Finding the center of a cluster

问题

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

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

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.

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

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

回答1:

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

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

回答2:

I may be about to have that frisson that comes just before displaying utter stupidity. But doesn't this yield easily to brute force? In Python:

distances = [
[ 0 , 2 , 1 , 3 , 2 , 3 , 3 , 2 , 3 , 4 , ],
[ 2 , 0 , 1 , 3 , 2 , 3 , 3 , 2 , 3 , 4 , ],
[ 1 , 1 , 0 , 2 , 0 , 1 , 2 , 1 , 2 , 3 , ],
[ 3 , 3 , 2 , 0 , 1 , 2 , 3 , 2 , 3 , 4 , ],
[ 2 , 2 , 1 , 1 , 0 , 1 , 2 , 1 , 2 , 3 , ],
[ 3 , 3 , 2 , 2 , 1 , 0 , 3 , 2 , 3 , 4 , ],
[ 3 , 3 , 2 , 3 , 2 , 3 , 0 , 1 , 2 , 3 , ],
[ 2 , 2 , 1 , 2 , 1 , 2 , 1 , 0 , 1 , 2 , ],
[ 3 , 3 , 2 , 3 , 2 , 3 , 2 , 1 , 0 , 1 , ],
[ 4 , 4 , 3 , 4 , 3 , 4 , 3 , 2 , 1 , 0 , ],
]

currentMinimum = 99999

for point in range ( 10 ) :
    distance_sum = 0
    for second_point in range ( 10 ) :
        if point == second_point : continue
        distance_sum += distances [ point ] [ second_point ]
    print '>>>>>', point, distance_sum 

    if distance_sum < currentMinimum :
        currentMinimum = distance_sum 
        centre = point

print centre

回答3:

find median or average values of all distances. = avgAll
For each p, find average distance to other machines. = avgP(i)
Pick the closer one as center. avgAll ~= avgP(i)

b) no idea for now..

maybe for each p, find the closer machine.

by this logic make a graph.

than somehow (i dont know yet) divide the graph

回答4:

What you're trying to do, or at least (b) belongs to Cluster Analysis. A branch of mathematics / statistics / econometrics where datapoints (e.g. points in n-dimensional space) are divided among groups or clusters. How to do this is not a trivial questions, there are many, many possible ways.

Read more at the wikipedia article on cluster analysis.

来源：https://stackoverflow.com/questions/1253801/finding-the-center-of-a-cluster

标签

algorithm

cluster-analysis

data-mining