计算 k 均值的方差度量百分比? [英] Calculating the percentage of variance measure for k-means?
问题描述
在:它有 1797 张从 0 到 9 的数字图像，每个图像的大小为 8×8 像素.我重复上面的实验，稍作修改: 的实现在 scikit-learn(以及许多其他聚类算法和各种聚类指标).这里是另一个类似的例子.
On the Wikipedia page, an elbow method is described for determining the number of clusters in k-means. The built-in method of scipy provides an implementation but I am not sure I understand how the distortion as they call it, is calculated.
More precisely, if you graph the percentage of variance explained by the clusters against the number of clusters, the first clusters will add much information (explain a lot of variance), but at some point the marginal gain will drop, giving an angle in the graph.
Assuming that I have the following points with their associated centroids, what is a good way of calculating this measure?
points = numpy.array([[ 0, 0],
[ 0, 1],
[ 0, -1],
[ 1, 0],
[-1, 0],
[ 9, 9],
[ 9, 10],
[ 9, 8],
[10, 9],
[10, 8]])
kmeans(pp,2)
(array([[9, 8],
[0, 0]]), 0.9414213562373096)
I am specifically looking at computing the 0.94.. measure given just the points and the centroids. I am not sure if any of the inbuilt methods of scipy can be used or I have to write my own. Any suggestions on how to do this efficiently for large number of points?
In short, my questions (all related) are the following:
- Given a distance matrix and a mapping of which point belongs to which cluster, what is a good way of computing a measure that can be used to draw the elbow plot?
- How would the methodology change if a different distance function such as cosine similarity is used?
EDIT 2: Distortion
from scipy.spatial.distance import cdist
D = cdist(points, centroids, 'euclidean')
sum(numpy.min(D, axis=1))
The output for the first set of points is accurate. However, when I try a different set:
>>> pp = numpy.array([[1,2], [2,1], [2,2], [1,3], [6,7], [6,5], [7,8], [8,8]])
>>> kmeans(pp, 2)
(array([[6, 7],
[1, 2]]), 1.1330618877807475)
>>> centroids = numpy.array([[6,7], [1,2]])
>>> D = cdist(points, centroids, 'euclidean')
>>> sum(numpy.min(D, axis=1))
9.0644951022459797
I guess the last value does not match because kmeans seems to be dividing the value by the total number of points in the dataset.
EDIT 1: Percent Variance
My code so far (should be added to Denis's K-means implementation):
centres, xtoc, dist = kmeanssample( points, 2, nsample=2,
delta=kmdelta, maxiter=kmiter, metric=metric, verbose=0 )
print "Unique clusters: ", set(xtoc)
print ""
cluster_vars = []
for cluster in set(xtoc):
print "Cluster: ", cluster
truthcondition = ([x == cluster for x in xtoc])
distances_inside_cluster = (truthcondition * dist)
indices = [i for i,x in enumerate(truthcondition) if x == True]
final_distances = [distances_inside_cluster[k] for k in indices]
print final_distances
print np.array(final_distances).var()
cluster_vars.append(np.array(final_distances).var())
print ""
print "Sum of variances: ", sum(cluster_vars)
print "Total Variance: ", points.var()
print "Percent: ", (100 * sum(cluster_vars) / points.var())
And following is the output for k=2:
Unique clusters: set([0, 1])
Cluster: 0
[1.0, 2.0, 0.0, 1.4142135623730951, 1.0]
0.427451660041
Cluster: 1
[0.0, 1.0, 1.0, 1.0, 1.0]
0.16
Sum of variances: 0.587451660041
Total Variance: 21.1475
Percent: 2.77787757437
On my real dataset (does not look right to me!):
Sum of variances: 0.0188124746402
Total Variance: 0.00313754329764
Percent: 599.592510943
Unique clusters: set([0, 1, 2, 3])
Sum of variances: 0.0255808508714
Total Variance: 0.00313754329764
Percent: 815.314672809
Unique clusters: set([0, 1, 2, 3, 4])
Sum of variances: 0.0588210052519
Total Variance: 0.00313754329764
Percent: 1874.74720416
Unique clusters: set([0, 1, 2, 3, 4, 5])
Sum of variances: 0.0672406353655
Total Variance: 0.00313754329764
Percent: 2143.09824556
Unique clusters: set([0, 1, 2, 3, 4, 5, 6])
Sum of variances: 0.0646291452839
Total Variance: 0.00313754329764
Percent: 2059.86465055
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7])
Sum of variances: 0.0817517362176
Total Variance: 0.00313754329764
Percent: 2605.5970695
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7, 8])
Sum of variances: 0.0912820650486
Total Variance: 0.00313754329764
Percent: 2909.34837831
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
Sum of variances: 0.102119601368
Total Variance: 0.00313754329764
Percent: 3254.76309585
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
Sum of variances: 0.125549475536
Total Variance: 0.00313754329764
Percent: 4001.52168834
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
Sum of variances: 0.138469402779
Total Variance: 0.00313754329764
Percent: 4413.30651542
Unique clusters: set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
The distortion, as far as Kmeans is concerned, is used as a stopping criterion (if the change between two iterations is less than some threshold, we assume convergence)
If you want to calculate it from a set of points and the centroids, you can do the following (the code is in MATLAB using pdist2 function, but it should be straightforward to rewrite in Python/Numpy/Scipy):
% data
X = [0 1 ; 0 -1 ; 1 0 ; -1 0 ; 9 9 ; 9 10 ; 9 8 ; 10 9 ; 10 8];
% centroids
C = [9 8 ; 0 0];
% euclidean distance from each point to each cluster centroid
D = pdist2(X, C, 'euclidean');
% find closest centroid to each point, and the corresponding distance
[distortions,idx] = min(D,[],2);
the result:
% total distortion
>> sum(distortions)
ans =
9.4142135623731
EDIT#1:
I had some time to play around with this.. Here is an example of KMeans clustering applied on the 'Fisher Iris Dataset' (4 features, 150 instances). We iterate over k=1..10, plot the elbow curve, pick K=3 as number of clusters, and show a scatter plot of the result.
Note that I included a number of ways to compute the within-cluster variances (distortions), given the points and the centroids. The scipy.cluster.vq.kmeans function returns this measure by default (computed with Euclidean as a distance measure). You can also use the scipy.spatial.distance.cdist function to calculate the distances with the function of your choice (provided you obtained the cluster centroids using the same distance measure: @Denis have a solution for that), then compute the distortion from that.
import numpy as np
from scipy.cluster.vq import kmeans,vq
from scipy.spatial.distance import cdist
import matplotlib.pyplot as plt
# load the iris dataset
fName = 'C:\Python27\Lib\site-packages\scipy\spatial\tests\data\iris.txt'
fp = open(fName)
X = np.loadtxt(fp)
fp.close()
##### cluster data into K=1..10 clusters #####
K = range(1,10)
# scipy.cluster.vq.kmeans
KM = [kmeans(X,k) for k in K]
centroids = [cent for (cent,var) in KM] # cluster centroids
#avgWithinSS = [var for (cent,var) in KM] # mean within-cluster sum of squares
# alternative: scipy.cluster.vq.vq
#Z = [vq(X,cent) for cent in centroids]
#avgWithinSS = [sum(dist)/X.shape[0] for (cIdx,dist) in Z]
# alternative: scipy.spatial.distance.cdist
D_k = [cdist(X, cent, 'euclidean') for cent in centroids]
cIdx = [np.argmin(D,axis=1) for D in D_k]
dist = [np.min(D,axis=1) for D in D_k]
avgWithinSS = [sum(d)/X.shape[0] for d in dist]
##### plot ###
kIdx = 2
# elbow curve
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(K, avgWithinSS, 'b*-')
ax.plot(K[kIdx], avgWithinSS[kIdx], marker='o', markersize=12,
markeredgewidth=2, markeredgecolor='r', markerfacecolor='None')
plt.grid(True)
plt.xlabel('Number of clusters')
plt.ylabel('Average within-cluster sum of squares')
plt.title('Elbow for KMeans clustering')
# scatter plot
fig = plt.figure()
ax = fig.add_subplot(111)
#ax.scatter(X[:,2],X[:,1], s=30, c=cIdx[k])
clr = ['b','g','r','c','m','y','k']
for i in range(K[kIdx]):
ind = (cIdx[kIdx]==i)
ax.scatter(X[ind,2],X[ind,1], s=30, c=clr[i], label='Cluster %d'%i)
plt.xlabel('Petal Length')
plt.ylabel('Sepal Width')
plt.title('Iris Dataset, KMeans clustering with K=%d' % K[kIdx])
plt.legend()
plt.show()
EDIT#2:
In response to the comments, I give below another complete example using the NIST hand-written digits dataset: it has 1797 images of digits from 0 to 9, each of size 8-by-8 pixels. I repeat the experiment above slightly modified: Principal Components Analysis is applied to reduce the dimensionality from 64 down to 2:
import numpy as np
from scipy.cluster.vq import kmeans
from scipy.spatial.distance import cdist,pdist
from sklearn import datasets
from sklearn.decomposition import RandomizedPCA
from matplotlib import pyplot as plt
from matplotlib import cm
##### data #####
# load digits dataset
data = datasets.load_digits()
t = data['target']
# perform PCA dimensionality reduction
pca = RandomizedPCA(n_components=2).fit(data['data'])
X = pca.transform(data['data'])
##### cluster data into K=1..20 clusters #####
K_MAX = 20
KK = range(1,K_MAX+1)
KM = [kmeans(X,k) for k in KK]
centroids = [cent for (cent,var) in KM]
D_k = [cdist(X, cent, 'euclidean') for cent in centroids]
cIdx = [np.argmin(D,axis=1) for D in D_k]
dist = [np.min(D,axis=1) for D in D_k]
tot_withinss = [sum(d**2) for d in dist] # Total within-cluster sum of squares
totss = sum(pdist(X)**2)/X.shape[0] # The total sum of squares
betweenss = totss - tot_withinss # The between-cluster sum of squares
##### plots #####
kIdx = 9 # K=10
clr = cm.spectral( np.linspace(0,1,10) ).tolist()
mrk = 'os^p<dvh8>+x.'
# elbow curve
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(KK, betweenss/totss*100, 'b*-')
ax.plot(KK[kIdx], betweenss[kIdx]/totss*100, marker='o', markersize=12,
markeredgewidth=2, markeredgecolor='r', markerfacecolor='None')
ax.set_ylim((0,100))
plt.grid(True)
plt.xlabel('Number of clusters')
plt.ylabel('Percentage of variance explained (%)')
plt.title('Elbow for KMeans clustering')
# show centroids for K=10 clusters
plt.figure()
for i in range(kIdx+1):
img = pca.inverse_transform(centroids[kIdx][i]).reshape(8,8)
ax = plt.subplot(3,4,i+1)
ax.set_xticks([])
ax.set_yticks([])
plt.imshow(img, cmap=cm.gray)
plt.title( 'Cluster %d' % i )
# compare K=10 clustering vs. actual digits (PCA projections)
fig = plt.figure()
ax = fig.add_subplot(121)
for i in range(10):
ind = (t==i)
ax.scatter(X[ind,0],X[ind,1], s=35, c=clr[i], marker=mrk[i], label='%d'%i)
plt.legend()
plt.title('Actual Digits')
ax = fig.add_subplot(122)
for i in range(kIdx+1):
ind = (cIdx[kIdx]==i)
ax.scatter(X[ind,0],X[ind,1], s=35, c=clr[i], marker=mrk[i], label='C%d'%i)
plt.legend()
plt.title('K=%d clusters'%KK[kIdx])
plt.show()
You can see how some clusters actually correspond to distinguishable digits, while others don't match a single number.
Note: An implementation of K-means is included in scikit-learn (as well as many other clustering algorithms and various clustering metrics). Here is another similar example.
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