Thanks Gael.

BTW, I modified my code to loop until it gets the same clustering twice in a row.  This yields more consistent results.  I don’t know if this is a general solution but it worked for my simple test case.  Code below.

James

import sys

import scipy

import warnings

from scipy.cluster.vq import *

print sys.version

vals = scipy.array((0.0,0.1,0.5,0.6,1.0,1.1))

print vals

white_vals = whiten(vals)

print white_vals.shape, white_vals

# Check for same clustering

def clustering_test(a,b):

    # have to create copies, then sort so we don't modify the original

    ea = a.copy()

    eb = b.copy()

    ea.sort()

    eb.sort()

    r = (ea == eb).all()

    print a,b,ea,eb,r

    return r

# try it until we get the same clustering twice in a row

found = False

prior_idx = None

while not found:

    with warnings.catch_warnings():

        warnings.simplefilter("ignore") # suppress the warning message (happens if it doesn't find the right number of clusters)

        res, idx = kmeans2(white_vals, 3) # changing iter doesn't seem to matter

    #print res, idx

    if prior_idx is not None:

        eq = clustering_test(idx, prior_idx)

        #print eq.all()

        if eq:

            found = True

    prior_idx = idx

print "result", res, idx

Thanks – sklearn works great!  I got exactly what I expected each time I ran it!

James

import sys

import numpy

from sklearn.cluster import *

print sys.version

vals = numpy.array([[0.0],[0.1],[0.5],[0.6],[1.0],[1.1]])

print vals

k_means_ex = KMeans(k=3)

x = k_means_ex.fit_predict(vals)

print x

print k_means_ex.cluster_centers_

print k_means_ex.score(vals)

2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)]

[[ 0. ]

[ 0.1]

[ 0.5]

[ 0.6]

[ 1. ]

[ 1.1]]

[1 1 0 0 2 2]

[[ 0.55]

[ 0.05]

[ 1.05]]

-0.015