAT_LON_CENTER.LATITUDE
        //       && (home.LAT_LON_CENTER.LATITUDE - latitudeSensitivity) < imposter.LAT_LON_CENTER.LATITUDE
        //       && (home.LAT_LON_CENTER.LONGITUDE + longitutdeSensitivity) > imposter.LAT_LON_CENTER.LONGITUDE
        //       && (home.LAT_LON_CENTER.LONGITUDE - longitutdeSensitivity) < imposter.LAT_LON_CENTER.LONGITUDE
        //   )
        //{
        //    return true;
        //}
        //return false;
    }

}
public class SimpleCluster
{
    #region Constructors
    public SimpleCluster(int id, string name, double latitude, double longitude)
    {
        ID = id;
        LAT_LON_CENTER = new LAT_LONG
        {
            LATITUDE = latitude,
            LONGITUDE = longitude
        };
        NAMES = new List<string>();
        NAMES.Add(name);
        LAT_LON_LIST = new List<LAT_LONG>();
        LAT_LON_LIST.Add(LAT_LON_CENTER);
    }

    public SimpleCluster(SimpleCluster old)
    {
        ID = old.ID;
        LAT_LON_CENTER = new LAT_LONG
        {
            LATITUDE = old.LAT_LON_CENTER.LATITUDE,
            LONGITUDE = old.LAT_LON_CENTER.LONGITUDE
        };
        LAT_LON_LIST = new List<LAT_LONG>(old.LAT_LON_LIST);
        NAMES = new List<string>(old.NAMES);
    }
    #endregion
    public int ID { get; set; }
    public List<string> NAMES { get; set; }
    public LAT_LONG LAT_LON_CENTER { get; set; }
    public List<LAT_LONG> LAT_LON_LIST { get; set; }
}

public class LAT_LONG
{
    public double LATITUDE { get; set; }
    public double LONGITUDE { get; set; }
}

Lat，Long的集群彼此接近

1 个答案: