Question

在Wikipedia page上，描述了用于确定k均值中的聚类数的肘方法。 The built-in method of scipy提供了一个实现，但我不确定我是否理解它们所称的失真是如何计算的。

更确切地说，如果您绘制由方法解释的方差百分比群集将反对群集的数量，第一个群集将添加很多信息（解释很多变化），但在某些时候边际增益将下降，在图中给出一个角度。

假设我的相关质心有以下几点，那么计算这个量度的好方法是什么？

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)

我特别关注只计算点和质心的0.94 ..计算。我不确定是否可以使用任何内置的scipy方法，或者我必须编写自己的方法。关于如何有效地为大量积分做这些的任何建议？

简而言之，我的问题（所有相关的）如下：

给定距离矩阵和哪个点所属的映射群集，计算可以使用的度量的好方法是什么绘制肘部情节？
如果使用不同的距离函数（如余弦相似度），方法会如何变化？

编辑2：失真

from scipy.spatial.distance import cdist
D = cdist(points, centroids, 'euclidean')
sum(numpy.min(D, axis=1))

第一组点的输出是准确的。但是，当我尝试不同的设置时：

>>> 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

我猜最后一个值不匹配，因为kmeans似乎是将数值乘以数据集中的总点数。

编辑1：差异百分比

到目前为止我的代码（应该添加到Denis的K-means实现中）：

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())

以下是k = 2的输出：

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

在我的真实数据集上（看起来不对我！）：

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])

Answer 1

就Kmeans而言，失真被用作停止标准（如果两次迭代之间的变化小于某个阈值，我们假设收敛）

如果你想从一组点和质心计算它，你可以执行以下操作（代码在MATLAB中使用pdist2函数，但在Python / Numpy / Scipy中重写应该很简单）：

% 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);

结果：

% total distortion
>> sum(distortions)
ans =
           9.4142135623731

编辑＃1：

我有时间玩这个..以下是应用于'Fisher Iris Dataset'的KMeans群集示例（4个功能，150个实例）。我们迭代k=1..10，绘制肘曲线，选择K=3作为聚类数，并显示结果的散点图。

请注意，在给定点和质心的情况下，我提供了许多计算群内方差（扭曲）的方法。 scipy.cluster.vq.kmeans函数默认返回此度量（使用Euclidean作为距离度量计算）。您还可以使用scipy.spatial.distance.cdist函数计算与您选择的函数的距离（假设您使用相同的距离度量获得了聚类质心：@Denis有解决方案），然后计算失真从那起。

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()

elbow_curve scatter_plot

编辑＃2：

在回应评论时，我在下面使用NIST hand-written digits dataset给出了另一个完整的例子：它有1797个0到9的数字图像，每个图像大小为8×8像素。我重复上面的实验略微修改：Principal Components Analysis用于将维度从64降低到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()

elbow_curve digits_centroids PCA_compare

您可以看到某些群集实际上如何与可区分的数字相对应，而其他群集与单个数字不匹配。

注意：K-means中包含scikit-learn的实现（以及许多其他群集算法和各种clustering metrics）。 Here是另一个类似的例子。

Answer 2

一个简单的集群措施：
1）从每个点到最近的聚类中心绘制“旭日”光线，
2）查看所有光线的长度 - 距离（点，中心，公制= ......）。

对于metric="sqeuclidean"和1个群集，平均长度平方是总方差X.var();对于2个集群，它更少......到N个集群，长度全部为0。 “解释的方差百分比”是100％ - 这个平均值。

此代码，在is-it-possible-to-specify-your-own-distance-function-using-scikits-learn-k-means下：

def distancestocentres( X, centres, metric="euclidean", p=2 ):
    """ all distances X -> nearest centre, any metric
            euclidean2 (~ withinss) is more sensitive to outliers,
            cityblock (manhattan, L1) less sensitive
    """
    D = cdist( X, centres, metric=metric, p=p )  # |X| x |centres|
    return D.min(axis=1)  # all the distances

与任何长数字列表一样，可以通过各种方式查看这些距离：np.mean（），np.histogram（）...绘图，可视化并不容易。
另见stats.stackexchange.com/questions/tagged/clustering，特别是
How to tell if data is “clustered” enough for clustering algorithms to produce meaningful results?

计算k均值的方差百分比？

2 个答案:

编辑＃1：

编辑＃2：