我有以下问题 - 作出抽象带出的关键问题

I have the following problem - made abstract to bring out the key issues.

我有10个点，每个从其他一些距离。我想

I have 10 points each which is some distance from the other. I want to

能够找到群集即点的量彼此点成对的距离被最小化的中心， 设p（J）〜P（k）的再present高炮点J和K成对的距离 P（I）是中心点群集当且仅当P（I）的圣分[之和（p（j）条〜P（k））的]所有0℃ J，K＆LT; = n，其中我们在集群中有n个点 确定如何分割群集中两个群集一次数据点的集群的数量必须高于某个阈值吨 be able to find the center of the cluster i.e. the point for which the pairwise distance to each other point is minimised, let p(j) ~ p(k) represent the pairwise distance beteen points j and k p(i) is center-point of the cluster iff p(i) s.t. min[sum(p(j)~p(k))] for all 0 < j,k <= n where we have n points in the cluster determine how to split the cluster in to two clusters once the number of data points in the cluster goes above some threshold t.

这是不是欧氏空间。但距离可以概括如下： - P（I）是点i

This is not euclidean space. But the distances can be summarised as follows - p(i) is point i:

       p(1)    p(2)    p(3)    p(4)    p(5)    p(6)    p(7)    p(8)    p(9)    p(10)
p(1)    0       2       1       3       2       3       3       2       3        4
p(2)    2       0       1       3       2       3       3       2       3        4
p(3)    1       1       0       2       0       1       2       1       2        3
p(4)    3       3       2       0       1       2       3       2       3        4      
p(5)    2       2       1       1       0       1       2       1       2        3   
p(6)    3       3       2       2       1       0       3       2       3        4   
p(7)    3       3       2       3       2       3       0       1       2        3  
p(8)    2       2       1       2       1       2       1       0       1        2 
p(9)    3       3       2       3       2       3       2       1       0        1
p(10)   4       4       3       4       3       4       3       2       1        0

我将如何计算这是本簇的中心点？

How would I calculate which is the center point of this cluster?

推荐答案

我据我所知，这看起来像K均值聚类，而你所追求的是通常被称为Medoids'是什么：

I far as I understand this looks like K Means Clustering, and what you are looking for is usually known as 'Medoids':

在这里看到： http://en.wikipedia.org/wiki/Medoids   或在这里： http://en.wikipedia.org/wiki/K-medoids

See here : http://en.wikipedia.org/wiki/Medoids or here: http://en.wikipedia.org/wiki/K-medoids