Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
484 views
in Technique[技术] by (71.8m points)

language agnostic - How Could One Implement the K-Means++ Algorithm?

I am having trouble fully understanding the K-Means++ algorithm. I am interested exactly how the first k centroids are picked, namely the initialization as the rest is like in the original K-Means algorithm.

  1. Is the probability function used based on distance or Gaussian?
  2. In the same time the most long distant point (From the other centroids) is picked for a new centroid.

I will appreciate a step by step explanation and an example. The one in Wikipedia is not clear enough. Also a very well commented source code would also help. If you are using 6 arrays then please tell us which one is for what.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Interesting question. Thank you for bringing this paper to my attention - K-Means++: The Advantages of Careful Seeding

In simple terms, cluster centers are initially chosen at random from the set of input observation vectors, where the probability of choosing vector x is high if x is not near any previously chosen centers.

Here is a one-dimensional example. Our observations are [0, 1, 2, 3, 4]. Let the first center, c1, be 0. The probability that the next cluster center, c2, is x is proportional to ||c1-x||^2. So, P(c2 = 1) = 1a, P(c2 = 2) = 4a, P(c2 = 3) = 9a, P(c2 = 4) = 16a, where a = 1/(1+4+9+16).

Suppose c2=4. Then, P(c3 = 1) = 1a, P(c3 = 2) = 4a, P(c3 = 3) = 1a, where a = 1/(1+4+1).

I've coded the initialization procedure in Python; I don't know if this helps you.

def initialize(X, K):
    C = [X[0]]
    for k in range(1, K):
        D2 = scipy.array([min([scipy.inner(c-x,c-x) for c in C]) for x in X])
        probs = D2/D2.sum()
        cumprobs = probs.cumsum()
        r = scipy.rand()
        for j,p in enumerate(cumprobs):
            if r < p:
                i = j
                break
        C.append(X[i])
    return C

EDIT with clarification: The output of cumsum gives us boundaries to partition the interval [0,1]. These partitions have length equal to the probability of the corresponding point being chosen as a center. So then, since r is uniformly chosen between [0,1], it will fall into exactly one of these intervals (because of break). The for loop checks to see which partition r is in.

Example:

probs = [0.1, 0.2, 0.3, 0.4]
cumprobs = [0.1, 0.3, 0.6, 1.0]
if r < cumprobs[0]:
    # this event has probability 0.1
    i = 0
elif r < cumprobs[1]:
    # this event has probability 0.2
    i = 1
elif r < cumprobs[2]:
    # this event has probability 0.3
    i = 2
elif r < cumprobs[3]:
    # this event has probability 0.4
    i = 3

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...