问题
I want to perform a k means clustering analysis on a set of 10 data points that each have an array of 4 numeric values associated with them. I'm using the Pearson correlation coefficient as the distance metric. I did the first two steps of the k means clustering algorithm which were:
1) Select a set of initial centres of k clusters. [I selected two initial centres at random]
2) Assign each object to the cluster with the closest centre. [I used the Pearson correlation coefficient as the distance metric -- See below]
Now I need help understanding the 3rd step in the algorithm:
3) Compute the new centres of the clusters:
where X, in this case is a 4 dimensional vector and n is the number of data points in the cluster.
How would I go about calculating C(S) for say the following data?
# Cluster 1
A 10 15 20 25 # randomly chosen centre
B 21 33 21 23
C 43 14 23 23
D 37 45 43 49
E 40 43 32 32
# Cluster 2
F 100 102 143 212 #random chosen centre
G 303 213 212 302
H 102 329 203 212
I 32 201 430 48
J 60 99 87 34
The last step of the k means algorithm is to repeat step 2 and 3 until no object changes cluster which is simple enough.
I need help with step 3. Computing the new centres of the clusters. If someone could go through and explain how to compute the new centre of just one of the clusters, that would help me immensely.
回答1:
Step 3 corresponds to calculating the mean for each cluster.
For cluster 1, you'd get as new cluster center (B+C+D+E) / 4, which is (35.25 33.75 29.75 21.75), i.e sum each component for all the points in the cluster separately, and divide it by the number of points in the cluster.
The cluster center (A for cluster 1) is usually not part of the calculation of the new cluster center.
回答2:
Don't throw in other distance functions into k-means.
K-means is designed to minimize the "sum of squares", not distances! By minimizing the sum of squares, it will coincidentially minimize Squared Eudlidean and thus Euclidean distance, but this may not hold for other distances and thus K-means may stop converging when used with arbitrary distance functions.
Again: k-means does not minimize arbitrary distances. It minimizes the "sum of squares" which happens to agree with squared Euclidean distance.
If you want an algorithm that is well-defined for arbitrary distance functions, consider using k-medoids (Wikipedia), a k-means variant. PAM is guaranteed to converge with arbitrary distance functions.
回答3:
For each cluster with n-dimensional points, calculate an n-dimensional center of mass to get the centroid. In your example, there are 4-dimensional points, so the center of mass is the mean along each of the 4 dimensions. For cluster 1, the centroid is: (30.20, 30.00, 27.80, 30.40). For example, the mean for the first dimension is calculated as (10+21+43+37+40)/5 = 30.20.
See the Wikipedia article on K-Means clustering for more information.
来源：https://stackoverflow.com/questions/15604647/k-means-clustering-algorithm